Answer
$155.78N$
Work Step by Step
We know that
$\Sigma \tau=0$
$\implies r_1F_Q sin29^{\circ}-r_2mgcos39^{\circ}=0$
We plug in the known values to obtain:
$(0.12m)F_Q sin 29^{\circ}=(0.35m)mg cos 39^{\circ}$
This simplifies to:
$F_Q=155.78N$
Physics Technology Update (4th Edition)
by Walker, James S.
Published by
Pearson
ISBN 10:
0-32190-308-0
ISBN 13:
978-0-32190-308-2
Chapter 11 - Rotational Dynamics and Static Equilibrium - Problems and Conceptual Exercises - Page 373: 104
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