Force needed to keep block from moving on frictionless triangular block

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Strukus
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Homework Statement


This is straight from the book.

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 Fig. 4-50. 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).

Free-body style diagram of a green triangular wedge of mass M on a frictionless surface, with a yellow block of mass m resting on its sloped face. A horizontal force F points right into the vertical face on the left, and the incline angle is marked theta at the bottom right.

(I forgot to draw the surface the triangular block is on and sorry for the big picture!)

The answer is:
(m + M) [tex]\ast[/tex] g[tex]\ast[/tex] tan([tex]\phi[/tex])


Homework Equations


Force along the x-axis: m [tex]\ast[/tex] g [tex]\ast[/tex] sin([tex]\phi[/tex])
Force along the y-axis: m [tex]\ast[/tex] g [tex]\ast[/tex] cos([tex]\phi[/tex])

The Attempt at a Solution


I know that there must be an equal and opposite force along the incline but I have no clue how to approach this problem.
 
Welcome to PF!

Hi Strukus! Welcome to PF! :smile:
Strukus said:
Force along the x-axis: m [tex]\ast[/tex] g [tex]\ast[/tex] sin([tex]\phi[/tex])
Force along the y-axis: m [tex]\ast[/tex] g [tex]\ast[/tex] cos([tex]\phi[/tex])

No, that's the wrong approach …

only one of those equations is correct …

start again … you know the vertical acceleration of m is zero, so what is N (the normal reaction force ) …

and then what is the horizontal component of N? :smile: