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25. The beam is in equilibrium: the sum of the forces and the sum of the torques acting
on it each vanish. As shown in the figure, the beam makes an angle of 60º with the
vertical and the wire makes an angle of 30º with the vertical.
(a) We calculate the torques around the hinge. Their sum is
TL
sin 30º –
W
(
L
/2) sin 60º = 0.
Here
W
is the force of gravity acting at the center of the beam, and
T
is the tension force
of the wire. We solve for the tension:
()
222N sin 60
sin60
=
=
=192 N.
2sin30
W
T
°
°
°°
(b) Let
F
h
be the horizontal component of the force exerted by the hinge and take it to be
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 Spring '08
 Any
 Physics, Force

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