5.5: American Modernism (1910-1935)
Introduction
American Modernism reflected the worldwide changes in art based on the unique American culture. The movement, along with others in this period, aligned with the experiences of modern industrial existence. Artists rejected the realistic approach of art and moved to more abstraction, experimentation with color and shapes. Artistic themes were based on social agendas, emotional feelings, and political changes. Art was varied, the personal expression of each artist as they created new visions.
Georgia O'Keeffe
Georgia O'Keeffe (1887-1986) was born in Wisconsin on a wheat farm. She was interested in art from an early age, studied art in high school, and attended the Art Institute of Chicago. After moving to New York, O’Keeffe continued to study traditional art methods until learning the concepts espoused by Arthur Dow. He believed art should be expressive and composition composed of harmonious lines, mass, and colors. These ideas influenced O'Keeffe to develop more abstracted drawings, a way to express her feelings. One of her friends took her drawings to show to Alfred Stieglitz, a famous photographer who owned his art gallery. In 1916, he started to exhibit her paintings, and by the mid-1920s, she was acknowledged as a very successful artist, well-known for her oversized depiction of flowers. Her work placed her as a significant practitioner of American abstraction, a difficult achievement for a female in a male-dominated art world. O'Keeffe developed her style, maintained her vision, and was not deterred through her life by other movements and trends of the art world.
O'Keeffe married Stieglitz in 1924 as he continued to promote her work with multiple shows and installations. She left New York to travel through New Mexico in 1929, a place she was enthralled with and inspired by; the stark desert landscape, the native cultures, and the region's architecture generated new ideas and directions for her art. O’Keeffe traveled between New York and New Mexico to work, permanently moving there in 1949 after Stieglitz died. She painted different scenes throughout New Mexico, traveling internationally to depict mountain regions in Peru and Japan. Although she suffered from macular degeneration, as her vision failed, she continued to paint images in her memory until she passed away at age 98.
O'Keeffe was first known for her incredible close-ups of flowers. Georgia O'Keeffe believed that people merely glance at flowers but never really observe their exquisiteness due to people's fast-paced lives. She wished to give such rushing people experience and the feel of the true beauty of flowers. In her words, "If I could paint the flower exactly as I see it, no one would see what I see because I would paint it small like the flower is small. So I said to myself - I'll paint what I see - what the flower is to me but I'll paint it big and they will be surprised into taking time to look at it - I will make even busy New Yorkers take time to see what I see of flowers." [1] She was exceptionally observant, noticing the subtle shapes, lines, and colors of her subjects. Flowers became the theme of her first major set of paintings, focusing on a small flower that filled a large canvas, appearing abstract although a perfect representation of the flower with enhanced bold color and texture.
Black Iris (5.5.2) is a close-up of an iris; the flower perfectly reproduced on canvas while engendering a possible abstracted image of a person with their hands overhead. The dark, rich colors draw the eye into the center of the flower as the petals unfold into an almost transparent pink.
On her early trips to New Mexico, O'Keeffe became enamored with the animal bones she found in the desert landscape, bones bleached by the sun and weather. In Cow's Skull: Red, White, and Blue (5.5.3), the jagged edges of the skull's bottom half contrasted with the smooth, weatherworn top half. She believed the skull represented the long-lasting and enduring beauty of the desert, the red, white, and blue background symbolizing the American spirit. She painted several different versions of skulls and bones throughout her time in New Mexico. The spectacular landscape and scenery of the rocks and mountains of the variable desert settings were a major inspiration for O'Keeffe. Near the house she purchased at Ghost Ranch were brightly colored hills of different hues.
Red and Yellow Cliffs (5.5.4) exemplify the incredible mountains rising from the desert floor. The striated precipices were eroded over centuries revealing the dynamic red and yellow colors. Sparse greenery dots the desert floor, growing where the minimal amount of water collected. These mountains were the subject of multiple paintings.
Thomas Hart Benton
Thomas Hart Benton (1889-1975) was born in Missouri in a family active in politics, his father a U. S. congressman and a great-uncle who was of the first senators from Missouri. Benton lived in both Missouri and Washington D. C. He went to a military school to hone his skills for a political career. However, Benton wanted to study art, his mother encouraged the concept, and he went to the Art Institute in Chicago, then to Paris for further study. During World War I, Benton was in the navy; his job was to draw camouflaged ships in the Virginia harbor, ensure the vessels were properly painted with the correct camouflage patterns, and create an inventory. He was married for over fifty years with two children. After the war, he moved to New York to paint, developing a frequently termed regionalism style, exalting rural America, and denigrating abstract art. Initially, he studied modern art and decided he did not like abstract work, preferring to create a representational and more naturalistic style. He traveled throughout the country, taught for a while in Kansas before moving back to New York. He trained many well-known artists, including Jackson Pollock, who started the Abstract Expressionist movement. Pollock always claimed the traditional work of Benton gave him a reason to rebel against the ideas of representational art.
Benton's most famous work was America Today , a set of ten oversized canvas panels from floor to ceiling covering all four walls, painted for the New School for Social Research boardroom in 1931. He used egg tempera and an oil glaze over Permalba and gesso on the linen-mounted wood panels. Benton agreed to create the murals without any compensation except for the materials he needed. He stated, "I'll paint you a picture in tempera if you finance the eggs." [2] He hoped the work would grow his reputation and obtain other commissions. Benton had traveled throughout the United States, making sketches as he journeyed to create an extensive view of American life. Some of the panels represented different regions of the United States including, Midwes t (5.5.5), Deep South (5.5.6), and Changing West (5.5.7). Other panels portrayed life in specific occupations, as seen in Steel (5.5.8). The smallest panel was Outreaching Hands (5.5.9), some reaching for food and others for money, relating to the extreme economic dire of some and the inequity of the moneygrubbers in the Great Depression.
Despite the Depression, Benton wanted to portray a vision of the country's progress based on the new technologies and industries. He also believed progress was generated by the work of the people, their physical participation, and struggles. Each panel is filled with workers, large muscular physiques dressed in everyday clothing, and exaggerated positions and expressions. He was a master of arrangement, using strong diagonals to partition sections and move the eye through the narrative of the painting. Benton also used wood moldings covered in aluminum leaf to create spatial breaks while unifying each panel, the moldings similar to the Art Deco décor of the building where they were installed. The mural Deep South was based on cotton, the king of industries in the south as the tall form of the black man looming over the scene empties his bag of cotton, the white man moving the cotton with a plow, the guard overseeing the chain gang, cotton loading onto the steamboat. Midwest depicts the lumberjacks as they clear-cut the forest, emptying the land to grow corn, the grain elevator occupying the skyline, Benton's Ford hidden in the background. Changing West is the image of the Texas oil boom with thick smoke and a dominating derrick, the vanishing cowboys smaller in the setting. Steel is based on a single industry, displaying the depth the workers performed, power, and strength a requirement. He also included murals of other discrete industries. Benton demonstrated his attention to detail as he traversed the United States, recording the workers, the complexity of their jobs, and the dominant industries of the country while adding the details of the changes of life he observed.
Edward Hopper
Edward Hopper (1882-1967) was born in New York; his parents were from an established Hudson River family. He began drawing at an early age and, after high school, went to the New York School of Art and Design. After working as an illustrator, Hopper went to Paris to study, trying different techniques, deciding he didn't want to use any particular style, only his own, uninterested in the abstract art of the time. Hopper returned to the United States, working for several years before he achieved success. He married, and his wife successfully helped promote his work. Hopper, well-known for his oil paintings, started any work with careful sketches, focusing on geometrical design, the balance of each element, and how any figure appeared in a painting. He also concentrated on light, whether natural or artificial, using light to cast specific shadows and create the mood of a painting. Hopper primarily painted images of American life and the structures inhabited by people, for example, a motel room, an office space, or the waiting room at the station; the buildings of ordinary life and how people were perceived in spaces along with the interaction of light in a space. Hopper incorporated emotion into his work with a limited number of figures, positioned in an environment to elicit feelings of loneliness or boredom, emotions enhanced by his use of light. He also painted sea things: crashing waves, lighthouses, boats on the water, or grasses on the shore.
Gas (5.5.10) represents the rural station, dusk setting in as the solitary attendant checks the pumps, perhaps ready to close for the night. Hopper used both the natural light of the darkening sky and the artificial light of the station. The road is empty, depicting the solitude of the country road, the gas station a refuge for the traveler who stops for gas or directions, the light a beacon on the road.
Hopper brought his emotion into seascapes, The Lighthouse at Two Lights (5.5.11) standing starkly in silhouette against the dimming sky. The sun is low on the horizon casting its light on the face of the walls with deep shadows. The sturdiness of the lighthouse demonstrates its resistance to nature, standing for decades. Hopper uses a limited palette, the gray building against the muted gray sky, the fields dried by late summer, adding to the feeling of loneliness and isolation.
