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and the force exerted there is the tension
T
. Computing torques about the hinge, we find
()
(
)
22
11
12
2
50.0 kg
9.8 m/s
1.00 m
(50.0 kg)(9.8m/s )(3.00 m)
=
sin
3.00 m 0.800
408 N.
mgx
mgx
T
x
θ
+
+
=
=
(b) Equilibrium of horizontal forces requires the horizontal hinge force be
F
x
=
T
cos
= 245 N.
(c) The direction of the horizontal force is rightward.
(d) Equilibrium of vertical forces requires the vertical hinge force be
F
y
=
mg
–
T
sin
= 163 N.
(e) The direction of the vertical force is upward.
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Force

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