## College Physics (4th Edition)

(a) $\mu_s = 0.41$ (b) $F = 40~N$
(a) The applied force is equal in magnitude to the force of static friction. We can find the coefficient of static friction: $mg~\mu_s = 12.0~N$ $\mu_s = \frac{12.0~N}{mg}$ $\mu_s = \frac{12.0~N}{(3.0~kg)(9.80~m/s^2)}$ $\mu_s = 0.41$ (b) We can assume that the required force $F$ is equal in magnitude to the force of static friction. We can find $F$: $F = F_N~\mu_s$ $F = (3.0~kg+7.0~kg)~g~\mu_s$ $F = (3.0~kg+7.0~kg)(9.80~m/s^2)(0.41)$ $F = 40~N$