Tables for Ladies (5.5.12) was based on the newly available ability for ladies to dine at a restaurant without a male in attendance. Previously, women dining out were presumed to be prostitutes. Hopper portrays each person in isolation without interaction; the cashier only focused on her register, the waitress rearranging the food as the couple sits in the background. The painting allows the viewer to peer through the window, watching the scene in front of them. Hopper's attention to detail is visible in the precise method the table is arranged. The foreground is brightly colored, grabbing the viewers' attention, the couple in muted, simple colors almost disappearing into the background.
Office in a Small City (5.5.13) conveys the isolation and seclusion of the man sitting in his office. The stark concrete building removes him from the outside world; a world only viewed through the soundless windows. Hopper's wife described the painting as "the man in concrete wall." [3] Hopper used the immensity of the corner office walls and window in juxtaposition to the small person bringing the isolation of the person from the humanity of the outside, others only visible through the glass.
Nighthawks (5.5.14) is one of Hopper's most famous paintings. The inspiration for the restaurant was in New York; however, the construction of the painting places the scene in anywhere America, the universality of the lonely late-night panorama. The painting is an outstanding demonstration of Hopper's use of light. Fluorescent lighting was new in the 1940s, giving an eerie luminosity to the diner's interior positioned without any visible entrance. The large window gave the viewer an intimate look at the remote activities of the people inside. Hopper's use of a green palette enhanced the glow of the light through the window.
Marion Beckett
Marion Beckett (1886-1949) was born in New York, receiving a large inheritance after her father's death and giving her the freedom to travel to Paris and study art with her life-long friends Agnes Meyer and Katharine Rhoades. Beckett was described as exceptionally beautiful and shy. She mainly painted portraits and exhibited at Alfred Stieglitz's gallery, along with her friends, where they were given the nickname "The Three Graces." Beckett and the other women were in a transitional period, the ideals of Victorian femininity replaced by new opportunities and expectations of behavior and dress. She was successful as an artist, showing in many galleries; however, she stopped painting in 1926. She and her friends frequently went to France, visiting photographer Edward Steichen and his wife to paint their portraits. As World War I started, the German army invaded Steichen's house, and Beckett and the Steichens escaped two days before. After the war, Beckett created a studio in Paris, where she supposedly had an affair with Eduard Streichen. In 1919. Beckett was sued by Clara Steichen for $200,000, claiming "alienation of affections." When testifying against Beckett in court, she asserted, "Miss Beckett, when my husband was near, would wear nothing but long, clinging gowns and roses…She dressed in a very artistic or theatrical manner to attract attention and exploit her physical charms." [4] Clara Steichen could not prove anything and lost the case. The paintings of the couple seem to reflect their feelings. Clara Steichen (5.5.15) is not facing the artist, looking dejectedly to the side. The paint colors are very grayed and muted tones respond to the saddened state of the woman.
Eduard Steichen (5.5.16) is looking directly at the artist, a slight smile on his face. The colors in the painting are more dynamic and alive.
Alfred Stieglitz
Alfred Stieglitz (1864-1946) was born in New Jersey; as a teenager, he and his family moved to Germany. When he studied engineering in Berlin, he became interested in photochemistry and traveled across Europe photographing the sights with a newly purchased camera. In 1890, Stieglitz returned to America, working in New York to improve the newly established community of photographers. He devoted his time to pioneer new techniques in photography, particularly improving methods to create photographs in poor weather, rain or snow, and the darkness of night, not just snapped images, rather compositional scenes. Stieglitz wanted to improve the perception of photography as an art form and opened the first of his galleries, including "291," and published the Camera Work magazine. He incorporated the new contemporary work created in Europe by artists in his gallery, expanding his reputation.
He also met Georgia O'Keeffe when a friend took her drawings to the gallery at 291 and, by 1917, developed a relationship with O'Keeffe, changing the focus of his work. O'Keeffe moved to New York to work in a studio Stieglitz provided. As Stieglitz changed his style, he moved from an encapsulated scene to the modern ideas of fragmented parts, only capturing partial sections of any scene. O'Keeffe was his primary model, and he took over three hundred images of her, often nude, focusing on her hands, neck, or other divided parts of her body, capturing emotional poses instead of the usual flat portrait. They also had what others described as a torrid affair. Stieglitz's wife divorced him, which took six years to settle after lengthy legal battles, then he married O'Keeffe. By 1921, he held a major exhibition of his photographs, including almost fifty of O'Keeffe, furthering his reputation as an artist-photographer. Stieglitz began showing and successfully selling American artists' work at the gallery, including O'Keeffe's work.
When Stieglitz died in 1964, O'Keeffe selected over 1,600 photographs she considered his major work and donated them to the National Gallery of Art, the most extensive collection of his photographs.
The Terminal (5.5.17) was one of the early photographs Stieglitz took, marking the beginning of elevating photography as an art. He used a smaller 4 x 5 camera instead of the larger sizes requiring a tripod. With the small camera, he could move around the city and capture images as they occurred. The techniques transformed photography from posed, still images to more painterly scenes. He captured the steam coming from the horses releasing heat in the cold, snowy weather giving the image the look of a painting instead of the pose.
The Steerage (5.5.18) depicts immigrants from Germany landing in New York. Stieglitz captured the people as they lived aboard the ship, jammed in the steerage decks, the lowest level on the ship with little ventilation. The early photographs of Stieglitz are generally of the people in the lower classes, workers, immigrants, those toiling in the factories. As he evolved, he became interested in line and form.
The two images of Georgia O'Keeffe were part of the three hundred photographs Stieglitz took of her from 1917 to 1937. In each image, Georgia O'Keeffe (5.5.19) and Georgia O'Keeffe—Neck (5.5.20), Stieglitz demonstrates his concept of a portrait composed of more than the image of a face. He wanted the portrait to capture the person's story and express their emotions, movements, feelings, life's essence. He stated, "To demand the portrait that will be a complete portrait of any person is as futile as to demand that a motion picture be condensed into a single still." [5] In the first image, O'Keeffe's hands and face against the dark sweater, two distinct segments of the person bring a complete feeling of her thoughtful emotion into the image. The second photograph of her neck and the partial face and hands presents a different feeling and set of expressions. O’Keeffe said of Stieglitz;
"Stieglitz photographed me first at his gallery "291" in the spring of 1917. … My hands had always been admired since I was a little girl—but I never thought much about it. He wanted head and hands and arms on a pillow—in many different positions. I was asked to move my hands in many different ways—also my head—and I had to turn this way and that. … Stieglitz had a very sharp eye for what he wanted to say with the camera. When I look over the photographs Stieglitz took of me—some of them more than sixty years ago—I wonder who that person is. It is as if in my one life I have lived many lives. … His idea of a portrait was not just one picture. His dream was to start with a child at birth and photograph that child in all of its activities as it grew to be a person and on throughout its adult life. As a portrait it would be a photographic diary." Georgia O'Keeffe, 1978 [6]
Frank Lloyd Wright
Frank Lloyd Wright (1867-1959) was born in Wisconsin, his father a musician and minister, his mother a teacher. As a young child, his mother was interested in the Froebel theory of education based on the concept that children should master drawing basic geometric forms before drawing the spontaneous images found in nature. She bought him a set of geometrically shaped educational blocks able to be assembled in different three-dimensional combinations. For several years of his early education, Wright sat and drew lines and shapes based on the wooden blocks, constructing patterns and designs. The wooden blocks had a significant impact on his future, and he wrote, "…the smooth cardboard triangles and maple-wood blocks were most important. All are in my fingers to this day." [7]
Before he left, Wright attended the University of Wisconsin, moved to Chicago, and worked as a draftsman with an architectural company. He worked for other firms, starting to design houses on his own at night. Wright was fired from his job with one firm because they discovered he was also working on commissions outside their company. He attributed his need for money to his poor financial acumen and expensive tastes. Starting his own company, Wright designed houses with geometric constructs and features with open floor plans or rows of windows for several houses in the Chicago area. Wright married and had six children. From 1900 to 1914, he designed homes identified as the Prairie Style, a more organic architecture. Others who wanted to improve smaller houses with better design noticed his concepts, allowing him to expand his scope and business. Wright's prairie designs usually included low-pitched roofs, strong lines on the horizontal planes, open floor plan instead of boxy rooms, the building with a longer form instead of the squarer bungalow types, and the use of stone and wood natural materials. Meanwhile, his personal life was spiraling, and he divorced his wife, marrying two more times as his second wife died after one year of marriage. He also lost his Taliesin studio in Wisconsin when a servant killed seven people in the living quarters or working outside, then set fire to the building.
