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

(a) $F = 57.1~N$ (b) $F_f = 146~N$ (up the ramp)
(a) We can find the angle $\theta$ of the ramp. $tan(\theta) = \frac{2.50~m}{4.75~m}$ $\theta = arctan(\frac{2.50~m}{4.75~m})$ $\theta = 27.76^{\circ}$ Since the speed is constant, the force $F$ we need to apply plus the force of kinetic friction (which is directed up the ramp) is equal in magnitude to the component of the weight directed down the ramp. $F+ mg~cos(\theta)~\mu_k = mg~sin(\theta)$ $F =mg~sin(\theta) - mg~cos(\theta)~\mu_k$ $F = (80.0~kg)(9.80~m/s^2)~sin(27.76^{\circ})- (80.0~kg)(9.80~m/s^2)~cos(27.76^{\circ})(0.444)$ $F = 57.1~N$ (b) The force of static friction on the upper box is directed up the ramp. This friction force is equal in magnitude to the component of the upper box's weight which is directed down the ramp. $F_f = mg~sin(\theta)$ $F_f = (32.0~kg)(9.80~m/s^2)~sin(27.76^{\circ})$ $F_f = 146~N$ (up the ramp)