# Newton's Laws: Wedge and block magnitude, acceleration

A.) A wedge with mass M rests on a frictionless horizontal tabletop. A block with mass m is placed on the wedge. There is no friction between the block and the wedge. The system is released from rest. Calculate:

i) the acceleration of the wedge

ii) the horizontal and vertical components of the acceleration of the block. (Do these answers reduce to the correct results when M is very large?)

b) The wedge and block now have a horizontal force P applied to the wedge. What magnitude must P have if the block is to remain at a constant height above the tabletop?

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#### Solution Preview

a.) let the slope of wedge with horizontal = Q

see freebody diagram in attached jpg file.

Let the acceleration of the wedge is a and the acceleration of block with respect to wedge along the slope is a'.

N = mg.cos(Q) - m.a.sin(Q)

horizontal component of N on wedge:

Nh = {mg.cos(Q) - ma.sin(Q)}.sin(Q)= mgcos(Q).sin(Q) -ma.sin^2(Q)

vertical component of N:

Nv = {mg.cos(Q) - ma.sin(Q)}.cos(Q) = mg.cos^2(Q) - ...

#### Solution Summary

The solution carefully explains the problem and solves for the answers.