Answer
$200lb$
Work Step by Step
We know that
$F_B=\mu_s N_B$
$\implies F_B=0.35N_B$
We also know that
$\Sigma M_B=0$
$\implies N_A(3)-P(3-1)=0$
$\implies N_A(3)-P(2)=0$
This simplifies to:
$N_A=200lb$
Engineering Mechanics: Statics & Dynamics (14th Edition)
by Hibbeler, Russell C.
Published by
Pearson
ISBN 10:
0133915425
ISBN 13:
978-0-13391-542-6
Chapter 8 - Friction - Section 8.2 - Problems Involving Dry Friction - Problems - Page 421: 19
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