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

AI Thread Summary
To keep a small block stationary on an inclined plane while resting on a triangular block, an external force F must be applied to the triangular block. The force required can be calculated using the equation (M + m)gtan θ, where M is the mass of the triangular block, m is the mass of the small block, and θ is the angle of inclination. The discussion highlights that without this force, the small block would slide down due to gravity. The reasoning involves understanding the balance of forces acting on the small block, including gravitational force and the reaction from the triangular block. Ultimately, the correct calculation of force ensures the small block remains fixed in position relative to the triangular 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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