5.120: a) There are many ways to do these sorts of problems; the method presented is fairly straightforward in terms of application of Newton’s laws, but involves a good deal of algebra. For both parts, take the x-direction to be horizontal and positive to the right, and the y-direction to be vertical and positive upward. The normal force between the block and the wedge is n ; the normal force between the wedge and the horizontal surface will not enter, as the wedge is presumed to have zero vertical acceleration. The horizontal acceleration of the wedge is A , and the components of acceleration of the block are x a and y a . The equations of motion are then . cos sin sin mg n ma n ma n MA y x-α = α = α-= Note that the normal force gives the wedge a negative acceleration; the wedge is expected to move to the left. These are three equations in four unknowns, A , y x a a , and n . Solution is possible with the imposition of the relation between A , x a and y a . An observer on the wedge is not in an inertial frame, and should not apply Newton’s
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