Wright designed a revolutionary style of a concrete block system in the 1920s. The blocks were premade with a set of internal bars for assembly. The prefabrication allowed builders to order any color or texture for the easy to assemble blocks. He updated the system after World War II when he was designing the Usonian-style houses. Usonian homes (5.5.21) were usually constructed in Wright's typical flat roofs, a radiant heating system incorporated into a concrete slab floor, and open floor plans. In the 1930s, Americans seldom had servants, so Wright designed the houses with a kitchen open to the dining room, a 'workspace' for a woman to monitor the children or her guests. The living area was the focal point with a large fireplace, the bedrooms small and isolated. After the war, the new design was popular for developers to build homes with mechanized and simpler construction methods.
In 1934, Wright struggled financially after the stock market crash, depression, and bank failures; private homes were no longer constructed. He was also becoming a relic, unwilling to follow the newer developing European trends. He did create an apprentice program attended by the son of a wealthy Pittsburgh businessman, Edgar Kaufmann. Upon recommendation from his son, Kaufmann hired Wright to build a vacation home in the woods of Pennsylvania. Wright designed Fallingwater , inspired by the natural environment surrounding the site; he incorporated all of nature's beauty in the organic architecture for the materials and colors. Fallingwater (5.5.22) was noted as one of the best structures of American architecture.
Wright's clients loved the natural setting, and he wanted to change as little of the surroundings as possible, including retaining the stream and the waterfall in its natural position. He attached reinforced concrete projections (5.5.23) to the natural rock to hold the cantilevered terraces (5.5.24) made from local stone, blending in with the existing sandstone, seemingly floating above the stream.
The first-floor entry, living, and dining rooms (5.5.25) are open in one contiguous space with broad glass windows. The living room houses the door to the stairway descending to the stream. Wright also incorporated a large rock outcropping into part of the living room, leaving the rock undisturbed. Guest rooms were located above the main house. Uniquely, no windows face the falls, all of the windows are oriented to the trees and horizon.
Wright went to Arizona in the winters, leaving the Wisconsin winters because of his health. In 1937, he bought land for Taliesin West, named after his studio in Wisconsin. When they were excavating the property for the building, they found Native American (Hohokam) petroglyphs appearing to be clasped hands. He used the interconnection of the lines for his building, now the symbol for Taliesin West (5.5.26). He wanted the design to compliment the desert and made the walls with desert rock inserted into the concrete, generally referred to as 'desert masonry.' The flatter surface of the stones faced outward, the larger part of the stone evident in the interior space. He used his concept of the long, low-hanging roofline to protect the walls from the hot desert sun. He added other facilities and rooms throughout the years, including the theatre's hexagonal shape and perfect acoustics.
The Garden Room (5.5.27) is a long meeting space with a glass roof open to the stars and seating along with expansive windows. Wright believed the view was key to the beauty and aesthetics of the site, and in the 1940s, when nearby power lines were going to be installed, he fought against it.
When he lost the battle, he rebuilt the entrance (5.5.28), moving it to the rear of the existing structure without the offensive view of lines.
Julia Morgan
Julia Morgan (1872-1957) had a long and prolific career as an architect and engineer, designing over 700 buildings, including the famous Hearst Castle, also known as San Simeon. She was born in California, the family living on the west coast. Her New York grandfather was wealthy and helped support the family. When he died, her grandmother moved to California to live with the Morgan family, bringing her wealth with her. Morgan excelled in school and went to the University of California at Berkeley, majoring in civil engineering. During this period, none of the schools on the West Coast had an architectural program, so she chose engineering. When she was a senior, one of her professors encouraged her to go to Paris and attend the L’École Nationale et Spéciale des Beaux-Arts. Morgan went to Paris and twice failed to pass the entrance exam, learning they failed her deliberately because they "did not want to encourage young girls,…Morgan wrote, "I'll try again next time anyway even without any expectations, just to show 'les jeunes filles' are not discouraged." [8] She finally was admitted, advancing through the first level in half the time. Morgan was the first female to obtain a certificate from Beaux-Arts in architecture.
After she received her certificate in 1904, she returned to California and opened an office in California. After the earthquake and extensive fire, she received commissions helping to rebuild San Francisco, including the Fairmont Hotel. In this job, she continued to receive comments about women who should not be architects. Her success at the Fairmont led to multiple other commissions. From the time she first opened her office until she retired in 1951, she designed and supervised the construction of over 700 projects, a few in other states. Still, the majority of the projects were in California. She designed sixteen YWCA buildings, private houses, Asilomar in Monterey, and many structures for the Hearst family. In 1919, she met with William Randolph Hearst as a potential client to build his concept of San Simeon; he wanted something different than the usual California style. The commission for Hearst Castle (5.5.29), which he called 'the ranch,' led to a long-time association between Morgan and Hearst running from 1919 to the mid-1940s.
Morgan designed La Casa Grande, as part of Hearst Castle, with 115 rooms as the main building; its construction started in 1922 and continued until 1947, a building still unfinished. Guesthouses (Casa del Mar, Casa del Sol, and Casa del Monte) were connected to Casa Grande (5.5.30) by walkways and terraces. The façade of Casa Grande is four stories, similar to a building in Seville. Although constructed with concrete, the façade was covered with stone. The large overhang was made with teak Morgan found in San Francisco, purchased initially for a ship. Casa Grande did not have an elegant staircase leading to the upper floors, only elevators, and stairs. Morgan used the latest technology; a sound system wired throughout the building, radio stations, and eighty telephones located around the buildings and grounds and interconnected with a PBX switchboard, all advanced concepts for the time period.
In addition to the major set of buildings with their rooms and guest suites, the site also included indoor and outdoor swimming pools, a theater for movies and live shows, tennis courts, a billiard room, a wine cellar, two libraries, a zoo, dog kennels, airplane landing strip and hanger, thoroughbred horse ranch, riding trails and a five-mile pergola. [9] Based on the continual requirements by Hearst, the site continued to grow as Morgan designed the buildings and surrounding environments. Hearst did not want the native oaks touched, so work was sited around them; broad walkways and gardens connected the buildings, and spaces were designed for Hearst's art and antiques. One of the impressive rooms was the library (5.5.31). The ceiling came from a 16th-century house in Spain. Morgan used ceiling parts from different rooms and combined them into a single ceiling for the grand room. The large fireplace was made from carved limestone. Hearst had collected books and antiques for a long time and incorporated over 5,000 books into this room, adding over 3,000 in his study. Many of the books were special collections, first editions, or books signed by the authors. Hearst's extensive collection of ancient Greek vases were incorporated into the décor of the room.
The ceiling in the billiard room (5.5.32) was from a 15th-century Spanish building adorned with images of the courtly life of the time. On the wall hung immense Flemish tapestry millefleurs (thousands of flowers) dated around the 1500s and woven with wool and silk. The billiard room was a popular gathering spot for guests to play billiards or pool, and instead of the room being the customary domain of male guests, both men and women congregated and played.
The site had two pools, the Roman pool (5.5.33) made for indoor swimming and the outside Neptune pool (5.5.32). The Roman pool was built underneath the tennis courts and adorned with mosaics made from Murano glass tiles and gold leaf. The elegant pool was located farther away and not used very much despite its extraordinary beauty. The Neptune pool was called the most luxurious and extravagant pool ever constructed. Located along the edge of the hill, the pool is bounded by a low wall—integrated concrete struts throughout the pool to allow movement during an earthquake. The pool started as a decorative pond and expanded as a swimming pool. Extensive statuary, temple, and columns were added to complete the image of ancient Greece. The immense pool holds almost 350,000 gallons of water. To accommodate the guests, seventeen changing rooms were constructed nearby.
American Modernism is uniquely an American concept and design, illustrating the people and the country in the industrialized world. The art-recognized the economic growth and cultural changes in the country and the growing entry of female artists. The artists illustrated scenes of life found in the cities and the countryside, encompassing the people's emotions in ordinary lives and circumstances.
[1] Retrieved from https://www.georgiaokeeffe.net/oriental-poppies.jsp
[2] Retrieved from https://www.smithsonianmag.com/arts-culture/story-behind-thomas-hart-bentons-incredible-masterwork-1-180953405/
[3] Retrieved from https://www.metmuseum.org/art/collection/search/488730?searchField=All&sortBy=Relevance&ft=edward+hopper&offset=0&rpp=20&pos=8
[4] Retrieved from https://human.libretexts.org/Bookshelves/Art/A_World_Perspective_of_Art_Appreciation_(Gustlin_and_Gustlin)/12%3A_The_Modern_Art_Movement_(1900_CE_%E2%80%93_1930_CE)/12.02%3A_American_Modernism_(1900_%E2%80%93_1930s)
[5] Retrieved from https://www.metmuseum.org/art/collection/search/271615
[6] Retrieved from https://www.metmuseum.org/art/collection/search/271570?searchField=All&sortBy=Relevance&ft=Alfred+Stieglitz&offset=0&rpp=20&pos=4
[7] Wright, F. L. (2010). The Essential Frank Lloyd Wright: Critical Writings on Architecture , Princeton University Press, p. 368.
