1. The problem statement, all variables and given/known data In the ﬂgure below, the incline has mass M and is fastened to the stationary horizontal tabletop. The block of mass m is placed near the bottom of the incline and is released with a quick push that sets it sliding upward. It stops near the top of the incline, as shown in the ﬂgure, and then slides down again, always without friction. Find the force that the tabletop exerts on the incline throughout this motion. http://i359.photobucket.com/albums/oo31/tanzl/PHYS1A_ProbSet2_08-09.jpg 2. Relevant equations 3. The attempt at a solution My answer is (M+m)g since the resultant downward force should be the sum of the weight of incline plane and the weight of the mass. But, it looks too simple to be correct. Please clear my doubts. 1)Although there is a normal force of magnitude mg*cos(theta) that will eventually balance the weight of the mass m since the mass is not moving downward o upward. 2)The initial push on the mass m is not necessary to be considered in the calculation because the force supplied is not constant and so the only force acting on mass m which parallel to the incline is mg*sin(theta). 3)Throughout the motion of mass, only two force were involved. The weight of mass m and the normal force of incline exerted on the mass.