# Friction Problem

1. Oct 10, 2006

### Fusilli_Jerry89

A 0.5 kg mass is on top of a 1.0 kg mass on a table. The static coefficient of friction between the blocks is 0.35, and the kinetic coefficient of friction between the table and1.0 kg block is 0.20. What's the maximum force that can be applied horizontally to the 1.0 kg block without letting the 0.5 kg block slip?

the acceleration of the small block has to be zero, so the friction between the two blocks cannot exceed 1.715 N.I got the equation for the first blockdown to F-3.675=1.0a but what do I do after that? Does the acceleration of the large block have to be zero as well, because if it does then the answer is 3.675 N.

2. Oct 10, 2006

### OlderDan

No it does not. Think about the motion that will occur if there is no slipping.

3. Oct 10, 2006

### Fusilli_Jerry89

well won'tthe accelration of the small block havet to match that of the big one? Wouldn't that mean it is 0 in relation to the big block?

4. Oct 10, 2006

### Fusilli_Jerry89

So is the answer 7.105 Newtons then?

5. Oct 10, 2006

### OlderDan

Yes they have to be the same. Yes that means it is zero relative to the big block, but the big block is accelerating. You don't know how to do problems in accelerating reference frames. Do the problem in the laboratory frame. Both blocks will be accelerating.

6. Oct 10, 2006

### OlderDan

I don't think so. How did you get it?

7. Oct 10, 2006

### Fusilli_Jerry89

well i found theat the equation for the large block was:
F-Ff-Ff=ma
F-1.96-1.715=1.0a
F-3.675=a

and for the small block:
Ff=ma
1.715=0.5a
a=3.43

then I substituted a into the first equation cuz they are accelerating at the same rate
F-3.675=3.43
F=7.105 N

8. Oct 10, 2006

### OlderDan

Why 1.96? What is the normal force acting on the bottom of the 1kg block?

9. Oct 11, 2006

### Fusilli_Jerry89

woops i forgot to add the 0.5 kg. So it would be 2.94 N instead?

10. Oct 11, 2006

### OlderDan

That looks better.