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

• rosyroguey
In summary: Normal force and summation force are not the same thing. You would need to use ∑F=ma to find the acceleration.
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.

Attachments

• block problem.bmp
90.1 KB · Views: 915
• small cube diagram.bmp
82.8 KB · Views: 838
rosyroguey said:
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.

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?

rosyroguey said:
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.)

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.

rosyroguey said:
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?

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

rosyroguey said:
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.

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.

rosyroguey said:
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?

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)

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

1. What is the small cube/large cube sliding problem with friction?

The small cube/large cube sliding problem with friction is a physics problem that involves two cubes of different sizes sliding on top of each other with the force of friction acting between them. The problem is to determine the acceleration and direction of the smaller cube as it slides on top of the larger cube.

2. How is friction involved in this problem?

Friction is the force that opposes the motion of objects in contact with each other. In this problem, the force of friction is acting between the two cubes, causing the smaller cube to slide on top of the larger cube instead of falling off.

3. What factors affect the outcome of this problem?

The outcome of this problem is affected by several factors such as the mass and size of the cubes, the coefficient of friction between the two cubes, and the angle at which the smaller cube is placed on top of the larger cube.

4. How can this problem be solved quickly?

To solve this problem quickly, one can use the laws of physics such as Newton's Second Law of Motion and the equations for frictional force to calculate the acceleration and direction of the smaller cube. One can also use online tools or computer simulations to solve the problem quickly.

5. What real-life applications does this problem have?

This problem has real-life applications in various fields such as engineering, robotics, and transportation. Understanding the factors that affect the motion of objects with friction can help in designing more efficient and stable structures and machines. It can also aid in improving the performance of vehicles on different surfaces with varying coefficients of friction.

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