Question

《 previous 1 12 12 I return to assignment A wedge with mass M rests on a frictionless horizontal table top. A block with mass m is placed on the wedge (the figure (Figure 1) (a)). There is no friction between the block and the wedge. The system is released from rest Part A Calculate the r component of the acceleration of the wedge Express your answer in terms of M, m, a, and constant g Figure 1 of 1 Submit My Answers Give Up In Part B Calculate the x component of the acceleration of the block. Express your answer in terms of M, m, a, and constant g abz- Submit My Answers Give Up

Answer 1

Let the normal force to the block by n.
n=mgcosα
Therefore the horizontal component of force on the wedge due to the block is given by F=-mgcosαsinα. The combined downward force due to the ramp and block is Mg+mgsinαsinα
Therefore (M+msinαsinα)a=-mgcosαsinα (By Newtons second law)
Hence a=(mgsinαcosα)/[M+m(sinα)²]

Before doing this problem, we look at two remarks first. If M is not moving then we have the trajectory Traglectary sketched in the righ .If M is moving to the left, then the actual trajectory is sketched as below, where ae is the acceleration of the block relative to the wedge of mass M. Actval Drayectory Now, we look at the force diagram of the wedge and the block. For A For nc N,sin-Ma -a 3) NE nng This is a set of simultaneous equations with 3 unkows After solving, we obeain 41 (M+gsine M + m sin θ Since the horizontal and vertical components of a are given by a-a cos-a, and a, -asin

M+msn . hence we have The reaction between the wedge and the block is given by the expression of a and equation (I) ie. N,ul'g cose

Replace $\theta$ by $\alpha$

You'll get the answer