How Is Force Calculated to Keep a Block Stationary on an Inclined Plane?

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SUMMARY

The discussion focuses on calculating the force required to keep a small block of mass m stationary on an inclined plane formed by a larger block of mass M. The correct formula for this force, given that all surfaces are frictionless, is (a) (M + m)gtan θ. The reasoning involves understanding that without an applied force, the small block would slide down due to gravity. The system must be accelerating to maintain the small block's position relative to the larger block.

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Homework Statement


A small block of mass m rests on the sloping side of a triangular block of mass M which itself rests on a horizontal table as shown in the attached figure. Assuming all surfaces are frictionless, determine the force F that must be applied to M so that m remains in a fixed position relative to M (that is, m doesn't move on the incline).

\mbox{(a)} \quad (M + m)gtan \theta

\mbox{(b)} \quad (Mtan \theta + m)g

\mbox{(c)} \quad \frac{(M + m)g}{tan \theta}

\mbox{(d)} \quad (M + m)g sin \theta

\mbox{(e)} \quad (M + m)g cos \theta

I know the answer is (a). Just don't understand how to arrive there.

Homework Equations





The Attempt at a Solution



My reasoning:

If there was no force being exerted on the system, i.e. F = F_{net} = 0, the small mass would slide down the triangular block due to the force of gravity (mgsin \theta). So, it would be safe to assume that if the system was at constant speed then the block would still slide down. Thus, the system must be accelerating in order for the small block to stay at rest relative to M.

However, how would the small block not move? Along the incline, there is only the force of gravity acting on the small block or would the force exerted by M on m somehow counteract that?
 

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