Calculating Acceleration with Friction and Applied Force

[bodensee9]

Hello:
Here is a picture:



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the top block has mass 10 kg, and the bottom slab has mass 40 kg. The bottom slab lies on a frictionless surface. Between the block and slab, the coefficient of static friction is 0.60 and the coefficient of kinetic friction is 0.40. The slab is pulled by a horizontal force F of magnitude of 100N to the left. What are the resulting acceleration of the block and the slab?

Since we have 100N pulling the block, would the acceleration of the block be:

100 - 0.40*10*9.8 = 10*a?

This 100 N is enough to move the block given the static friction.

But I'm not sure about the slab? Would it be 100 = (40 + 10)*a? But that seems too simple... Many thanks.

 

[bodensee9]

Sorry, picture got mangled. Just pretend that there is a block and a slab, the block is lying on top of the larger slab. Thanks.

 

[Doc Al]

The slab is pulled by a horizontal force F of magnitude of 100N to the left.

I assume you meant that the force F is pulling on the block, not the slab.

Since we have 100N pulling the block, would the acceleration of the block be:

100 - 0.40*10*9.8 = 10*a?

Yes.

This 100 N is enough to move the block given the static friction.

Not sure what you mean here. (Static friction is not enough to prevent slipping between block and slab.)

But I'm not sure about the slab? Would it be 100 = (40 + 10)*a?

What forces act on the slab?

 

[RoyalCat]

Hrrmph, as I was typing this I noticed the reply above mine. My solution assumes that the external force is pulling on the slab, and not on the block. Either the thread starter tried solving my problem and got mixed up, or he just copied it wrong when posting it here. Either way, he should be on his way to a solution between the both of us.

You got to the right result, but you didn't consider the two possible solutions.
Your first attempt at a solution,

100 - 0.40*10*9.8 = 10*a?

Assumed that the block slides on top of the slab. And then, you calculated the acceleration of the slab, and not that of the block. Take note that the slab is acted upon by two forces (In the horizontal axis): The external pull, and the friction of the block on top of it, while the block is only acted upon by the friction from the slab. The friction is the conveyor of the external pull to the block, if you wish to perceive it like that.

What we need to do first, is determine whether the block slides on top of the slab or not. If it does not, then we can treat the bodies as one body.
If the block does slide on top of the slab, then we must treat each body separately and calculate the individual accelerations of the slab and the block, based on the forces acting upon them in the horizontal axis.

To find out, we'll check to see if the slab is trying to "provide" more force to the block, than static friction can enact upon it.

Static friction "tops out" at f_{s max}=µ_s * N
In our case, the normal is mg, where m is the mass of the top block.
f_{s max}=µ_s * N = µ_s * mg

In our case, it is equal to 60N. It is the maximal force that static friction can enact on the block. How would you calculate the critical acceleration, then?

Spoiler

That means that the maximal acceleration that static friction can give the block is:
a_с = f_{s max}/m = 60/10 = 6 m/s²

If the slab were to move at an acceleration greater than the critical acceleration, then the static friction simply would not be able to accelerate the block to match, and the block would start sliding on top of the slab, and the force we'd have to be considering would be kinetic friction instead of static friction.

The problem is pretty short, I can't give another hint without solving it, so if you want to check your answer against mine, open the spoiler.

Spoiler

Assuming static friction applies, the acceleration the external force grants the two bodies would be:
a = F_ext/(m+M) where M is the mass of the slab.
a = 100/(10+40) = 2 m/s² which is below the threshold for static friction, meaning that the bodies will move as one, and that this calculation is valid.

 

[bodensee9]

Oh okay, so the force that act on the slab is friction from pulling the block. This friction would oppose the direction of the pull. And I think it's kinetic friction since the block is already moving? So would it be 0.4*10*9.8 = (40)*a?
Thanks.

 

[Doc Al]

Oh okay, so the force that act on the slab is friction from pulling the block.

Right. The only horizontal force on the slab is the friction force exerted by the block.

This friction would oppose the direction of the pull.

No, the block drags the slab in the same direction as the pull.

And I think it's kinetic friction since the block is already moving? So would it be 0.4*10*9.8 = (40)*a?

Right.

The very first thing to do in this kind of problem is determine whether static friction is sufficient to prevent slipping. What must the static friction force be if no slipping occurred? Can the surfaces provide that amount of static friction?

 

[bodensee9]

If no slipping occurred, then the slab must have the same acceleration as the block.
So since a calculated above = 6.08 m/s^2, then the static friction would need to equal 40*6.08 = 243N.
But since the coefficient of friction is only .60, and .60*10*9.8 = 58.8, this isn't enough to prevent the block from slipping?
Thanks.

 

[Doc Al]

If no slipping occurred, then the slab must have the same acceleration as the block.

Right.

So since a calculated above = 6.08 m/s^2, then the static friction would need to equal 40*6.08 = 243N.

Careful: If no slipping occurs, then the acceleration will be different. Calculate that acceleration.

But since the coefficient of friction is only .60, and .60*10*9.8 = 58.8, this isn't enough to prevent the block from slipping?

Your reasoning is correct, but you are using the wrong acceleration.

 

[bodensee9]

Oh right, since no slipping occurred, you can treat the slab and the block as a single entity of 10 + 40 = 50 kg, and treat the 100N as applied to that whole entity. So a = 2 m/s^2.
Thanks.