# Forces and Newton's Laws of Motion

1. Feb 8, 2015

### kaspis245

1. The problem statement, all variables and given/known data
The drawing shows a large cube (mass = 25 kg) being accelerated across a horizontal frictionless surface by a horizontal force P. A small cube (msmall = 4 kg) is in contact with the front surface of the large cube and will slide downward unless P is sufficiently large. The coefficient of friction between the cubes is f = 0.71 . What is the smallest magnitude that P can have in order to keep the small cube from sliding down?

2. Relevant equations
Ffriction= f*mg

3. The attempt at a solution
Ffriction=msmall*g
f*P=msmall*g
P=56,34N

But the answer must be P = 400 N.

2. Feb 8, 2015

### Staff: Mentor

P has to accelerate both the small and the large block, that will take more force than the force between the two blocks.

3. Feb 8, 2015

### kaspis245

Great answer but still not clear.

4. Feb 8, 2015

### Staff: Mentor

What you found is the force that must push against the small mass, not P, which is the force that pulls the large mass. Use what you found to figure out what P must be.

5. Feb 8, 2015

### kaspis245

I kinda figured that out. Can somebody show me how to do it?

6. Feb 8, 2015

### Staff: Mentor

Here's a hint: Consider the acceleration.

7. Feb 8, 2015

### kaspis245

You mean P = ma ?

8. Feb 8, 2015

### Staff: Mentor

Yes. Well, F = ma. Consider the acceleration of the small mass first.

9. Feb 8, 2015

### kaspis245

The acceleration of the small mass is equal to the large mass.

10. Feb 8, 2015

### Staff: Mentor

Exactly! What's the acceleration of the small mass?

11. Feb 8, 2015

### kaspis245

The acceleration of the small mass a = P/msmall

12. Feb 8, 2015

### kaspis245

Got it. asmall = 14 m/s2

Plarge = (m + msmall)*a = 400 N

13. Feb 8, 2015

### Staff: Mentor

No. It's the force acting on the small mass divided by the small mass.

14. Feb 8, 2015

### Staff: Mentor

Good!

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