Answer
$\sum \tau = -28.0~N~m$
Work Step by Step
$\sum \tau = \tau_1+\tau_2$
$\sum \tau = -F_1~r_1~sin(\theta_1) + F_2~r_2~sin(\theta_2)$
$\sum \tau = -(8.00~N)(5.00~m)~sin(90.0^{\circ})+(12.0~N)(2.00~m)~sin(30.0^{\circ})$
$\sum \tau = -28.0~N~m$
University Physics with Modern Physics (14th Edition)
by Young, Hugh D.; Freedman, Roger A.
Published by
Pearson
ISBN 10:
0321973615
ISBN 13:
978-0-32197-361-0
Chapter 10 - Dynamics of Rotational Motion - Problems - Exercises - Page 329: 10.2
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