Answer
(a) The maximum possible traction force is 577 N.
(b) The tension in each cable is 318 N.
Work Step by Step
(a) The maximum possible traction force is equal in magnitude to the maximum possible force of static friction.
$F_f = mg~\mu_s = (78.5~kg)(9.80~m/s^2)(0.75)$
$F_f = 577~N$
The maximum possible traction force is 577 N.
(b) Let $T$ be the tension in each cable.
$2T~sin(65^{\circ}) = 577~N$
$T = \frac{577~N}{2~sin(65^{\circ})}$
$T = 318~N$
The tension in each cable is 318 N.
