Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 2 - Force Vectors - Section 2.3 - Vector Addition of Forces - Problems: 16

Answer

$F_{BC}=434lb$ $\phi=33.5^{\circ}$

Work Step by Step

We are asked to determine the force in member BC and the required angle $\phi$ Given that the force on member AB is 650lb and the force F is 850lb, we can apply the law of cosines to find force in member BC, $F_{BC}$ $c=\sqrt{a^2+b^2-2*a*b*\cos(C)}$ $c=\sqrt{850^2+650^2-2*850*650*\cos(30^{\circ})}$ $c=433.64$ $F_{BC}=434lb$ Using this result, we can apply the law of sines to find the$ \phi$ $\sin(45^{\circ}+\phi) / 850 = \sin 30^{\circ} / 433.64$ Solving for $\phi$, we obtain: $\phi=33.5^{\circ}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.