# An apparently tricky problem on Friction/Forces.

• cantgetaname
In summary: 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
cantgetaname

## Homework Statement

Two bodies of mass m1 and m2 are placed on a horizontal table with coefficient of friction $$\mu$$ and are joined by a spring. Initially, the spring has its natural length.
If F is the minimum force which, when applied on m1, will make the other block m2 just move, find F. (k is the spring constant)

Here is an artistic illustration of the situation.

## Homework 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)

## 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.

Does the spring act only on m2?

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

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.

cantgetaname said:
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 $$\frac{\mu m_{2}g}{k}$$.
So far, so good!

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's certainly true.

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.
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.

Doc Al said:
So far, so good!

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.

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

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

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:

## 1. What is friction and how does it affect forces?

Friction is a force that opposes the motion of an object when in contact with a surface. It can affect forces by either increasing or decreasing the amount of force needed to move an object.

## 2. How can friction be reduced?

Friction can be reduced by using lubricants, such as oil or grease, between two surfaces to reduce the amount of contact and friction between them.

## 3. What is the difference between static and kinetic friction?

Static friction is the resistance between two surfaces that are not moving relative to each other, while kinetic friction is the resistance between two surfaces that are moving against each other.

## 4. How does the weight of an object affect friction?

The weight of an object affects friction by increasing the force between two surfaces, which in turn increases the amount of friction between them.

## 5. How does surface texture affect friction?

Surface texture can affect friction by either increasing or decreasing the amount of surface area in contact between two surfaces, which can change the amount of friction between them.

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