Physics (10th Edition)

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

Chapter 9 - Rotational Dynamics - Problems - Page 247: 47


The torque produced by the engine is $1700N.m$

Work Step by Step

We call the tension in the cable between the drum and pulley $T_1$ and the tension in the other cable part $T_2$ The acceleration of the crate, when translated to the pulley and drum, becomes tangential acceleration. Therefore, the angular acceleration of the drum and the pulley is $\alpha_d=\frac{a}{r_d}$ and $\alpha_p=\frac{a}{r_p}$ 1) At the pulley, the force from the engine $E$ and $T_1$ produce torques in opposite direction: $$\sum\tau_d=\tau_E-\tau_{T_{1}}=Er_{d}-T_1r_{d}$$ From Newton's 2nd law, we also have $\sum\tau_d=I\alpha_d=m_dr^2_{d}\Big(\frac{a}{r_d}\Big)=m_dar_d$ So we have $$E-T_1=m_da$$ The drum's mass $m_d=150kg$ and $a=1.2m/s^2$. Therefore, $E-T_1=180N (1)$ 2) At the pulley, $T_1$ produces positive torque while $T_2$ produces negative torque: $$\sum\tau_p=\tau_{T_1}-\tau_{T_{2}}=T_1r_{p}-T_2r_{p}$$ From Newton's 2nd law, we also have $\sum\tau_p=I\alpha_p=\frac{1}{2}m_pr^2_{p}\Big(\frac{a}{r_p}\Big)=\frac{1}{2}m_par_p$ So we have $$T_2-T_1=\frac{1}{2}m_pa$$ The pulley's mass $m_p=130kg$ and $a=1.2m/s^2$. Therefore, $T_1-T_2=78N (1)$ 3) $T_2$ can be calculated when we examine the crate. $T_2$ opposes the weight of the crate, so $$T_2-W=m_{crate}a$$ $$T_2=m_{crate}a+W=180kg(9.8m/s^2+1.2m/s^2)=1980N$$ (1): $T_1=2058N$ (2): $E=2238N$ The torque produced by $E$ is $Er_{d}=2238N\times0.76m=1700N.m$
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