Thus,agm mmmcom22m/sggg=−+=−+=×−2121222229 8520480480520 g16 10bgbgch...The acceleration is downward. 105. (a) We place the origin of a coordinate system at the center of the pulley, with the xaxis horizontal and to the right and with the yaxis downward. The center of mass is halfway between the containers, at x= 0 and y= A, where Ais the vertical distance from the pulley center to either of the containers. Since the diameter of the pulley is 50 mm, the center of mass is at a horizontal distance of 25 mm from each container. (b) Suppose 20 g is transferred from the container on the left to the container on the right. The container on the left has mass m1= 480 g and is at x1= –25 mm. The container on the right has mass m2= 520 g and is at x2= +25 mm. The xcoordinate of the center of mass is then xmxcomggg 520 gmm.=++=−++=112 2480255202548010bgbg.Theycoordinate is still A. The center of mass is 26 mm from the lighter container, along the line that joins the bodies.
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