## 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: 2

#### Answer

$T=5.98kip$

#### Work Step by Step

We can determine the required force as follows: First, we apply Newton's second law to car A $\Sigma F_y=0$ $\implies N_A-20000cos\theta=0$ $\implies N_A=19923.894lb$ and $F_x=ma_x$ $\implies \mu_k N_A-20000sin\theta-T=m_A a$ We plug in the known values to obtain: $(0.5)(19923.894)-20000sin 5^{\circ}-T=\frac{20000}{32.2}a$ $\implies T+621.118a=8218.832$..eq(1) Now we apply Newton's second law to car A and B together $\Sigma F_x=ma_x$ $\implies \mu_k N_A-(W_A+W_B)sin\theta=(W_A+W_B)a$ We plug in the known values to obtain: $(0.5)(19923.894)-(20000+30000)sin 5^{\circ}=\frac{20000+30000}{32.2}a$ This simplifies to: $a=3.609ft/s^2$ Now from eq(1), we obtain: $T=8218.832-(621.118)(3.609)$ $\implies T=5977.168lb=5.98kip$

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