# Homework Help: Small mass (m) resting on a larger slanted mass (M)

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1. Feb 22, 2016

### TroyP

1. The problem statement, all variables and given/known data
A small object with mass m sits on a ramp with mass M which itself sits on the ground (ignore friction between the ramp and the ground).
Given the coefficient of static friction μ , compute the minimum horizontal force F that when pushing on M will just barely start the small object m sliding up the ramp.

2. Relevant equations
F = ma
Ffriction = μmgcos(Θ)

3. The attempt at a solution
I started the problem by setting up two different force formulas:
1) Fpush = (M+m)a
2) -mgsin(Θ) - μmgcos(Θ) = ma
*This is where I went wrong and cannot solve the prblem. The acceleration of the system as a whole is due to Fpush, but this acceleration is not equal to the system of mass m itself sliding up the ramp. I do not know how to solve/ approach a problem with this complexity.

I then solved for a in both and set them equal to each other:

Fpush/(M+m) = -mgsin(Θ) - μmgcos(Θ)/m

Finally I solved for Fpush:

Fpush = (-mgsin(Θ) - μmgcos(Θ))(M+m)/m

But this is not correct. Any guidance in this problem would be appreciated!

2. Feb 22, 2016

### haruspex

You cannot assume the normal force is mg cos(θ). Let it be N and see what you get.
You can get another equation by considering vertical components of forces.

3. Feb 23, 2016

### kent davidge

For this problem, should we ignore the mass M of the ramp?

4. Feb 23, 2016

### TroyP

I don't think I understand what your'e saying fully. Do you mean setting up a coordinate system with the x axis parallel to the ramp?

5. Feb 23, 2016

### TroyP

I wouldn't think so. The push Force is acting on the large mass M and the small mass m.

6. Feb 23, 2016

### Biker

All you need to do here. Is to ensure that the net force on object m is just in the same direction and magnitude as force F
How can you achieve that? Notice Haruspex said about not calculating N as mg cos because it is not.
Make a diagram and think about it for a bit

7. Feb 23, 2016

### haruspex

No. Your equations 1) and 2) both relate to horizontal components of forces. You can also consider the vertical components of the forces on m. What equation can you write for those?

Edit: I accidentally wrote vertical twice instead of horizontal, then vertical.

Last edited: Feb 23, 2016
8. Feb 23, 2016

### ehild

The problem asks the force when the small object is in rest with respect to the slope, but applying a bit bigger force, it starts sliding up.
So the acceleration is the same both for the small object and the wedge, horizontal, of magnitude a=F/(M+m).
As the acceleration is horizontal, choose a coordinate system with horizontal and vertical axes. Make the FBD for the small mass, and write the equations for the horizontal and vertical components of the forces.

9. Feb 23, 2016

### kent davidge

What about my solution, F = mg(μ - tgθ)?

10. Feb 23, 2016

### haruspex

Doesn't look right. What is t? Where is M?

11. Feb 23, 2016

### ehild

It is wrong, You equated a horizontal force with a force along a steady slope, but the slope is accelerating and the small mass is in rest with respect to it. And you can compare only parallel force components.

12. Feb 23, 2016