Condition satisified when the body does not slide

[Eugen]

Homework Statement

A sledge of mass m1 is pulled horizontally with a force F. On the sledge there is a body of mass m2 that can slide on the horizontal platform of the sledge with the friction coefficient μ. Another sledge of mass m3 is tied with a horizontal string of the body m2. Between the sledges and the snow the friction is negligible. Find the condition that is satisfied when the
m2 body does not slide.

Homework Equations

f_s = μN

T = [m2/(m1 + m2)]*F

The Attempt at a Solution

I assumed that μ is the coefficient of static friction. If the m2 body is not to slide, the static friction between this body and the m1 sledge must be bigger than the tension in the string.
The tension in the string should be:

T = [m3/(m1 + m2 + m3)] * F

and the static friction:

f = m₂gμ

m₂gμ > [m3/(m1 + m2 + m3)] * F

This gives:

μ > F/m₂g * [m3/(m1 + m2 + m3)]

but the book answers say that:

μ > F/m₂g * [(m3 + m2)/(m1 + m2 + m3)]

What is wrong?

 

[TSny]

If the m2 body is not to slide, the static friction between this body and the m1 sledge must be bigger than the tension in the string.

Draw a free body diagram for m2 and set up the second law for m2. This will give you the correct relation between the static friction force, the tension, and the acceleration.

 

[Eugen]

The acceleration of the system is F/(m1 + m2 + m3). This means that horizontally, in the positive direction, on m2 acts the force:

F' = m₂ F/(m1 + m2 + m3)

The tension acts horizontally in the negative direction:

T = - m₃F/(m1 + m2 + m3)

So, on m2 acts horizontally this force:

F_r = F(m2 - m3)/(m1 + m2 + m3)

which can be positive or negative. The friction should compensate for the resultant force, so it should be positive and equal in magnitude with the tension.

But, the solution says that
μ > F/m₂g * [(m2 + m3)(m1 + m2 + m3)]

which in my opinion means that the friction should be bigger that the F' + T. I don't understand where this sum comes.

 

[TSny]

The acceleration of the system is F/(m1 + m2 + m3). This means that horizontally, in the positive direction, on m2 acts the force:

F' = m₂ F/(m1 + m2 + m3)

Yes, that represents m₂a. So it must equal the net force acting horizontally on m₂. That is, it equals ∑F_x.

The tension acts horizontally in the negative direction:

T = - m₃F/(m1 + m2 + m3)

Yes.

So, on m2 acts horizontally this force:

F_r = F(m2 - m3)/(m1 + m2 + m3)

There is a sign error here. Start with ∑F_x = m₂a_x. On the left side, add all the x-components of the forces acting on m₂, taking into account the signs of the x-components. Then solve for F_r.

The friction should compensate for the resultant force, so it should be positive and equal in magnitude with the tension.

If the friction force is equal in magnitude to the tension force (but opposite in direction), would m₂ have any acceleration?

 

[Eugen]

Oh. Now I get it. The force acting on m2 is the resultant of tension and friction, the friction being in the positive direction:

-m₃F/(m1 + m2 + m3) + m₂gμ = m₂F/(m1 + m2 + m3)

Thank you.