Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 5 - Force and Motion-I - Problems - Page 122: 67

Answer

The tension in the cord at the right is $~~81.7~N$

Work Step by Step

Since $m_C \gt m_A$, then block C will accelerate downward, block B will accelerate to the right, and block A will accelerate upward. We can find the magnitude of the acceleration: $m_C~g - m_A~g = (m_A+m_B+m_C)~a$ $a = \frac{(m_C - m_A)~g}{m_A+m_B+m_C}$ $a = \frac{(10.0~kg - 6.00~kg)~(9.8~m/s^2)}{6.00~kg+8.00~kg+10.0~kg}$ $a = 1.63~m/s^2$ We can consider the forces on block C to find the tension $F_T$ in the cord at the right: $m_C~g-F_T = m_C~a$ $F_T = m_C~(g-a)$ $F_T = (10.0~kg)~(9.8~m/s^2-1.63~m/s^2)$ $F_T = 81.7~N$ The tension in the cord at the right is $~~81.7~N$
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