# Sliding blocks and friction

## Homework Statement

The two blocks (m = 12 kg and M = 98 kg) in Fig. 6-38 are not attached to each other. The coefficient of static friction between the blocks is μs = 0.37, but the surface beneath the larger block is frictionless. What is the minimum magnitude F of the horizontal force required to keep the smaller block from slipping down the larger block?

F=ma
Fs=Us(Fn)

## The Attempt at a Solution

Not sure how to do this one. So far I got that Sum of F = 0 because the block is ont accelerating.
The big block is moving because it is frictionless at the bottom and it is being pushed by the horizontal force F.
Fs has to > gravity so that gravity doesn't move the small block
So Fs > mg
Fn*Us > mg
that's all the thoughts I have so far... but it can't be that simple because the big block is moving, so more force has to be applied. But the more force that is applied, the more the big block moves... and my thoughts just end up in circles.
Also, I'm not sure if the horizontal force F includes the normal force already, or do I have to add the normal force when doing sum of forces = 0 ?

#### Attachments

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## Answers and Replies

SammyS
Staff Emeritus
Homework Helper
Gold Member
Both blocks accelerate together. F ≠ 0 .

How much force must be exerted on the small block in the vertical direction so that its acceleration in the vertical direction is zero?

take the bigger block as your frame
apply pseudo force
then find the normal force b.w bigger and smaller block and thus find friction
for equilibrium, frictional force = mg

How do I find the normal force between the big and the small block
I got Friction = mg
Fs= mg
Fn(Us) = mg
Fn(0.37)=(12)(9.8)
Fn = 317.83
and the total Fn would be F + normal force from the big block exerted on the small block. But I don't know how to find that.
Also, do you have to compensate for the fact that the two systems are moving?

that is why i said take the blocks as frame. blocks are at rest in that frame. then apply pseudo force.

and for normal force ... think of newton 3rd law i.e. action rxn pair. the force F on m is the force transferred to M. So M will apply a force of ___ on m

SammyS
Staff Emeritus
Homework Helper
Gold Member
How do I find the normal force between the big and the small block
I got Friction = mg
Fs= mg
Fn(Us) = mg
Fn(0.37)=(12)(9.8)
Fn = 317.83
and the total Fn would be F + normal force from the big block exerted on the small block. But I don't know how to find that.
Also, do you have to compensate for the fact that the two systems are moving?

Yes, you now have that you need a normal force, Fn ≥ 317.84 N. (This is in the x direction.)

F - Fn = ma, for the small block.

The large block exerts a force, Fn, on the small block in the negative x direction.

Therefore, the small block exerts a force, Fn, on the large block in the positive x direction. This force must be enough so that the acceleration of the large block is equal to the acceleration of the small block. Newton's 2nd Law should give you what that acceleration is.

For F-Fn = ma
wouldn't F be the total force of Fn + F horizontal?
I am only looking for F horizontal so instead can I do
F horizontal + Fn = (m1+m2)a
and since Fn is the force that is pushing the big block, then Fn=M2a
M2 being the big block.

SammyS
Staff Emeritus
Homework Helper
Gold Member
F and Fn ARE horizontal. (The frictional force, Fs, is vertical, the magnitude of which is given by Fs = μsFn.)

The net force on the pair of blocks is just F, so F = (m1+m2)a. (This equation has two unknowns in it.) Fn is an internal force.

As I said in my last post, "The large block exerts a force, Fn, on the small block in the negative x direction." This gives: F ‒ Fn = m1a .

Using Newton's 3rd Law, the small block exerts a force, Fn, on the large block in the positive x direction. This gives: F + Fn = m2a .

Use any two of these last three equations to find F and a.

I use m1 & m2 in this post with the understanding that m1 = m from your diagram & m2 = M.