Two blocks kept side by side and friction between them

[Hamiltonian]

Homework Statement

in the given figure a force F I applied on a block of mass M and the coefficient of friction(u) between the two blocks of mass M and m is u. the surface on which the block of mass M is kept is smooth. for a force F > g(M+m)/u will the block of mass m move upward

Relevant Equations

f = ma

if the applied force F >(M+m)g/$\mu$ the force of friction between the two blocks should become f=N$\mu$ and since the normal reaction will be greater than what it should be to keep the block m at rest with respect to M, will the mass m accelerate upwards.

in short,for let's say a very large F can the block of mass m overcome the force of gravity and move upwards due to the force of friction between them. as f=N$\mu$ and as the force F increase N also increases and at some time the force of friction f will become grater than mg.

 

[PeroK]

Friction always acts to oppose motion between surfaces. If there were enough friction to move the mass upwards, there would be enough friction to resist that motion!

 

[haruspex]

@Hamiltonian299792458 , You appear to be making the classic error of confusing the maximum static frictional force, $\mu_sN$, with the actual frictional force.

 

[Hamiltonian]

Friction always acts to oppose motion between surfaces. If there were enough friction to move the mass upwards, there would be enough friction to resist that motion!

so its impossible for block m to move upward

 

[PeroK]

so its impossible for block m to move upward

Why wouldn't friction move it downward instead? Or sideways?

 

[haruspex]

so its impossible for block m to move upward

There is no reason for it to move upward. The actual frictional force will never exceed the force needed to prevent relative motion of the surfaces, in this case mg.

 

[etotheipi]

As mentioned, friction will not do the trick. But you could alter the problem so that the right face of the block is slightly less steep than vertical. In the box frame, the inertial force will have a small component parallel to the plane, which must oppose both the components of friction and gravity parallel to the plane if it is to slide upward.

 

[Hamiltonian]

ok i think i understand my mistake, as N increases that's just increase the maximum frictional force and not the force that is acting in the given case

 

[haruspex]

As mentioned, friction will not do the trick. But you could alter the problem so that the right face of the block is slightly less steep than vertical. In the box frame, the inertial force will have a small component parallel to the plane, which must oppose both the components of friction and gravity parallel to the plane if it is to slide upward.

.. and the question posed might then be what the maximum angle of the face to the vertical is such that, at sufficient applied force, m will slide upwards.

 

[PeroK]

ok i think i understand my mistake, as N increases that's just increase the maximum frictional force and not the force that is acting in the given case

You need to remember this because you've gone wrong on several problems now with this idea.