Engineering Mechanics: Statics & Dynamics (14th Edition)

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

Chapter 15 - Kinetics of a Particle: Impulse and Momentum - Section 15.2 - Principle of Linear Impulse and Momentum for a System of Particles - Problems - Page 247: 6


$F=19.444KN; T=12.5KN$

Work Step by Step

The required forces $F$ and $T$ can be determined as follows: $mv_1+\Sigma \int_{t_1}^{t_2} F_x dt=mv_2 $ We plug in the known values to obtain: $(50000+3\times 30000)(0)+F\Delta t=(50000+3\times 30000)(\frac{40\times 1000}{3600})$ $\implies (80-0)=1.556\times 10^6$ $\implies F=19.444KN$ Similarly, $mv_1+\Sigma \int_{t_1}^{t_2} F_x dt=mv_2$ We plug in the known values to obtain: $(3\times 30000)(0)+T\Delta T=(3\times 30000)(\frac{40\times 1000}{3600})$ $\implies T=12.5KN$
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