372 CHAPTER 11 Equilibrium and Elasticity

11.101 ... CP An angler hangs a 4.50-kg fish from a verticalsteel wire 1.50 m long and in cross-sectionalarea. The upper end of the wire is securely fastened to a support.(a) Calculate the amount the wire is stretched by the hanging fish.The angler now applies a force to the fish, pulling it very slowlydownward by 0.500 mm from its equilibrium position. For this

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5.00 * 10-3 cm2downward motion, calculate (b) the work done by gravity; (c) thework done by the force (d) the work done by the force the wireexerts on the fish; and (e) the change in the elastic potential energy(the potential energy associated with the tensile stress in the wire).Compare the answers in parts (d) and (e).

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Chapter Opening Question ?Each stone in the arch is under compression, not tension. This isbecause the forces on the stones tend to push them inward towardthe center of the arch and thus squeeze them together. Compared toa solid supporting wall, a wall with arches is just as strong yetmuch more economical to build.

Test Your Understanding Questions11.1 Answer: (i) Situation (i) satisfies both equilibrium conditionsbecause the seagull has zero acceleration so and notendency to start rotating so Situation (ii) satisfies thefirst condition because the crankshaft as a whole does not acceler-ate through space, but it does not satisfy the second condition; thecrankshaft has an angular acceleration, so is not zero. Situa-tion (iii) satisfies the second condition (there is no tendency torotate) but not the first one; the baseball accelerates in its flight(due to gravity), so is not zero.11.2 Answer: (ii) In equilibrium, the center of gravity must be atthe point of support. Since the rock and meter stick have the samemass and hence the same weight, the center of gravity of the sys-tem is midway between their respective centers. The center ofgravity of the meter stick alone is 0.50 m from the left end (that is,at the middle of the meter stick), so the center of gravity of thecombination of rock and meter stick is 0.25 m from the left end.11.3 Answer: (ii), (i), (iii) This is the same situation described inExample 11.4, with the rod replacing the forearm, the hinge replac-ing the elbow, and the cable replacing the tendon. The only differ-ence is that the cable attachment point is at the end of the rod, so thedistances D and L are identical. From Example 11.4, the tension is

T =Lw

L sinu=

w

sinu

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gTS

gTS � 02.1gF

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Since is less than 1, the tension T is greater than the weight w.The vertical component of the force exerted by the hinge is

In this situation, the hinge exerts no vertical force. You can see thiseasily if you calculate torques around the right end of the horizon-tal rod: The only force that exerts a torque around this point is thevertical component of the hinge force, so this force componentmust be zero.11.4 Answers: (a) (iii), (b) (ii) In (a), the copper rod has 10 timesthe elongation of the steel rod, but it also has 10 times the orig-inal length Hence the tensile strain is the same for bothrods. In (b), the stress is equal to Young’s modulus Y multiplied bythe strain. From Table 11.1, steel has a larger value of Y, so agreater stress is required to produce the same strain.11.5 In (a) and (b), the bumper will have sprung back to its origi-nal shape (although the paint may be scratched). In (c), the bumperwill have a permanent dent or deformation. In (d), the bumper willbe torn or broken.

Bridging Problem

Answers:

(a)

(b)

(c) (d) 4¢l =2mgl0

3AY tanu

F =2mg

3 sinu3 cos 2u + 1

4 sin 2u, f = arctan A12 tanu BT =

2mg

3 sinu

¢l>l0l0.¢l

Ey = -1L - L2w

L= 0

sinu

Answers