37. (a) We assume the center of mass is closer to the right end of the rod, so the distance from the leftend to the center of mass is`=0.60 m. Four forces act on the rod: the upward force of the leftropeTL, the upward force of the right ropeTR, the downward force of gravitymg,andtheupwardbuoyant forceFb. The force of gravity (e±ectively) acts at the center of mass, and the buoyantforce acts at the geometric center of the rod (which has lengthL.80 m). Computing torquesabout the left end of the rod, we ²ndTRL+FbµL2¶−mg`=0 =⇒TR=mg`−FbL/2L.Now, the buoyant force is equal to the weight of the displaced water (where the volume of displace-
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