Force needed to keep block from moving on frictionless triangular block

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SUMMARY

The discussion focuses on calculating the force required to keep a small block of mass m stationary on a frictionless triangular block of mass M. The key equations provided are the force components along the x-axis, represented as m * g * sin(φ), and along the y-axis as m * g * cos(φ). However, the correct approach emphasizes the normal reaction force (N) and its horizontal component, which is crucial for solving the problem accurately.

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  • Understanding of Newton's laws of motion
  • Familiarity with trigonometric functions in physics
  • Knowledge of normal force concepts
  • Basic principles of static equilibrium
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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).

Chapter4problem53.jpg

(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.
 
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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:
 

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