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

Published by Pearson

# Chapter 2 - Force Vectors - Section 2.3 - Vector Addition of Forces - Problems - Page 29: 3

#### Answer

$F_R=393lb$ $\theta=353^{\circ}$

#### Work Step by Step

We are asked to determine the magnitude of the force, $F_R$, and its direction, measured counterclockwise from the positive x axis. We can treat the forces like a triangle and use the law of cosines: $c^2=a^2+b^2−2a∗b∗\cos C$ $c=\sqrt{a^2+b^2−2a∗b∗\cos C}$ $c=\sqrt{(250lb)^2+(375lb)^2−2∗250lb∗375lb∗\cos(75^{\circ})}$ $c=393lb$ $F_R=393lb$ Now we can use the law of sines to find the angle between $F$ and the x-axis. $\sin\alpha/250lb=\sin75^{\circ}/393lb$ $\alpha=\arcsin(\sin75^{\circ}/393lb∗250lb)$ $\alpha=37.89^{\circ}$ This is measured from $F_2$. We need to subtract 45 from 365 and then add our angle to get the measurement from the x-axis $\theta=360^{\circ}-45^{\circ}+37.89^{\circ}=353^{\circ}$

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