## Physics (10th Edition)

Published by Wiley

# Chapter 9 - Rotational Dynamics - Problems - Page 244: 18

#### Answer

1st design: $F=189.2N$ 2nd design: $F=27.7N$

#### Work Step by Step

1) Design 1: Let's examine the torque created by the forces in this exercise. The axis of rotation is located at the point where the tire contacts the ground. - The weight of the wheels, axle and handles $W_1=60N$ produces a clockwise torque, so $\tau_1\lt0$. We have $l_1=0.6m$ - The weight of the chamber and contents $W_2=525N$ produces a clockwise torque, so $\tau_2\lt0$. We have $l_2=0.4m$ - The man's force $F$ produces a counterclockwise torque, so $\tau_F\gt0$. We have $l_F=0.4+0.2+0.7=1.3m$ The system is in equilibrium, so the net torque is zero $$1.3F-0.6W_1-0.4W_2=0$$ $$F=\frac{0.6W_1+0.4W_2}{1.3}=189.2N$$ 2) Design 2: - The weight of the wheels, axle and handles $W_1=60N$ produces a clockwise torque, so $\tau_1\lt0$. We have $l_1=0.6m$ - The weight of the chamber and contents goes through the axis of rotation, so it produces no torque. - The man's force $F$ produces a counterclockwise torque, so $\tau_F\gt0$. We have $l_F=1.3m$ The system is in equilibrium, so the net torque is zero $$1.3F=0.6W_1=36$$ $$F=27.7N$$

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