[8] Retrieved from https://lib.calpoly.edu/search-and-find/collections-and-archives/architectural/julia-morgan/
[9] Retrieved from https://lib.calpoly.edu/search-and-find/collections-and-archives/architectural/julia-morgan/


7.1: Work
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Review Exercises
- Section A1.3 on the scalar product.
- Section A2.3 on integrals.
We introduce the concept of work as the starting point for building models using energy instead of forces. Work is a scalar quantity that is meant to represent how a force exerted on an object over a given distance results in a change in speed of that object. We will first introduce the concept of work done by a force on an object, and then look at how work can change the kinematics of the object. This is analogous to how we first defined the concept of force, and then looked at how force affects motion (by using Newton's Second Law, which connected the concept of force to the acceleration of the object).
The work done by a force, \(\vec F\), on an object over a displacement, \(\vec d\), is defined to be:
\[W = \vec F \cdot \vec d = Fd\cos\theta = F_xd_x+F_yd_y+F_zd_z\]
where \(\theta\) is the angle between the vectors when they are placed tail to tail, as in Figure 7.1.1 . The dimension of work, force times displacement, is also called "energy''. The S.I. unit for energy is the Joule (abbreviated \(\text{J}\)) which is equivalent to \(\text{Nm}\) or \(\text{kg m}^{2}/\text{s}^{2}\) in base units.
The work "done'' by the force is the scalar product of the force vector and the displacement vector of the object. We say that the force "does work'' if it is exerted while the object moves (has a displacement vector) and in such a way that the scalar product of the force and displacement vectors is non-zero. A force that is perpendicular to the displacement vector of an object does no work (since the scalar product of two perpendicular vectors is zero). A force exerted in the same direction as the displacement will do positive work (\(\cos\theta \) positive), and a force in the opposite direction of the displacement will do negative work (\(\cos\theta \) negative). As we will see, positive work corresponds to increasing the speed of the object, whereas negative work corresponds to decreasing its speed. No work corresponds to no change in speed (but could corresponds to a change in velocity).
Exercise \(\PageIndex{1}\)
A pendulum of length \(R\) consists of a mass connected to a string (Figure \(\PageIndex{2}\)). The string exerts a force of tension \(\vec F_T\) on the mass. What is the work done by tension when the pendulum swings through an angle \(\theta \)?
- \(W=F_{T}R\theta\)
- \(W=F_{T}R(1-\cos\theta)\)
- Tension does no work on the mass.
- Answer
- 3.
You may be tempted to ask, "Why work? Why not something else? Why that scalar product in particular? How could we possibly have thought of that?''. In general, it seems arbitrary that we introduce the quantity "work'' and then find that it leads to a convenient way of building models. However, we did not just pull this quantity out of thin air! Many theorists, over many years, tried all sorts of quantities and ways to rephrase Newton's Theory that were not helpful. The quantities that make it into textbooks are the ones that turned out to be useful. You should also keep in mind that, just like force, work is a "made-up'' mathematical tool that is helpful in describing the world around us. There is no such thing as work or energy; they are just useful mathematical tools.
Work in one dimension.
Work involves vectors, so we can first examine the concept in one dimension, before extending this to two and three dimensions. We can choose \(x\) as the coordinate in one dimension, so that all vectors only have an \(x\) component. We can write a force vector as \(\vec F=F\hat x\), where \(F\) is the \(x\) component of the force (which could be positive or negative). A displacement vector can be written as \(\vec d = d \hat x\), where again, \(d\) is the \(x\) component of the displacement, and can be positive or negative. In one dimension, work is thus:
\( W = \vec F \cdot \vec d = (F\hat x) \cdot ( d\hat x ) = Fd (\hat x\cdot\hat x)=Fd\)
where \(\hat x \cdot \hat x = 1\). Consider, for example, the work done by a force, \(\vec F\), on a box, as the box moves along the \(x\) axis from position \(x=x_0\) to position \(x=x_1\), as shown in Figure \(\PageIndex{3}\).
We can write the length of the displacement vector as \(||\vec d|| =d= \Delta x = x_1-x_0\). The work done by the force is given by:
\(W = \vec F \cdot \vec d = F\hat x\cdot \Delta x\hat x =F\Delta x =F(x_1-x_0)\)
which is a positive quantity, since \(x_1 > x_0\), with our choice of coordinate system.
Exercise \(\PageIndex{2}\)
A constant force in the positive \(x\) direction, \(\vec F\), acts on a box, as in Figure \(\PageIndex{3}\) . Consider the work done by \(\vec F\) as the box moves from \(x_1\) to \(x_0\). How does it compare to the work done by \(\vec F\) when moving from \(x_0\) to \(x_1\) (that we calculated above)?
- \(\vec F\) does no work on the box when it moves from \(x_0\) to \(x_1\).
- The work has the same magnitude as before, but the work is now negative.
- The work done by \(\vec F\) is the same in both cases.
- Answer
- B.
Work in one dimension - varying force
Suppose that instead of a constant force, \(\vec F\), we have a force that changes with position, \(\vec F(x)\), and can take on three different values between \(x=x_0\) and \(x=x_3\):
\begin{align*} \vec F (x)= \begin{cases} F_1\hat x & x<\Delta x \\[4pt] F_2\hat x & \Delta x \leq x< 2\Delta x \\[4pt] F_3\hat x & 2\Delta x \leq x \end{cases} \end{align*}
as illustrated in Figure \(\PageIndex{3}\), which shows the force on an object as it moves from position \(x=x_0\) to position \(x=x_3$\), along three (equal) displacement vectors, \(\vec d_1=\vec d_2=\vec d_3=\Delta x \hat x\)
The total work done by the force over the three separate displacements is the sum of the work done over each displacement:
\begin{align*} W^{tot}&=W_1+W_2+W_3\\[4pt] &=\vec F_1\cdot \vec d_2+\vec F_2\cdot \vec d_2+\vec F_3\cdot \vec d_3\\[4pt] &= F_1\Delta x +F_2\Delta x + F_3\Delta x \end{align*}
If instead of \(3\) segments we had \(N\) segments and the \(x\) component of the force had the \(N\) corresponding values \(F_i\) in the \(N\) segments, the total work done by the force would be:
\begin{align*} W^{tot} = \sum_{i=0}^N\vec F_i \cdot \Delta \vec x \end{align*}
where we introduced a vector \(\Delta \vec x\) to be the vector of length \(\Delta x\) pointing in the positive \(x\) direction. In the limit where \(\vec F(x)\) changes continuously as a function of position, we take the limit of an infinite number of infinitely small segments of length \(dx\), and the sum becomes an integral:
\begin{align} W^{tot} = \int_{x_0}^{x_f}\vec F(x) \cdot d\vec x \end{align}
where the work was calculated in going from \(x=x_0\) to \(x=x_f\), and \(d\vec x=dx\hat x\) is an infinitely small displacement vector (of length \(dx\)) in the positive \(x\) direction.
Example \(\PageIndex{1}\)
A block is pressed against the free end of a horizontal spring with spring constant, \(k\), so as to compress the spring by a distance \(D\) relative to its rest length, as shown in Figure \(\PageIndex{5}\) . The other end of the spring is fixed to a wall. What is the work done by the spring force on the block in going from \(x=-D\) to \(x=0\)? What is the work done by the block on the spring over the same displacement?
Solution
The force exerted by the spring on the block changes continuously with position, according to Hooke's law: \begin{align*} \vec F(x) = -kx \hat x \end{align*}
and points in the positive \(x\) direction when the end of the spring has a negative $x$ position (with our coordinate choice illustrated in Figure \(\PageIndex{5}\), where the origin is located at the rest length of the spring). To calculate the work done by the force, we sum the work done by the force over many infinitesimally small displacements \(d\vec x\) (using an integral):
\begin{align*} W &= \int_{-D}^0 \vec F(x) \cdot d\vec x\\[4pt] &=\int_{-D}^0 (-kx \hat x) \cdot (dx \hat x)\\[4pt] &=\int_{-D}^0 -kxdx (\hat x \cdot \hat x)\\[4pt] &=-\int_{-D}^0 kx dx\\[4pt] &=-\left[\frac{1}{2}kx^2 \right]_{-D}^0\\[4pt] &=\frac{1}{2}kD^2 \end{align*}
In order to determine the work that was done by the block on the spring, we need to determine the force, \(\vec F'(x)\), exerted by the block on the spring. By Newton's Third Law, this is equal in magnitude but opposite in direction to the force exerted by the spring on the block:
\begin{align*} \vec F'(x) = -\vec F(x) = kx \hat x \end{align*}
The work done by the block on the spring over the same displacement is:
\begin{align*} W' &= \int_{-D}^0 \vec F'(x) \cdot d\vec x\\[4pt] &=\int_{-D}^0 (kx \hat x) \cdot (dx \hat x)\\[4pt] &=\int_{-D}^0 kx dx=-\frac{1}{2}kD^2\\[4pt] \end{align*}
which is negative. This makes sense because the force exerted by the block on the spring is in the direction opposite to the direction of displacement, so the work should be negative.
