29. The bar is in equilibrium, so the forces and the torques acting on it each sum to zero.
Let
T
l
be the tension force of the left–hand cord,
T
r
be the tension force of the right–hand
cord, and
m
be the mass of the bar. The equations for equilibrium are:
vertical force components
horizontal force components
torques
cos
cos
0
sin
sin
0
cos
0.
lr
r
TTm
g
TT
mgx T L
θ
φ
θφ
+−
=
−+
=
−=
The origin was chosen to be at the left end of the bar for purposes of calculating the
torque. The unknown quantities are
T
l
,
T
r
, and
x
. We want to eliminate
T
l
and
T
r
, then
solve for
x
. The second equation yields
T
l
=
T
r
sin
/sin
and when this is substituted
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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, Mass

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