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 15.3 - Conservation of Linear Momentum for a System of Particles - Problems - Page 263: 42


$v_{w}=3.48ft/s$ $N=504lb$ $t=0.216s$

Work Step by Step

According to the law of conservation of linear momentum $\Sigma m_1v_1=\Sigma m_2v_2$ $\implies W_bv_{b_1}(\frac{12}{13})+W_w v_{w_1}=W_b v_{b_2}(\frac{4}{5})+W_w v_{w_2}$ We plug in the known values to obtain: $(0.03)(1300)(\frac{12}{13})+0=(0.03)(50)(\frac{4}{5})+10v_w$ Solving the above equation, we obtain: $v_w=3.48ft/s$ Now we determine the normal force on the block as $mv_1+\Sigma \int Fdt=mv_2$ $(\frac{W_b}{g})v_{b_1}(\frac{5}{13})-W_w t+Nt=(\frac{W_b}{g})v_{b_2}(\frac{5}{3})$ $\implies (\frac{0.03}{32.2})(1300)(\frac{5}{13})-(10)(1\times 10^{-3})+N(1\times 10^{-3})=(\frac{0.03}{32.2})(50)(\frac{5}{3})$ $\implies N=504lb$ To find the required time, we use the principle of impulse and momentum in the x-direction $\implies mv_1+\Sigma \int Fdt=mv_2$ $\implies (\frac{W_w}{g})v_{w_1}-\mu_k Wt=(\frac{W_w}{g})v_{w_2}$ We plug in the known values to obtain: $(\frac{10}{32.2})(3.48)-(0.5)(10)t=0$ This simplifies to: $t=0.216s$
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