Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 12 - Exercises and Problems - Page 216: 32



Work Step by Step

As we know that the car is in equilibrium so $\sum T_p=0$ $\implies F\times 1.8sin34^{\circ}-Mg\times (2.4-1.8)cos34^{\circ}=0$ This simplifies to: $F=\frac{Mg\times 0.6cos34^{\circ}}{1.8sin34^{\circ}}$ We plug in the known values to obtain: $F=\frac{1250\times 9.8\times cot34^{\circ}}{3}$ $F=6kN$
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