Work in multiple dimensions
First, consider the work done by a force \(\vec F\) in pulling a crate over a displacement \(\vec d\), in the case where the force is directed at an angle \(\theta\) above the horizontal, as shown in Figure \(\PageIndex{6}\), and the displacement is along the \(x\) axis (or rather, we chose the \(x\) axis to be parallel to the displacement).
The work done by the force is given by:
\begin{align*} W = \vec F \cdot \vec d &= Fd\cos\theta\\[4pt] &= F_{\parallel}d\\[4pt] &= Fd_{\parallel}\\[4pt] \end{align*}
where we highlighted the fact that the scalar product "picks out'' components of vectors that are parallel to each other. \(F_{\parallel} = F\cos\theta\) is the component of \(\vec F\) that is parallel to \(\vec d\), and \(d_{\parallel}=d\cos\theta\) is the component of \(\vec d\) that is parallel to \(\vec F\). These are also shown in Figure \(\PageIndex{6}\).
Exercise \(\PageIndex{3}\)
Brent and Dean pull two crates by using ropes that make the same angle above the horizontal and with the same force. The magnitude of the crates' displacement is the same, but Dean's crate moves horizontally on the ground while Brent's crate moves up a frictionless ramp that is parallel to the rope used to pull the crate. Who did more work on the crate?
- Dean because there is friction between his crate and the ground.
- Brent.
- They did the same amount of work.
- Answer
- B.
In general, if an object is moving along an arbitrary path, we cannot choose the $x$ axis to be parallel to the displacement or to the force. If the path can be sub-divided into straight segments over which the force is constant, as in Figure \(\PageIndex{7}\), we can calculate the work done by the force over each segment and add the work done in each segment together to obtain the total work done by the force. Note that, in general, the work done by a force as an object moves from one position to another depends on the particular path that was taken between the two positions, since different paths will have difference lengths.
Example \(\PageIndex{2}\)
Compare the work done by the force of kinetic friction in sliding a crate along a horizontal surface from position \(A\) (coordinates \(x_A, y_A\)) to position \(B\) (coordinates \(x_B, y_B\)) using the two different paths depicted in Figure \(\PageIndex{8}\). Assume that the mass of the crate is \(m\) and that the coefficient of kinetic friction between the crate and the ground is \(\mu_k\).
Solution
The force of kinetic friction is always in the direction opposite to that of motion. Thus, regardless of the path taken, the force of friction will do negative work.
Let us first calculate the work done by the force of kinetic friction along the first path (the straight line). The force of kinetic friction will have a magnitude: \begin{align*} f_k = \mu_k N = \mu_k mg \end{align*}
The normal force will have the same magnitude as the weight because the crate is not moving (accelerating) in the direction perpendicular to the \(xy\) plane. The displacement vector from \(A\) to \(B\) can be written as: \begin{align*} \vec d &= (x_B-x_A)\hat x + (y_B-y_A) \hat y\\[4pt] \therefore ||\vec d|| &=d= \sqrt{(x_B-x_A)^2 - (y_B-y_A)^2} \end{align*}
The force of kinetic friction will be in the opposite direction of the displacement vector, so the angle between the two vectors is \(180^{\circ}(\cos\theta=-1)\). The work done by the force of kinetic friction is thus: \begin{align*} W = \vec f_k \cdot\vec d = f_k d \cos\theta = -\mu_k mg\sqrt{(x_B-x_A)^2 - (y_B-y_A)^2} \end{align*} and is negative, as expected.
For path 2, we break up the motion into two segments, with displacements vectors \(\vec d_1\) (along \(y\)) and \(\vec d_2\) (along \(x\)). We can write the two displacement vectors as:
\begin{align*} \vec d_1 &= 0\hat x + (y_B-y_A) \hat y\\[4pt] \therefore ||\vec d_1||&=d_1=(y_B-y_A)\\[4pt] \vec d_2 &= (x_B-x_A)\hat x + 0 \hat y\\[4pt] \therefore ||\vec d_2||&=d_2=(x_B-x_A)\\[4pt] \end{align*}
Along each segment, the force of kinetic friction is anti-parallel to the displacement (note that the force of friction changes direction over the two segments), but the magnitude is \(f_k=\mu_kmg\). The work done along the first segment is thus: \begin{align*} W_1 = \vec f_k \cdot \vec d_1 = f_k d_1 \cos\theta = -\mu_k mg(y_B-y_A) \end{align*} The work done along the second segment is: \begin{align*} W_2 = \vec f_k \cdot \vec d_2 = f_k d_2 \cos\theta = -\mu_k mg(x_B-x_A) \end{align*} And the total work done by the force of kinetic friction over the second path is: \begin{align*} W^{tot} = W_1 + W_2 = -\mu_k mg \left((x_B-x_A) + (y_B-y_A)\right) \end{align*} which is more work than was done along path 1. This makes sense because for both paths, the force of friction has the same magnitude and is always in the opposite direction of motion; thus, the longer the path, the more work will be done by the force.
Example \(\PageIndex{3}\)
A box of mass \(m\) is moved from the floor onto a table using two different paths, as shown in Figure \(\PageIndex{9}\). The table is a horizontal distance \(L\) away from where the box starts and a height \(H\) above the floor. Compare the work done by the weight of the box along the two possible paths.
Solution
We can use a coordinate system such that the origin coincides with the initial position of the box. \(x\) is horizontal and \(y\) is vertical, as shown in Figure \(\PageIndex{9}\). The weight of the box can be written as:
\begin{align*} \vec F_g = -mg \hat y \end{align*}
and points in the negative \(y\) direction with a magnitude of \(mg\). To calculate the work done by the weight along the first path, we first determine the corresponding displacement vector, \(\vec d\):
\begin{align*} \vec d = L\hat x + H\hat y \end{align*}
and we can then determine the work:
\begin{align*} W &= \vec F_g \cdot \vec d = (-mg \hat y) \cdot (L\hat x + H\hat y)\\[4pt] &=F_xd_x+F_yd_y= (0)(L) + (-mg)(H)\\[4pt] &= -mgH \end{align*}
Along path 1, the work done by the weight is negative, and does not depend on the horizontal distance \(L\). Let us now calculate the work done along the second path, which we break up into two segments with displacement vectors \(\vec d_1\) (vertical) and \(\vec d_2\) (horizontal). The displacement vectors are:
\begin{align*} \vec d_1 &= H\hat y\\[4pt] \vec d_2 &= L\hat x \end{align*}
The work done along the vertical segment is:
\begin{align*} W_1 &= \vec F_g \cdot \vec d_1 = (-mg \hat y) \cdot (H\hat y)\\[4pt] &=-mgH \end{align*}
The work done along the horizontal segment is:
\begin{align*} W_2 &= \vec F_g \cdot \vec d_2 = (-mg \hat y) \cdot (L\hat x)\\[4pt] &=0 \end{align*}
which is zero, because the force of gravity is always vertical and thus perpendicular to the displacement vector of the horizontal segment. The total work done by the weight along the second path is:
\begin{align*} W^{tot} = W_1 + W_2 = -mgH \end{align*}
which is the same as the work done along path 1. As we will see, when a force is constant in magnitude and direction, the work that it does on an object in going from one position to another is independent of the path taken. This was not the case in Example 7.1.2 , because the direction of the force of kinetic friction depends on the direction of the displacement.
Exercise \(\PageIndex{4}\)
Clare and Amelia go down two different slides, as shown in Figure \(\PageIndex{10}\). Clare and Amelia have the same mass and the slides have the same non-zero coefficients of friction.
For each of the following forces, decide whether the force: does more work on Clare, does more work on Amelia, or does the same amount of work on both.
- The force of gravity...
- The force of friction...
- The normal force from the slide...
- Answer
-
Gravity does the same amount of work on both, friction does more work on Amelia, and the normal force does the same amount of work on both (the normal force does zero work, since it is always perpendicular to the displacement).
The most general case for which we can calculate the work done by a force is the case when the force changes continuously along a path where the displacement also changes direction continuously. This is illustrated in Figure \(\PageIndex{11}\) which shows an arbitrary path between two points \(A\) and \(B\), and a force, \(\vec F(\vec r)\), that depends on position (\(\vec r\)). In general, the work done by the force on an object that goes from \(A\) to \(B\) will depend on the actual path that was taken.
