## 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 130: 12

#### Answer

$F=\frac{m(g+a_B)\sqrt{4y^2+d^2}}{4y}$

#### Work Step by Step

We can determine the magnitude of the force as follows: $\Sigma F_y=ma_y$ $\implies 2Fcos\theta-mg=ma_B$ But from the given figure, $cos\theta=\frac{y}{\sqrt{y^2+(d/2)^2}}$ Now the above equation becomes $2F(\frac{y}{\sqrt{y^2+(d/2)^2}})-mg=ma_B$ $\implies F=\frac{mg+ma_B}{2(\frac{y}{\sqrt{y^2+(d/2)^2}})}$ $\implies F=\frac{m(g+a_B)}{4(\frac{y}{\sqrt{4y^2+d^2}})}$ This simplifies to: $F=\frac{m(g+a_B)\sqrt{4y^2+d^2}}{4y}$

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