## University Physics with Modern Physics (14th Edition)

(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.