1) The forces exerted by wires A (FA) and B (FB) on a log are related by the stretching of the wires based on their original lengths (LA and LB), cross-sectional areas, and Young's modulus. 2) Equating the total force on the log (FA + FB - mg) to 0 allows solving for FA and FB in terms of the known parameters. 3) FA is calculated to be 866 N and FB is then calculated from the total force equation to be 143 N.

1. 39. (a) Let FA and FB be the forces exerted by the wires on the log and let m be the mass of the log.
Since the log is in equilibrium FA + FB − mg = 0. Information given about the stretching of the
wires allows us to find a relationship between FA and FB . If wire A originally had a length LA and
stretches by ∆LA , then ∆LA = FA LA /AE, where A is the cross-sectional area of the wire and E
2
is Young’s modulus for steel (200 × 109 N/m ). Similarly, ∆LB = FB LB /AE. If is the amount
by which B was originally longer than A then, since they have the same length after the log is
attached, ∆LA = ∆LB + . This means
FA LA FB LB
= + .
AE AE
We solve for FB :
FA LA AE
FB = − .
LB LB
We substitute into FA + FB − mg = 0 and obtain
mgLB + AE
FA = .
LA + LB
The cross-sectional area of a wire is A = πr2 = π(1.20 × 10−3 m)2 = 4.52 × 10−6 m2 . Both LA and
LB may be taken to be 2.50 m without loss of significance. Thus
(103 kg)(9.8 m/s2 )(2.50 m) + (4.52 × 10−6 m2 )(200 × 109 N/m )(2.0 × 10−3 m)
2
FA =
2.50 m + 2.50 m
= 866 N .
(b) From the condition FA + FB − mg = 0, we obtain
2
FB = mg − FA = (103 kg)(9.8 m/s ) − 866 N = 143 N .
(c) The net torque must also vanish. We place the origin on the surface of the log at a point directly
above the center of mass. The force of gravity does not exert a torque about this point. Then, the
torque equation becomes FA dA − FB dB = 0, which leads to
dA FB 143 N
= = = 0.165 .
dB FA 866 N