The strategy for calculating the work in the general case is the same: we break up the path into small straight segments with displacement vectors \(d\vec l\) (Figure \(\PageIndex{12}\)) where we assume that the force is constant over the segment. The total work is the sum of the work over each segment:
\begin{align} W = \int_A^B \vec F(\vec r) \cdot d\vec l \end{align}
As usual, we use the integral symbol to indicate that you need to take an infinite number of infinitely small segments \(d\vec l\) in order to calculate the sum.
You should note that this is not an integral like any other that we have seen so far: the integral is not over a single integration variable (usually we use \(x\)), but it is the integral (the sum!) over the specific path that we have chosen in going from \(A\) to \(B\). This is called a "path integral'', and is generally difficult to evaluate.
Example \(\PageIndex{4}\)
A force, \(\vec F(\vec r) = \vec F(x,y) = F_x\hat x + F_y \hat y\), is exerted on an object. The object starts at position \(A\) and ends at position \(B\), along a parabolic path, \(y(x) = a+bx^2\), as depicted in Figure \(\PageIndex{13}\). What is the work done by the force, \(\vec F\), along this trajectory?
Solution
In this case, the force can change with position (if \(F_x\) and \(F_y\) are not constant), and the direction of the path changes continuously. When we break up the path into small segments \(d\vec l\), we need to incorporate the equation of the parabola to include the fact that \(d\vec l\)$ must always be tangent to the parabola. Consider one small segment along the trajectory and the infinitesimal displacement vector \(d\vec l\) at that point, as in Figure \(\PageIndex{14}\).
We can write the \(x\) and \(y\) components of the vector as infinitesimal distances, \(dx\) and \(dy\), along the \(x\) and \(y\) axes, respectively. The vector \(d\vec l\) can thus be written:
\begin{align*} d\vec l = dx \hat x + dy \hat y \end{align*}
The total work done by the force is then:
\begin{align*} W &= \int_A^B \vec F(\vec r) \cdot d\vec l\\[4pt] &=\int_A^B (F_x\hat x + F_y \hat y) \cdot (dx \hat x + dy \hat y)\\[4pt] &=\int_A^B (F_x dx + F_ydy)\\[4pt] \therefore W&= \int_A^B F_x dx + \int_A^B F_ydy \end{align*}
where in the last line, we simply used the property that the integral of a sum is the sum of the corresponding integrals. At this point, we have two integrals over integration variables (\(x\) and \(y\)) that are meaningful. However, we have not yet used the fact that our path is a parabola, and in general, we expect that the shape of the path is important. By saying that we are integrating (or calculating the work) over a specific path, we are really saying that \(x\) and \(y\) are not independent; that is, if we know the value of \(x\) at some point on the path, we know the corresponding value of \(y\) (\(y = a+bx^2\)).
Since \(x\) and \(y\) are not independent, we can use a "substitution of variables'' in order to express \(y\) in terms of \(x\), and \(dy\) in terms of \(dx\)$:
\begin{align*} y(x) &= a + bx^2\\[4pt] \frac{dy}{dx} &= 2bx\\[4pt] \therefore dy &= 2bxdx \end{align*}
This allows us to convert the integral over \(y\) to an integral over \(x\), which also allows us to be explicit for the limits of the integral (in our example, the integral goes from \(x=0\) to \(x=x_1\)):
\begin{align*} W&= \int_A^B F_x dx + \int_A^B F_ydy\\[4pt] &=\int_0^{x_1} F_x dx + \int_0^{x_1} F_y(2bxdx)\\[4pt] &=\int_0^{x_1} (F_x + 2bxF_y)dx \end{align*}
where we would need to know how \(F_x\) and \(F_y\) depends on \(x\) and \(y\) in order to actually evaluate the integral.
For example, if the force were constant (\(F_x\) and \(F_y\) constant), then the work done along the parabolic path would be:
\begin{align*} W &= \int_0^{x_1} (F_x + 2bxF_y)dx\\[4pt] &=\left[F_x x + bF_yx^2 \right]_0^{x_1}\\[4pt] &=F_x x_0 + bF_yx_0^2 \end{align*}
As we mentioned earlier, if the force is constant in magnitude and direction , then the work done is independent of path. We can easily check this, using the displacement vector \(\vec d = x_1\hat x + bx_1^2 \hat y\):
\begin{align*} W &= \vec F \cdot \vec d = (F_x\hat x+ F_y\hat y) \cdot (x_1\hat x + bx_1^2 \hat y)\\[4pt] &=F_x x_1 + bF_yx_1^2 \end{align*}
as we found above.
Net work done
So far, we have considered the work done on an object by a single force. If more than one force is exerted on an object, then each force can do work on the object, and we can calculate the "net work'' done on the object by adding together the work done by each force. We will show that this is equivalent to first calculating the net force on the object, \(F^{net}\) (i.e. the vector sum of the forces on the object), and then calculating the work done by the net force.
Suppose that three forces, \(\vec F_1\), \(\vec F_2\), and \(\vec F_3\) are exerted on an object as it moves such that its displacement vector is \(\vec d\). The net work done on the object is easily shown to be equivalent to the work done by the net force::
\begin{align*} W^{net} &= W_1 + W_2 + W_3 \\[4pt] &= \vec F_1 \cdot \vec d + \vec F_2 \cdot \vec d + \vec F_3 \cdot \vec d \\[4pt] &=(F_{1x}d_x+F_{1y}d_y+F_{1z}d_z)+ (F_{2x}d_x+F_{2y}d_y+F_{2z}d_z) + (F_{3x}d_x+F_{3y}d_y+F_{3z}d_z)\\[4pt] &=(F_{1x} + F_{2x} + F_{3x})d_x+(F_{1y} + F_{2y} + F_{3y})d_y+(F_{1z} + F_{2z} + F_{3z})d_z\\[4pt] &=\vec F^{net} \cdot \vec d \end{align*}
where \(\vec F^{net} = \vec F_1 + \vec F_2 + \vec F_3\) is the net force. The result is easily generalized to any number of forces, including if those forces change as a function of position:
\begin{align*} W^{net} = \int_A^B F^{net}(\vec r) \cdot d\vec l \end{align*}
Example \(\PageIndex{5}\)
You push with an unknown horizontal force, \(\vec F\), against a crate of mass \(m\) that is located on an inclined plane that makes an angle $\theta$ with respect to the horizontal, as shown in Figure \(\PageIndex{15}\). The coefficient of kinetic friction between the crate and the incline is \(\mu_k\). You push in such a way that that crates moves at a constant speed up the incline. What is the net work done on the crate if it moves up the incline by a distance \(d\)?
Solution
Although the answer may be obvious, let's go the long way about it and calculate the work done by each force, and then sum them together to get the total work done. We start by identifying the forces exerted on the crate:
- \(\vec F\), the applied force, of unknown magnitude, \(\vec F\).
- \(\vec F_g\), the weight of the crate, with magnitude \(mg\).
- \(\vec N\), a normal force exerted by the incline.
- \(\vec f_k\), a force of kinetic friction, with magnitude \(\mu_k N\), that points in the direction opposite of \(\vec d\).
These are shown in the free-body diagram in Figure \(\PageIndex{16}\), along with our choice of coordinate system, and the displacement vector.
