 in the work equation is the linear distance over which the force acts. When you use a wrench to turn a bolt, less effort is used (greater torque is produced) as the distance you place your hand from the bolt increases (Fig. 7.29). Plumbers often use a wrench with a long torque arm to loosen or tighten large bolts and fittings.
so that your fingers circle it in the direction that the torque tends to induce rotation. Your thumb will point in the direction of the torque vector [Fig. 7.32(b)]. Thus, the torque vector in Fig. 7.32(a) is perpendicular to and points out of the page. Figure 7.32 Torque is a vector quantity that acts along the axis of rotation according to the right-hand rule.
hammers and Screwdrivers torque is an essential component of most hand tools. Drive a nail into a piece of wood by holding a hammer near its head. Count the number of hits it takes to drive the nail into the wood. then, drive a like nail into the wood by holding the hammer near the end of the handle. Count the number of hits it takes. Using physics terminology, explain which handle grip is better for driving nails into wood. Use two screwdrivers with different diameter handles to screw two similar screws into a board. Which screwdriver is able to apply more torque to a screw? If one screwdriver applies more torque to a screw than the other, why would anyone want to use a screwdriver that cannot exert the maximum torque?
Assume that each force is applied perpendicular to the torque arm. 1. 4.
Given: F = 16.0 lb 2. Given: F = 100 N 3. Given: t = 60.0 N m r = 6.00 ft r = 0.420 m F = 30.0 N t=? t=? r=? 5. Given: t = 65.4 N m 6. Given: F = 630 N Given: t = 35.7 lb ft r = 35.0 cm r = 0.0240 ft r = 74.0 cm F=? F=? t=? If the torque on a shaft of radius 2.37 cm is 38.0 N m (Fig. 7.33), what force is applied to the shaft? If a force of 56.2 lb is applied to a torque wrench 1.50 ft long (Fig. 7.34), what torque is indicated by the wrench?
A motorcycle head bolt is torqued to 25.0 N m. What length shaft do we need on a wrench to exert a maximum force of 70.0 N? A force of 112 N is applied to a shaft of radius 3.50 cm. What is the torque on the shaft? A torque of 175 lb ft is needed to free a large rusted-on nut. The length of the wrench is 1.10 ft. What force must be applied to free it? A torque wrench reads 14.5 N m. If its length is 25.0 cm, what force is being applied to the handle? The torque on a shaft of radius 3.00 cm is 12.0 N m. What force is being applied to the shaft? An engine bolt is torqued to 30.0 N m. If the length of the wrench is 29.0 cm, what force is applied to the wrench? A mower bolt is torqued to 65.0 N m. If the length of the wrench is 30.0 cm, what force is applied to the wrench? An automobile bolt is torqued to 27.0 N m. If the length of the wrench is 30.0 cm, what force is applied to the wrench? A torque wrench reads 25 lb ft. (a) If its length is 1.0 ft, what force is being applied to the wrench? (b) What is the force if the length is doubled? explain the results. If 13 N m of torque is applied to a bolt with an applied force of 28 N, what is the length of the wrench? If the torque required to loosen a nut on the wheel of a pickup truck is 40.0 N m, what minimum force must be applied to the end of a wrench 30.0 cm long to loosen the nut? How is the required force to loosen the nut in Problem 19 affected if the length of the wrench is doubled? A truck mechanic must loosen a rusted lug nut. If the torque required to loosen the nut is 60.0 N m, what force must be applied to a 35.0-cm wrench? An agricultural mechanic tries to loosen a nut on a tractor wheel with a wrench that is 32.5 cm long. If the torque required to loosen the nut is 55.0 N m, what force must she apply to the wrench?
FirST CondiTion oF equiliBrium the sum of all parallel forces on a body in equilibrium must be zero.
If the vector sum is not zero (forces up unequal to forces down), we have an unbalanced force tending to cause motion. Now consider this situation: one end of the scaffold remains firmly in place, supporting the man, and the other is removed. What happens to the painter? The plank, supported only on one end, falls (Fig. 7.37), and the painter has a mess to clean up!
Figure 7.37 The position of the supporting force is important!
A sign of weight 1500 lb is supported by two cables (Fig. 7.38). If one cable has a tension of 600 lb, what is the tension in the other cable?
Sketch: Draw the force diagram. T2 =?
Not only must the forces balance each other (vector sum = 0), but they must also be positioned so that there is no rotation in the system. To avoid rotation, we can have no unbalanced torques. Sometimes there will be a natural point of rotation, as in our painter problem. We can, however, choose any point as our center of rotation as we consider the torques present. We will soon see that one of any number of points could be selected. What is necessary, though, is that there be no rotation (no unbalanced torques). Again, using engineering notation, we write Σtany point = 0 where Σtany point is the sum of the torques about any chosen point.
Data: Fw = 150 lb plank = 6.00 ft Fw is 2.00 ft from one end Basic Equations: 1.
to solve parallel force problems: 1. 2. 3. 4. 5. 6. 7. 8.
Sketch the problem; position the parallel forces vertically when possible. Write an equation setting the sums of the opposite forces equal to each other. Choose a point of rotation. Eliminate a variable if possible (by making its torque arm zero). Write the sum of all clockwise torques. Write the sum of all counterclockwise torques. Set Σtclockwise = Σtcounterclockwise. Solve the equation Σtclockwise = Σtcounterclockwise for the unknown quantity. Substitute the value found in step 7 into the equation in step 2 to find the other unknown quantity.
Fb = 40.0 lb F2 =?
Choose a point of rotation. Choose either end to eliminate one of the variables F1 or F2. let us choose the right end and label it A.
Σtcounterclockwise = (40.0 lb)(3.00 ft) + (175 lb)(5.00 ft) Note that there are two counterclockwise torques.
The physics of Window Washing Window washers on tall buildings typically stand on a platform that is suspended from the top of the building by ropes or cables. Designers of these devices must make sure that the cables can safely suspend the workers while they are all working on one side of the platform. to simulate this activity, take the two strings, the scales, and the mass from the previous try this Activity and use a metre stick or a wooden dowel as the platform. Set up the activity as indicated in Fig. 7.42, making sure both suspending strings are the same length and that the metre stick is horizontal. Place the weight directly in the middle of the metre stick and note the readings in the scales. then gradually shift the mass over to one side and observe what happens to the readings in the spring scales. What happens to the sum of the tension forces as the mass is shifted to one of the sides?
Figure 7.42 After suspending the mass from the middle of the platform, shift the mass toward one side and observe what happens to the tension in the strings.
Find the force F that will produce equilibrium for each force diagram. Use the same procedure as in example 1. 1.
450 N F=?
100 lb 250 lb F=?
2.50 m from the larger person. What weight must be supported by each of the ropes secured at the ends of the scaffold? An auto differential with a mass of 76.0 kg is 1.00 m from the end of a 2.22-m workbench. What mass must each end of the workbench support? A chop saw with a mass of 12.60 kg is 0.75 m from one end of a worktable that is 2.00 m long. What mass must each end of the table support?
7.5 Center of gravity In Section 7.4 we disregarded the weight of the plank in the painter example. In practice, the weight of the plank or bridge is extremely important. The weight of a bridge being designed must be known in order to use materials of sufficient strength to support the bridge and the traffic and not collapse