## Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson

# Chapter 13 - Kinetics of a Particle: Force and Acceleration - Section 13.4 - Equations of Motion: Rectangular Coordinates - Problems - Page 129: 4

#### Answer

$P=224~N$

#### Work Step by Step

We can determine the magnitude of the required force $P$ as follows: As $v=v_{\circ} +at$ $\implies 4=0+at$eq(1) and $s=s_{\circ}+v_{\circ}t+\frac{1}{2}at^2$ $\implies 5=0+0+\frac{1}{2}at^2$ $at^2=10$.eq(2) Solving eq(1) and eq(2), we obtain: $t=2.5s$ and $a=1.6m/s^2$ Now we apply Newton's second law $\Sigma F_y=0$ $\implies N-W+Psin\theta=0$ $\implies N=W-Psin\theta$ We plug in the known values to obtain: $N=(50)(9.81)-Psin30=490.5-0.5P$ and $\Sigma F_x=ma_x$ $\implies -\mu_k N+Pcos\theta=ma$ We plug in the known values to obtain: $-0.3(490.5-0.5)+Pcos30=(50)(1.6)$ This simplifies to: $P=223.567N\approx224~N$

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