Engineering Mechanics: Statics & Dynamics (14th Edition)

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

Chapter 3 - Equilibrium of a Particle - Section 3.3 - Coplanar Force Systems - Problems - Page 98: 4


$N_C=163N$ $N_B=105N$

Work Step by Step

First we will sum the forces horizontally $(\rightarrow +) \sum \overrightarrow{F}_{x} =0 $ $125-N_C\cos40^{\circ}=0$ $N_C=125/\cos40^{\circ}=163.18N=163N$ Now we will sum the forces vertically $(\uparrow +) \sum \overrightarrow{F}_{y} =0 $ $N_B-N_C\sin40^{\circ}=0$ $N_B=N_C\sin40^{\circ}=163.18\sin40^{\circ}=105N$
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.