ch12-p054 - 54. The beam has a mass M = 40.0 kg and a...

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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 right-hand 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.

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