Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 7 - Newton's Third Law - Exercises and Problems - Page 187: 30

Answer

$200\;\rm kg$

Work Step by Step

First of all, we need to draw a force diagram of the two objects, the boy and the crate, as we see below. We know that the boy barely keeps the crate moving so we can assume that the system is moving at a constant speed. Thus, the net force exerted on the boy and the net force exerted on the crate is zero. $$\sum F_{x,boy}=0,\;\;\;\sum F_{x,crate}=0$$ And since the boy's feet occasionally slip while he pushes the crate, he pushes the ground backward by his feet with the maximum force he could, and the ground pushed his feet forward by the same amount of force, which is a static friction force between his shoes and the floor. Noting that the friction force between the crate and the ground is a kinetic friction force. Thus, from all the above; $$\sum F_{x,crate}=F-f_k=m_{crate}a_x=m_{crate}(0)=0$$ Thus, $$F=f_k$$ and we know that the kinetic friction force exerted on the crate is given by $f_k=\mu_kF_{n}$; $$F=\mu_kF_{n,crate}\tag 1$$ $$\sum F_{y,crate}=F_{n,crate}-m_{crate}g=m_{crate}(0)=0$$ Hence; $$F_{n,crate}=m_{crate}g$$ Plugging into (1); $$F=\mu_km_{crate}g $$ Solving for $m_{crate}$; $$m_{crate}=\dfrac{F}{\mu_k g}\tag 2$$ Now we need to find the pushing force $F$ exerted by the boy on the crate which is the same force exerted by the crate on the boy, as we know from Newton's third law. $$\sum F_{x,boy}=f_s-F=m_{boy}a_x=m_{boy}(0)$$ Thus, $$f_s=F$$ and we know that the static friction force is given by $\mu_s F_n$ $$\mu_s F_{n,boy}=F\tag 3$$ $$\sum F_{y,boy}=F_{n,boy}-m_{boy}g=m_{boy}a_y=0$$ $$F_{n,boy}=m_{boy}g$$ Plugging into (3); $$\mu_s m_{boy}g=F $$ Plugging into (2); $$m_{crate}=\dfrac{\mu_s m_{boy}\color{red}{\bf\not}g}{\mu_k \color{red}{\bf\not}g}$$ Plugging the given; $$m_{crate}=\dfrac{0.8\times 50}{0.2}=\color{red}{\bf 200}\;\rm kg$$
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