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 131: 14


$T=11.25kN$, $F=33.75kN$

Work Step by Step

We can determine the required force as follows: $v^2=v_{\circ}^2+2a(s-s_{\circ})$ We plug in the known values to obtain: $0=(15)^2+2a(10-0)$ This simplifies to: $a=-11.25m/s^2$ According to Newton's second law $\Sigma F_x=ma_x$ $\implies -T=1000\times (-11.25)$ $\implies T=11250N=11.25kN$ Similarly, $\Sigma F_x=ma_x$ $T-F=ma$ $\implies 11250-F=2000\times (-11.25)$ This simplifies to: $F=33750N=33.75kN$
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