Engineering Mechanics: Statics & Dynamics (14th Edition)

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

Chapter 13 - Kinetics of a Particle: Force and Acceleration - Section 13.4 - Equations of Motion: Rectangular Coordinates - Problems - Page 134: 34



Work Step by Step

The velocity of the cylinder can be calculated as $F_s-mg=0$ $\implies 120x_{\circ}-4\times 9.81=0$ $\implies x_{\circ}=0.327m$ We know that $\Sigma F_y=ma_y$ $F_s-mg-60=ma_y$ $\implies 120(0.327+s)-4\times 9.81-60=-4a$ $\implies a=15-30s$ As $a=\frac{dv}{dt}=v\frac{dv}{ds}$ $\implies ads=vdv$ $\implies \int_0 ^s (15-30)ds=\int_0^v vdv$ $\implies 15s-15s^2=\frac{v^2}{2}$ $\implies 15\times 0.2-15(0.2)^2=\frac{v^2}{2}$ This simplifies to: $v=2.19m/s$
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