# Homework Help: An apparently tricky problem on Friction/Forces.

1. Sep 29, 2010

### cantgetaname

1. The problem statement, all variables and given/known data

2. Relevant equations
1. Limiting Frictional Force = $$\mu$$ x Normal/reaction offered by the surface.
2. Force due to spring = k * Elongation of spring (from its natural length)

3. The attempt at a solution
If m2 is to move, it needs to experience a force of $$\mu m_{2}g$$ (to oppose the friction), which can be only experienced from the spring.
This means the spring's elongation at that instant is atleast $$\frac{\mu m_{2}g}{k}$$.
An equal and opposite force acts on m1. For m1 to move, net force on it must be greater than $$\mu m_{1}g$$.
That gives $$F=\mu m_{1}g+\mu m_{2}g$$, which is incorrect.

The answer is $$F=\mu m_{1}g+\frac{\mu m_{2}g}{2}$$, which I have no idea of where it could possibly come from.

2. Sep 29, 2010

### fuzzywolf

Does the spring act only on m2?

3. Sep 29, 2010

### cantgetaname

Well, it obviously acts on both.
It can't possibly act on just one object.

4. Sep 29, 2010

### fuzzywolf

That is precisely correct. F = -kx assumes that one end of the spring is fixed, which is not true in this case. Try a special coordinate system where one direction is toward the center of the spring, and one direction is away from the center of the spring.

5. Sep 29, 2010

### Staff: Mentor

So far, so good!

That's certainly true.

Hint: Realize that the applied force just has to be enough to get m1 moving and stretch the spring the right amount. Think in terms of energy.

6. Sep 29, 2010

### ehild

No, the problems asks the force which makes m2 just move. I do not see why the minimal force is not (m1+m2)g .

ehild

7. Sep 29, 2010

### Staff: Mentor

I assume you meant μ(m1+m2)g.

Ask yourself: If you applied such a force through a distance x (the required elongation), how much energy will m1 have? If m1 has extra energy beyond that needed to stretch the spring, then you've pushed too hard. Note that the spring force is not constant.

8. Sep 29, 2010

### ehild

Sorry, i meant ug(m1+m2). Does the problem ask the minimum constant force? If so, it is really
ug(m1+m2/2), and anyway, it is less than ug(m1+m2), so you are right...

So we apply a constant force F along a path Δx, first accelerating m1, than letting it decelerate, stopping at the end. According to the Work-Energy Theorem, the work of all forces equals to the change of KE which is zero now. So
W=(F-μm1g)Δx-kΔx2/2 =0, F=μm1g+kΔx/2

ehild

Last edited: Sep 29, 2010