Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 8 - Friction - Section 8.2 - Problems Involving Dry Friction - Problems - Page 419: 5



Work Step by Step

We know that $\Sigma M_A=0$ $\implies N_B(2.5)-G(1.0)-F(cos 30^{\circ}(0.3)+sin30^{\circ}(0.75))=0$ $\implies N_B=0.04608F+7.848$ Now the friction forces under B are $F_B=0.3N_B$ $\implies F_B=(0.3)(0.04608F+7.848)$ $\implies F_B=2.3544+0.013824F$ As the sum of forces in the x-direction is zero, then: $\implies \Sigma F_x=0$ $\implies F.cos 30^{\circ}-F_B=0$ $\implies F.cos 30^{\circ}-(2.3544+0.013824F)=0$ This simplifies to: $F=2.76KN$
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