Conceptual Problem with Friction FBD.

[Sarin]

here is the scenario:

there are 2 blocks A & B, both have the same mass 'm'.
block A is kept on an endless surface.
there is no friction between the block A and the surface.
now the other block, B, is kept on top of the block A.
the coeff of friction between A & B is say 'u'.
the limiting friction between A & B is therefore = umg = 'f'.
now say i apply a force 'F', equal to 'f', on the block A, i.e F = f.

What will be the FBD's for A and B.

I did the FBD for B & A and i am getting f = 0 = F , which is not possible since f = umg.
So I think i have made some mistake.

And I am not saying that A and B are necessarily stationary initially, they could be moving with a uniform velocity.

Is this suggesting that since there is no relative motion between A & B therefore there is no friction existing. If this is the case then isn't this contradicting the fact that B moves with A
(with a relative velocity = 0) due to friction when a force is applied to A and that if friction was not present A would just slide past underneath B and B would just fall to the ground.
kindly help me out with this!

Thanks.

 

[Doc Al]

the limiting friction between A & B is therefore = um = 'f'.
now say i apply a force 'F', equal to 'f', on the block A, i.e F = f.

Realize that the limiting friction is not necessarily equal to the actual friction force.

If you push A with a force equal to μmg, then the friction acting between the masses will be only half that.

 

[Sarin]

Realize that the limiting friction is not necessarily equal to the actual friction force.

If you push A with a force equal to μmg, then the friction acting between the masses will be only half that.

not sure i understand what you're saying.

 

[Doc Al]

not sure i understand what you're saying.

Try this: Assume that you push block A with a force equal to μmg. What will be the acceleration of the two-block system? Use that to figure out the friction force.

 

[PJC01]

I would like to clarify some concepts regarding friction and its effects in this scenario. First, it is important to note that friction is a force that opposes motion between two surfaces in contact. In this case, the friction force between block A and the surface is zero because there is no contact between them. However, when block B is placed on top of block A, there is a contact surface between them and therefore the friction force can exist.

Now, let's consider the FBD for block A and B. For block A, the only forces acting on it are its weight (mg) and the applied force (F). Since there is no friction between block A and the surface, the net force on block A is equal to F-mg. This means that block A will accelerate in the direction of the applied force.

For block B, the forces acting on it are its weight (mg) and the friction force (f) between block A and B. The friction force will be in the opposite direction to the applied force (F) on block A. Therefore, the net force on block B is equal to mg-f. This means that block B will also accelerate in the direction of the applied force, but at a slower rate than block A due to the opposing friction force.

Now, let's consider the case where the applied force (F) is equal to the limiting friction force (f). In this case, the net force on block A will be zero (F-mg=0), meaning that block A will not accelerate. However, the net force on block B will be mg-f, which is also equal to zero. This does not mean that the friction force is zero, it means that the friction force is balanced by the weight of block B. Therefore, block B will not slide off block A, but it will also not accelerate.

In summary, the FBD for block A and B are correct. The fact that the friction force is equal to the applied force does not mean that the net force on block B is zero, it means that it is balanced by the weight of block B. This does not contradict the fact that block B moves with block A due to friction, as the net force on block B is still non-zero (mg-f). Friction is present and is necessary for block B to move with block A, even though the net force on block B is zero in this particular scenario. I hope this helps to clarify