## College Physics (7th Edition)

a). $(2)$ pulling at the same angle, but upward. b). $849N$, $289N$
a). Pushing the crate downwards implies that the reaction force in increased so, the friction force is increased. Similarly, pulling it at an angle upwards means that the reaction force is reduced hence reducing the friction force. At a reduced friction force, a smaller force would need to be applied to set the crate into motion. So option (2) is the correct answer. b). $F_{y}=Fsin30^{\circ}=0.5F$ So, $N=mg+0.5F$ or $N=mg-0.5F$ depending on pushing or pulling. Thus, $F_{f}=coefficient\, of\, friction \times (mg+0.5F)$ or, $F_{f}=coefficient\, of\, friction \times (mg-0.5F)$ Now, $F_{x}=Fcos30^{\circ}=F_{f}$ Solving for both cases, $F_{1}=849N$, and $F_{2}=289N$