With our choice of coordinate system, the displacement vector is given by:
\begin{align*} \vec d = d (\cos\theta \hat x + \sin\theta \hat y) \end{align*}
Before calculating the work done by each force, we need to determine the magnitude of the normal force (and thus of the force of kinetic friction). Since the crate is moving at a constant velocity, its acceleration is zero , so the sum of the forces must be zero. Writing out the \(y\) component of Newton's Second Law allows us to find the magnitude of the normal force:
\begin{align*} \sum F_y &= N\cos\theta -F_g - f_k\sin\theta = 0\\[4pt] \therefore mg &= N\cos\theta-\mu_kN\sin\theta = N(\cos\theta-\mu_k\sin\theta)\\[4pt] \therefore N &= \frac{mg}{\cos\theta-\mu_k\sin\theta} \end{align*}
Writing out the \(x\) component of Newton's Second Law allows us to find the magnitude of the unknown force \(F\):
\begin{align*} \sum F_x &= F - N\sin\theta - f_k\cos\theta = 0\\[4pt] \therefore F &= N\sin\theta+\mu_kN\cos\theta = N(\sin\theta+\mu_k\cos\theta)\\[4pt] &=mg\frac{\sin\theta+\mu_k\cos\theta}{\cos\theta-\mu_k\sin\theta} \end{align*}
We now proceed to calculate the work done by each force. The work done by the normal force is identically zero, since it is perpendicular to the displacement vector. The work done by the applied force, \(\vec F = F\hat x\), is:
\begin{align*} W_F &= \vec F \cdot \vec d = (F\hat x)\cdot(d (\cos\theta \hat x + \sin\theta \hat y))\\[4pt] &=Fd\cos\theta=mg\frac{\sin\theta+\mu_k\cos\theta}{\cos\theta-\mu_k\sin\theta}d\cos\theta \end{align*}
The work done by the force of gravity, \(\vec F_g = -mg \hat y\), is:
\begin{align*} W_g &= \vec F_g \cdot \vec d = (-mg \hat y)\cdot(d (\cos\theta \hat x + \sin\theta \hat y))\\[4pt] &=-mgd\sin\theta \end{align*}
The work done by the force of friction, \(\vec f_k\), noting that \(\vec f_k\) and \(\vec d\) are antiparallel:
\begin{align*} W_f &= \vec f_k \cdot \vec d = -f_kd = -\mu_kNd\\[4pt] &=-\mu_k\frac{mg}{\cos\theta-\mu_k\sin\theta}d \end{align*}
The net work done on the crate is thus:
\begin{align*} W^{net}&=W_F + W_g + W_f\\[4pt] &=mg\frac{\sin\theta+\mu_k\cos\theta}{\cos\theta-\mu_k\sin\theta}d\cos\theta-mgd\sin\theta -\mu_k\frac{mg}{\cos\theta-\mu_k\sin\theta}d\\[4pt] &=mgd \left( \frac{\sin\theta+\mu_k\cos\theta}{\cos\theta-\mu_k\sin\theta}\cos\theta - \sin\theta - \mu_k\frac{1}{\cos\theta-\mu_k\sin\theta} \right)\\[4pt] &=mgd \left( \frac{(\sin\theta+\mu_k\cos\theta)\cos\theta - \sin\theta(\cos\theta-\mu_k\sin\theta) - \mu_k}{\cos\theta-\mu_k\sin\theta} \right)\\[4pt] &=mgd \left( \frac{\sin\theta\cos\theta+\mu_k\cos^2\theta - \sin\theta\cos\theta+\mu_k\sin^2\theta - \mu_k}{\cos\theta-\mu_k\sin\theta} \right)\\[4pt] &=mgd \left( \frac{\mu_k(\cos^2\theta+\sin^2\theta) - \mu_k}{\cos\theta-\mu_k\sin\theta} \right)\\[4pt] &=0 \end{align*}
where we used the fact that \(\cos^2\theta+\sin^2\theta=1\). Thus we find that the net work done on the crate is zero!
Discussion
Of course, this makes sense, because the net force on the crate is zero, since it is not accelerating, so the net work done is also zero. As a consequence, or rather, by construction, we have the condition that if the net work done on an object is zero, then that object does not accelerate. We thus have a scalar quantity (work) that can tell us something about whether an object is changing speed. In the next section, we introduce a new quantity, "kinetic energy'', to describe how an object's speed changes when the net work done is not zero. \end{example}
Olivia's Thoughts
Pay close attention to the words "on'' and "by.'' There are a few things about this that can be tricky:
- In Example 7.1.5 , we were asked to find the net work done on the crate. Sometimes, the question won't specify that it wants you to find the net work, and will just say "What is the work done on the crate?'' When you are just asked for the work done "on'' an object, the question is implicitly asking for the net work done on the object.
- Just because the net work done on an object is zero doesn't mean that the work done by each of the forces is zero. This may seem obvious, but it's easy to get tripped up on a test or exam. If you are reading a question about work and it says that the object is moving at a constant speed, it's tempting to just jump ahead and say that the work must be equal to zero. However, you can only say this if it's asking you for the net work done on the object. For instance, in Example 7.1.5 , we concluded that since the crate was moving at a constant speed, the net work was equal to zero. But if the question asked you to find the work done on the crate by gravity, that would mean something different. The work done by gravity in this case is not equal to zero (it's actually negative).
- The work done "on'' an object is not the same as the net work done "by'' that object. For example, say you are in a tug-of-war and you pull the other team towards you, but you yourself do not move. The net work done on you is zero, but the work done by you is not zero. So, when you are talking about work, you should always state explicitly whether the work is being done "on'' the object or "by'' the object.
Note : The wording won't always be like this - sometimes it will say "How much work do you do on the box?'' instead of "How much work is done by you on the box,'' so always be careful. Still, looking for key words like "by'' and "on'' is a good place to start.
Exercise \(\PageIndex{1}\)
A \(2\text{kg}\) box sits on a horizontal surface. A constant horizontal force of \(6\text{N}\) is applied to the box. The box moves with a constant acceleration of \(2\text{m/s}^{2}\). Which of the following has the greatest magnitude?
- The work done by the applied force.
- The work done by friction.
- The net work done on the box.
- Answer
- A.


6.9A: Growth Rate and Temperature
Bacteria may grow across a wide range of temperatures, from very cold to very hot.
LEARNING OBJECTIVES
Describe how the growth of bacteria is affected by temperature and how bacterial growth can be measured
Key Takeaways
Growth Rate and Temperature
Bacterial growth is the division of one bacterium into two daughter cells in a process called binary fission. Providing no mutational event occurs the resulting daughter cells are genetically identical to the original cell. Hence, local doubling of the bacterial population occurs. Both daughter cells from the division do not necessarily survive. However, if the number surviving exceeds unity on average, the bacterial population undergoes exponential growth. The measurement of an exponential bacterial growth curve in batch culture was traditionally a part of the training of all microbiologists. The basic means requires bacterial enumeration (cell counting) by direct and individual (microscopic, flow cytometry), direct and bulk (biomass), indirect and individual (colony counting), or indirect and bulk (most probable number, turbidity, nutrient uptake) methods. Models reconcile theory with the measurements.
Bacteria may grow across a wide range of temperatures, from very cold to very hot. A mesophile is an organism that grows best in moderate temperature, neither too hot nor too cold. All human pathogens are mesophiles. Organisms that prefer extreme environments are known as extremophiles: those that prefer cold environments are termed psychrophilic, those preferring warmer temperatures are termed thermophilic or thermotrophs and those thriving in extremely hot environments are hyperthermophilic.
For example, in molecular biology, the cold-shock domain (CSD) is a protein domain of about 70 amino acids which has been found in prokaryotic and eukaryotic DNA-binding proteins. Part of this domain is highly similar to the RNP-1 RNA-binding motif. When Escherichia coli is exposed to a temperature drop from 37 to 10 degrees Celsius, a four to five hour lag phase occurs and then growth is resumed at a reduced rate. During the lag phase, the expression of around 13 proteins, which contain cold shock domains is increased two- to ten-fold. These so-called cold shock proteins are thought to help the cell survive in temperatures lower than optimum growth temperature, by contrast with heat shock proteins, which help the cell survive in temperatures greater than the optimum, possibly by condensation of the chromosome and organization of the prokaryotic nucleoid.
Key Points
- The basic means of measuring growth requires bacterial enumeration (cell counting).
- Methods for bacterial cell counting include: 1. direct and individual (microscopic, flow cytometry), 2. direct and bulk (biomass), 3. indirect and individual (colony counting), or 4. indirect and bulk (most probable number, turbidity, nutrient uptake).
- A mesophile is an organism that grows best in moderate temperature, neither too hot nor too cold. All human pathogens are mesophiles.
- Cold shock proteins help the cell to survive in temperatures lower than optimum growth temperature.
- Heat shock proteins help the cell to survive in temperatures greater than the optimum, possibly by condensation of the chromosome and organization of the prokaryotic nucleoid.
Key Terms
- mesophile : An organism, especially a microorganism, that lives and thrives at moderate temperatures.
- psychrophile : An organism that can live and thrive at temperatures much lower than normal; a form of extremophile.
- thermophile : An organism that lives and thrives at relatively high temperatures; a form of extremophile; many are members of the Archaea.


10.2: Soldering Tools
Historically, heating the host metals and melting the solder was done by firing in a forge or by pouring molten solder onto the metals and wiping it into place with leather pads. Later, heat was applied by means of an iron bit that was heated in the forge. The name soldering iron has carried forward to this day. Today, heat is applied by various electrically heated soldering tools called irons or pencils or guns . Figure 3 shows two examples. The majority of electronic soldering done during electronic repair, for example, is completed with a low-wattage soldering pencil.
- Some common soldering tools
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Soldering tip shapes
Regardless of the shape, size, or design of the iron, the tip must be tinned. Tinning the tip is the process of applying a thin layer of solder to the tip to keep atmospheric oxygen and other contaminants off the soldering surface and help with the flow of the solder. A poorly tinned tip will make it virtually impossible to achieve a sound solder joint.
Soldering iron manufacturers specify an operating temperature range for each type of tip. This requires mating the heating capability of iron and tip. Insufficient iron capability will result in tip temperatures that are lower than needed for quality solder work or in rapid drop in tip temperature during soldering. Excessive heat will quickly deteriorate the tip and may possibly damage the components and printed circuit board being soldered.
You can estimate tip temperature by following this two-step process:
- Apply a small amount of solder to the flat surface of the tip and immediately wipe the tip with a damp cellulose sponge or paper towel.
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Observe tip colour immediately after wiping.
