Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 6 - Work and Energy - Problems - Page 166: 11



Work Step by Step

We have $\sum W=\sum F\times s$, so $\sum W=0$ when $\sum F=0$, or the forces cancel each other out. There are only 2 horizontal forces acting in the opposite direction and influencing the horizontal movement of the crate: $P\cos30^o$ and $f_k$. Therefore $$\sum F=P\cos30^o-f_k=0.866P-f_k=0 (1)$$ We now take a look at $f_k$ $$f_k=\mu_kF_N=0.2F_N$$ There are 3 vertical forces in play here: $F_N$ pointing upward while $mg$ and $P\sin30$ point downward. As there is no vertical acceleration, the upward forces balance the downward ones: $$F_N=mg+P\sin30=100\times9.8+P\sin30=980+0.5P$$ Therefore, $$f_k=0.2(980+0.5P)=196+0.1P$$ Plug this into (1): $$0.866P-196-0.1P=0$$ $$0.766P=196$$ $$P=256N$$
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