Small cube/large cube sliding problem with friction quickly :/

[rosyroguey]

(I know this type of problem has been discussed on here before, but I still don't understand what to do next)

The attached drawing shows a large cube (mass=55kg) being accelerated across a horizontal frictionless surface by a horizontal force P. A small cube (mass=4.5kg) is in contact with the front surface of the large cube and will slide down unless P is sufficiently large. The coefficient of static friction between the cubes is 0.55. What is the smallest magnitude that P can have in order to keep the small cube from sliding downward?

So far I have found that:
Normal force=ma, (4.5 x 9.81)= 44.1 N
F sub s=Mu sub s times mg, (.55 x 44.1)=22.05N

I've also attatched a diagram of all the forces acting on the small cube, and concluded that an additional force of 22.05N is necessary to keep the small cube on the large one. I just don't know where to go from here to find P.

Thanks for your help.

 

[Doc Al]

So far I have found that:
Normal force=ma,

Good. (Where does that force act?)

(4.5 x 9.81)= 44.1 N

That's the weight of the small cube. OK.

F sub s=Mu sub s times mg, (.55 x 44.1)=22.05N

Say what? We're talking about the friction force between the two cubes, so what does mg have to do with it?

Hint: If the small cube doesn't slide down, what can you say about its vertical acceleration and net force? What are the only vertical forces acting on it?

Once you answer those questions, see what you can deduce about: the horizontal force on the small cube, the acceleration of the small cube, the force P which must accelerate the whole thing.

 

[rosyroguey]

If it's not sliding, that means the vertical acceleration has to be zero, which means the net force is zero? So the horizontal force on the small cube has to be great enough to compensate for the lack of vertical force? Am I going in the right direction?

 

[Doc Al]

If it's not sliding, that means the vertical acceleration has to be zero, which means the net force is zero?

Yes, the net vertical force must be zero. So what does that tell you about the friction force acting on the small cube?

Also: What's the general relationship for static friction? (In terms of the coefficient of friction.)

 

[rosyroguey]

The formula my teacher gave me for static friction is FsubS=μsubS x FsubN but didn't say anything involving normal force besides that it is opposite weight... this is the first problem he's ever given us of this kind and it's on a take-home quiz which he won't help us on. Anyway... I don't know what the net force equaling zero says about the friction force.. that it's large? I feel really ridiculous right now but this is all I can come up with... thanks for your help so far.

 

[Doc Al]

The formula my teacher gave me for static friction is FsubS=μsubS x FsubN

Right. The maximum static friction force is μN, where N is the normal force.

but didn't say anything involving normal force besides that it is opposite weight...

The normal force will only equal the weight under certain conditions: Such as if we were sliding the cube on a horizontal surface. In that case, the normal force would equal mg.

But in this problem the normal force is the force between the two cubes, which is not directly related to the weight of the smaller cube. (Note that the normal force is the only horizontal force acting on the small cube.)

Anyway... I don't know what the net force equaling zero says about the friction force.. that it's large?

Hint: There are two vertical forces; one acts down, the other acts up: what are those forces? (Check your diagram.) And if they add to zero, what can you say about them?

 

[rosyroguey]

The Kinetic friction has to equal the weight? So Fk=44.1?

 

[Doc Al]

The Kinetic friction has to equal the weight? So Fk=44.1?

The friction is static, not kinetic--remember that the small cube does not slide. But yes, the upward static friction force must equal the weight.

Next step: Use that fact to deduce the minimum normal force needed to produce that friction.

 

[rosyroguey]

So Normal force=ma: 44.1 divided by .55 equals 801.8N- the normal force. Now I know the normal force acting on the smaller block.

 

[Doc Al]

So Normal force=ma: 44.1 divided by .55 equals 801.8N- the normal force.

Recheck that arithmetic. (You are off by a factor of ten.)

Now I know the normal force acting on the smaller block.

Yes. Now use that to find the acceleration.

Then you can ask: What force P must push on both cubes to give them the needed acceleration?

 

[rosyroguey]

Can I use ∑F=ma to find the acceleration, or is normal force and summation force not the same thing?
(Haha thanks for the arithmetic check. I'm so bad at that)

 

[EthanB]

You'll always use ∑F=ma to find the acceleration. "Summation force" just means adding the forces, which includes normal force.