- If the color of the surface is silver, the temperature is between 315°C and 370°C (600°F and 698°F).
- If the tip shows gold streaks, the temperature is approaching 425°C (797°F).
The copper tips of irons and pencils are progressively dissolved by solder, and they soon become pitted and corroded. This is particularly true of the continuous-heat types. It is virtually impossible to do quality soldering with a corroded tip. Corroding can be slowed down by keeping the tip clean and well tinned.
Clean the tip by wiping it frequently with a damp cloth or cellulose sponge. The damp wipe will shock the built-up burned flux from the tip. Immediately re-tin the tip and leave a thin coating to keep atmospheric oxygen off the soldering surface. Wipe the excess solder off the tip before the next use. Reclean and recoat with solder when you finish each soldering task.
When pitting becomes significant, you should dress the tip and reshape it to clean metal with a fine file. This can be done with the tip hot so that the refreshed tip can immediately be re-tinned. Excess solder should be wiped away after re-tinning. Steel-clad tips suffer much less corrosion and should never be dressed with a file. Like copper tips, however, they should be cleaned frequently when hot.


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21.1: Decimals Fractions and Percents
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21.1.1: Comparing Fractions, Decimals, and Percents
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In this section, we will go over techniques to compare two numbers. These numbers could be presented as fractions, decimals or percents and may not be in the same form. For example, when we look at a histogram, we can compute the fraction of the group that occurs the most frequently. We might be interested in whether that fraction is greater than 25% of the population. By the end of this section we will know how to make this comparison.
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21.1.2: Converting Between Fractions, Decimals and Percents
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In this section, we will convert from decimals to percents and back. We will also start with a fraction and convert it to a decimal and a percent. In statistics we are often given a number as a percent and have to do calculations on it. To do so, we must first convert it to a percent. Also, the computer or calculator shows numbers as decimals, but for presentations, percents are friendlier. It is also much easier to compare decimals than fractions, thus converting to a decimal is helpful.
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21.1.3: Decimals- Rounding and Scientific Notation
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In this section, we will go over how to round decimals to the nearest whole number, nearest tenth, nearest hundredth, etc. In most statistics applications that you will encounter, the numbers will not come out evenly, and you will need to round the decimal.
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21.1.4: Using Fractions, Decimals and Percents to Describe Charts
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Charts, such as bar charts and pie charts are visual ways of presenting data. You can think of each slice of the pie or each bar as a part of the whole. The numerical versions of this are a list of fractions, decimals and percents. By the end of this section we will be able to look at one of these charts and produce the corresponding fractions, decimals, and percents.


1.9: Conjugation, Color, and the Chemistry of Vision
After completing this section, you should be able to
- explain why some organic compounds have different colors based on compound structure and our perception of light.
- state the relationship between frequency of light absorbed and the extent of conjugation in an extended pi electron system.
Introduction
An obvious difference between certain compounds is their color. Thus, quinone is yellow; chlorophyll is green; and aspirin is colorless. In this respect the human eye is functioning as a spectrometer analyzing the light reflected from the surface of a solid or passing through a liquid. Although we see sunlight (or white light) as uniform or homogeneous in color, it is actually composed of a broad range of radiation wavelengths in the ultraviolet (UV), visible and infrared (IR) portions of the spectrum. As shown on the image below, the component colors of the visible portion can be separated by passing sunlight through a prism, which acts to bend the light in differing degrees according to wavelength.
Visible wavelengths cover a range from approximately 400 to 800 nm. The longest visible wavelength is red and the shortest is violet. The wavelengths of what we perceive as particular colors in the visible portion of the spectrum are displayed and listed below.
- Violet: 400 - 420 nm
- Indigo: 420 - 440 nm
- Blue: 440 - 490 nm
- Green: 490 - 570 nm
- Yellow: 570 - 585 nm
- Orange: 585 - 620 nm
- Red: 620 - 780 nm
When white light passes through or is reflected by a colored substance, a characteristic portion of the mixed wavelengths is absorbed. The remaining light will then assume the complementary color to the wavelength(s) absorbed. This relationship is demonstrated by the color wheel shown below. Here, complementary colors are diametrically opposite each other. Thus, absorption of violet (400-440 nm) light renders a substance yellow, and absorption of 490-560 nm (green) light makes it red. Green is unique in that it can be created by absorption close to 400 nm as well as absorption near 800 nm.
Early humans valued colored pigments, and used them for decorative purposes. Many of these were inorganic minerals, but several important organic dyes were also known. These included the crimson pigment, kermesic acid, the blue dye, indigo, and the yellow saffron pigment, crocetin. A rare dibromo-indigo derivative, punicin, was used to color the robes of the royal and wealthy. A common feature of all these colored compounds, displayed below, is a system of extensively conjugated \(\pi\)-electrons .
By understanding visible light, complementary colors, and adsorption the color of organic compounds can be understood. Beta-carotene, a compound found in carrots, is a deep orange color. Beta-carotene has 11 conjugated double bonds which places its λ max at 455 nm which is within the blue region of the visible spectrum. The Beta-carotene compound absorbs blue from white light so it appears orange which is the the complementary color of blue. (Figure \(\PageIndex{1}\)).
Another example is seen in the absorption spectrum of another food coloring, Blue #1:
Blue #1 absorbs at a λ max at 630 nm which is the color red in the visible spectrum. Blue is the complementary color of red.
Mechanism of Vision
Conjugation is also important in light sensitive compounds used for vision. Beta carotene, found in carrots and other natural products is cleaved into the liver and converted into Vitamin A, also known as retinol. Vitamin A is critical for vision because it is needed by the retina of eye. Retinol can be oxidized to an aldehyde called retinal which is also important for vision. (Figure \(\PageIndex{2}\)).
The eye is an extraordinarily sensitive instrument. Although its wavelength response is restricted to 400-800 nm, but its degree of sensitivity is such that a fully dark-adapted eye can clearly detect objects in light so dim as to correspond to a light input over the retina of only about 10,000 quanta per second - one light quantum per three minutes per receptor cell in the retina!
The retina is made up of two kinds of light-sensitive (photoreceptor) cells, known as rods and cones. The rods are the more sensitive and are responsible for vision in dim light. The cones are much fewer in number than the rods and provide detail and color vision in good light. The part of the retina that corresponds to the center of the visual field contains only cones. A red pigment called rhodopsin is the photosensitive substance in the rod cells of the retina. It absorbs most strongly in the blue-green region of the visible spectrum (λ max = 500nm) and is essentially unaffected by the far-red end of the spectrum. Cone vision appears to involve a different pigment called iodopsin , which absorbs farther toward the red than does rhodopsin.
Rhodopsin, is made up of a protein (opsin) and retinal. Opsin does not absorb visible light, but when it is bonded with 11-cis-retinal to from rhodopsin, the new molecule has a very broad absorption band in the visible region of the spectrum. Rhodopsin is formed by an an imine (Schiff base) functional group formed between the aldehyde group of the retinal and the side-chain amino function of a lysine unit of opsin.
Opsin itself is colorless, whereas 11-cis-retinal absorbs strongly at 370 nm. The combination of opsin with 11-cis-retinal produces a remarkable shift of λ max to longer wavelengths (430 nm to 620 nm, depending on the species). Light striking the retina changes the color of rhodopsin from red to yellow. The primary photochemical event in this process was established by G. Wald (Nobel Laureate in Physiology and Medicine, 1967), who showed that light absorption led to a change of configuration about the C 11 -C 12 double bond of the retinal moiety fo rhodopsin from cis to trans to form a compound called metarhodopsin II. The new form of trans-retinal does not fit as well into the opsin protein, and so a series of geometry changes in the protein begins. As the protein changes its geometry, it initiates a cascade of biochemical reactions that result in changes in charge so that a large potential difference builds up across the neuron membranes. This potential difference is passed along to an adjoining nerve cell as an electrical impulse. The nerve cell carries this impulse to the brain, where the visual information is interpreted.
Metarhodopsin II can then be recycled back to rhodopsin by first cleaving to form all- trans -retinal and the isomerization back to 11- cis -retinal by the enzyme retinal isomerase. Finally, 11- cis -retinal is once again coupled with opsin to form rhodopsin.
Lycopene is conjugated compound found in tomatoes. What colors correspond to the UV-Vis absorption maxima of lycopene? What color do we see when we look at lycopene?
- Answer
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The absorption maxima of lycopene are regions of blue and green in the visible spectrum. Lycopene would be expected to appears orange and red.
Aspirin has a λ max of 220 nm. Briefly explain why Aspirin appears white.
- Answer
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Aspirin absorbs in the ultraviolet portion of the electromagnetic spectrum. Aspirin cannot remove any colors from white light by absorption so the compound itself appears white. Many compounds with a relatively small amount of conjugation appear white because they only absorb in the ultraviolet region and not the visible region.