College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 4 - Force and Motion - Learning Path Questions and Exercises - Exercises - Page 139: 72


a). $(2)$ pulling at the same angle, but upward. b). $849N$, $289N$

Work Step by Step

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 $F_{2}=289N$
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