## University Physics with Modern Physics (14th Edition)

(a) Tension left of pulley $= 32.7 \, \mathrm{N}$ Tension below pulley $= 35.4 \, \mathrm{N}$ (b) $a = 2.72 \, \mathrm{m/s^2}$ (c) $\Sigma F_y = 55.0 \, \mathrm{N}$ (straight up) $\Sigma F_x = 32.7 \, \mathrm{N}$ (to the right)
I'll call the tension in the wire between the pulley and the box $T_1$, and the tension between pulley and weight $T_2$. (a) and (b) For the pulley, with positive rotation clockwise: $$\Sigma \tau = T_2r - T_1r = r(T_2 - T_1) = I\alpha = \frac{Ia}{r} \Rightarrow$$ $$Ia = r^2(T_2 - T_1) \Rightarrow \frac{m_pa}{2} = T_2 - T_1$$ For the block, with positive $x$-direction to the right: $$\Sigma F_x = m_ba = T_1$$ For the weight, positive $y$-direction down: $$\Sigma F_y = m_wa - m_wg-T_2 \Rightarrow m_wg - m_wa = T_2$$ Subtracting the second equation (block) from the third (weight): $$m_wg - m_wa - m_ba = T_2 - T_1$$ Inserting into the first (pulley) equation and solving for $a$: $$\frac{m_pa}{2} = m_wg - m_wa - m_ba \Rightarrow$$ $$a = \frac{m_wg}{m_p/2 + m_w + m_b} = \frac{(5.00 \, \mathrm{kg})(9.80 \, \mathrm{N/kg})}{(2.00 \, \mathrm{kg})/2 + 5.00 \, \mathrm{kg} + 12.0 \, \mathrm{kg}} = 2.72 \, \mathrm{m/s^2} \, (2.7222...)$$ Calculating $T_1$ and $T_2$: $$T_1 = (12.0 \, \mathrm{kg})(2.722 \, \mathrm{N/kg}) = 32.7 \, \mathrm{N}$$ $$T_2 = (5.00 \, \mathrm{kg})(9.80 \, \mathrm{N/kg} - 2.722 \, \mathrm{N/kg}) = 35.4 \, \mathrm{N}$$ (c) For the pulley, with positive $x$-direction to the right and positive $y$-direction up: $$\Sigma F_x = T_1 = 32.7 \, \mathrm{N}$$ $$\Sigma F_y = T_2 + m_pg = 35.4 \, \mathrm{N} + (2.00 \, \mathrm{kg})(9.80 \, \mathrm{N/kg}) = 55.0 \, \mathrm{N}$$