# Friction, Newton's Laws

1. May 16, 2010

### Maybe_Memorie

1. The problem statement, all variables and given/known data
The two blocks shown in the diagram are at rest on a horizontal surface when a force P is applied to block B. Blaocks A and B have masses 20kg and 35kg respectively. The coefficient of friction between the two blocks is 0.35 and between the horizontal surface and Block B is 0.3.
Determine the maximum force, P, before A slips on B.

2. Relevant equations
F = ma , Friction max = uR

3. The attempt at a solution
Okay this thing has been killing me for months.
Friction between B and the ground is (20 + 35)g(0.3) = 16.5g and between B and A is
(0.3)(35)g = 10.5g. This produces anti-friction which propells A forward. Friction between A and B is (20)(0.3)g = 6g.
So, P should equal 16.5g +10.5g + (10.5g - 6g) = 31.5g N.
I'm supposed to be getting 35.75g N.

Where am I going wrong?

#### Attached Files:

• ###### problem.bmp
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2. May 16, 2010

### Staff: Mentor

Anti-friction??? The only force propelling A forward is the friction from B.

Hint: When the static friction between A and B is at its maximum value, what is the acceleration?

3. May 16, 2010

### Maybe_Memorie

Sorry, that's what I mean.

The acceleration of A or B?
The acceleration of A relative to B will be 0 if slipping doesn't occur.

4. May 16, 2010

### Staff: Mentor

Consider A first. What's the acceleration of A when slipping is just about to occur?
Exactly!

5. May 16, 2010

### Maybe_Memorie

Well if the force pushing A forward is (10.5)g, the acceleration should be (10.5)g/20, =
(0.525)g m/s^2.

6. May 16, 2010

### Staff: Mentor

What's the maximum static friction force acting on A?

7. May 16, 2010

### Maybe_Memorie

The normal reaction by the Coefficient of Friction. (20g)(0.3) = 6g N.

8. May 16, 2010

### Staff: Mentor

Good! Except that here μ = 0.35.

9. May 16, 2010

### Maybe_Memorie

Okay, I did it again, and got the right answer.
So let me see if I've understood.

I've included a force diagram.

The frictional forces acting on B are [(55g)(0.3) + (35)(0.35)] = 28.75g N

The force pushing A forward is (20g)(0.35) = 7g N.

To make B move, P must be 28.75g, and to ensure that there is no relative acceleration between A and B it must be (28.75g + 7g) = 35.75g N.

#### Attached Files:

• ###### force diagram.bmp
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267.5 KB
Views:
69
10. May 16, 2010

### Staff: Mentor

The friction force of A on B is equal and opposite to the friction force of B on A (which you have calculated correctly).
Good. So what must be the acceleration of A?

11. May 16, 2010

### Maybe_Memorie

If the friction force on B is the coefficient of friction by the normal reaction, should it not be
(0.35)(20g)?

F = ma, ma = 7g, so a = 7g/20 = 0.35g m/s^2.

12. May 16, 2010

### Staff: Mentor

Exactly.

Good! So what must be the acceleration of both A and B at that point? What force P is required to produce such an acceleration?

13. May 16, 2010

### Maybe_Memorie

If the acceleration of A is 0.35g, the acceleration of B must also be 0.35g, so that there is no relative acceleration between them and slipping will not occur.

F = ma, so P = (35)(0.35g).
Evidently I've gone wrong somewhere. Haha.

14. May 16, 2010

### Staff: Mentor

Good!

P is not the only force acting on B.

Hint: Try treating both blocks as a single object.

15. May 16, 2010

### Maybe_Memorie

P minus the two frictional forces acting on B has to produce an acceleration of 0.35g.

So [P - 16.5g-7g = 35(0.35g)]
P = 35.75g N

And if both blocks were a single object, the 7g N forces would not affect the system, and the mass of the system would be 55kg.

So [P -16.5g = 55(0.35g)]
P = 35.75g N

16. May 16, 2010

### Staff: Mentor

There you go.

17. May 16, 2010

### Maybe_Memorie

Thank you very much for your help.