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54. The beam has a mass
M
= 40.0 kg and a length
L
= 0.800 m. The mass of the package
of tamale is
m
= 10.0 kg.
(a) Since the system is in static equilibrium, the normal force on the beam from roller
A
is
equal to half of the weight of the beam:
F
A
=
Mg
/2 = (40.0 kg)(9.80 m/s
2
)/2 = 196 N.
(b) The normal force on the beam from roller
B
is equal to half of the weight of the beam
plus the weight of the tamale:
F
B
=
Mg
/2 +
mg
= (40.0 kg)(9.80 m/s
2
)/2 + (10.0 kg)(9.80 m/s
2
)= 294 N.
(c) When the righthand end of the beam is centered over roller
B
, the normal force on the
beam from roller
A
is equal to the weight of the beam plus half of the weight of the
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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, Static Equilibrium

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