# Homework Help: Friction and tension in a string

1. Oct 21, 2014

### Jennings

1. The problem statement, all variables and given/known data
Mass m1 = 5.00kg is tied to the wall with a horizontal piece of string and is at rest onto of mass m2 = 10.00kg. A horizontal force F= 50.0N is applied to m2. The static coefficient of friction between all surfaces is μs = .350 and the kinetic coefficient of friction is μk = .250. Find the tension in the rope attached to m1 and the acceleration of m2

2. Relevant equations
Fk = μkN
Fsmax = μsN
ΣF = ma

3. The attempt at a solution
Trying to draw a free body diagram for the m1 and the tension, i know that the sum of forces must be 0 because m1 does not move I know that the weight and normal act on it in the y direction. But im confused about the x direction. Is friction and tension the only forces in the x direction? Or is there another force?

I am just confused

2. Oct 21, 2014

### Staff: Mentor

Yes.
No.

Chet

3. Oct 21, 2014

### Jennings

Does friction oppose the direction of tension? . The force of friction and tension must cancel to equal 0 right?

4. Oct 22, 2014

### Jennings

How to determine how much is the friction, since there are two frictions and we do not even know if thr mass m2 will move

5. Oct 22, 2014

### Jennings

Is it kinetic or static friction? How can we determine this?

6. Oct 22, 2014

### Staff: Mentor

In this problem, the problem statement implies that you are dealing with kinetic friction and that m2 is moving. However, this may not be the case. See my next post below.

Chet

Last edited: Oct 22, 2014
7. Oct 22, 2014

### Staff: Mentor

You solve the problem assuming that static friction prevails, and that you are just on the verge of exceeding the coefficient of static friction on both surfaces. Then you check to see whether the frictional forces are sufficient to balance the applied force F. Make sure you do a free body diagram on each of the two masses; this is very important.

Chet

8. Oct 22, 2014

### Staff: Mentor

I have given this problem significantly more thought, and, in my judgement, it is extremely tricky. It really requires a great deal of thought and "what if" calculations (even though it looks deceptively simple). I'm not sure the person who posed this problem realized how tricky it is.

Chet

9. Oct 22, 2014

### Jennings

Hmm.. well i hope ill be able to figure it out regardless.. with your help of course.

10. Oct 22, 2014

### Jennings

The normal force between the mass 1 and the mass 2 i will call N1 , and the maximum static friction ill call it fsmax1

The normal force between mass 2 and the surfacd i will call N2 and likewise the maximum static friction ill call it fsmax2

Therefore

N1 = mg = (5.0kg)(9.81m/s^2) =49.1N
N2 = mg = (15kg)(9.81m/s^2) = 147N

fsmax1 = (.305)(49.1N) = 17.2N
fsmax2 = (.305)(147N) = 51.5N

Im thinking this means that the force applied is not enough therefore the tension of the string and acceleration mass 2 will both be zero.

Please correct me if im wrong but i believe that this is what it is ? Right ?

11. Oct 22, 2014

### Jennings

meant for Ms to be .35 not .305. My fault

12. Oct 22, 2014

### Staff: Mentor

As you said, the frictional forces you calculated are certainly the maximum values that can be tolerated at each surface without slippage. However, if we assume that m2 is totally rigid, we don't know what the actual frictional forces will be prior to slippage. This is because the system is statically indeterminate. Prior to slippage, we only know that their sum is 50 N. We would have to determine how the frictional forces are distributed between the upper and lower surfaces in order to ascertain whether slippage would occur. In order to do that, we would have to solve a problem in elasticity. For the present problem, with the 50 N force applied at the middle of m2, if we assumed no slippage at the interfaces, the forces at the two interfaces would be 25 N each. And 25 N at the upper surface would be enough for slippage to occur there, but 25 N at the bottom surface would not be sufficient for slippage to occur there. So, with m2 being elastically deformable (not totally rigid), the top surface of m2 would begin to slip in the x direction while the bottom surface of m2 would still be non-slipping. Continuing the solution to this problem would involve elastic deformation with friction and dynamics. This would be a very complicated problem. Do you see now what I meant by this all being very tricky?

Chet

13. Oct 22, 2014

### Jennings

I do see now how this simple question has turned into a devastatingly complex problem when you look into it in depth. Wow. We havent even gotten to elasticity yet. I cant imagine my professor would expect this from me at this point in time. I think you were correct when you said that the one who wrote the problem had no idea how tricky this was. I have only a basic knowledge of friction and forces at the current time. Is there any way we could dumb it down to a level where we can disregard all this complexity while still being theoretically correct with our answer?

14. Oct 22, 2014

### Jennings

Or is that not possible with this

15. Oct 22, 2014

### Staff: Mentor

I would assume that he didn't recognize the complexity and just continue solving the problem assuming kinetic friction at both interfaces, based on the rationale that you presented.

Chet

16. Oct 22, 2014

### Jennings

So we cannot solve the problem if it were to be static friction? I understand that if static friction were to occur then the acceleration of M2 would be 0, but your saying in order to find out what occurs with mass 1 would require more complex calculculations if i understood you correctly

17. Oct 22, 2014

### Jennings

Assuming the friction is kinetic .. do you have any hints you can give to me to set me up on the right path. I am assuming that value for kinetic friction will be less with M1 since M1 has a smaller normal, but M1 itself does not move, so its kind of confusing on what exactly to do

18. Oct 22, 2014

### haruspex

It is a bit tricky, but I think it is all quite reasonable in the end. Suppose one surface just starts to slip. Is kinetic friction there plus static friction at the other interface enough to hold it? Try both cases. I think you'll find it holds.

19. Oct 22, 2014

### Staff: Mentor

If it's rigid, you can't have one surface slipping while the other is not.

Chet

20. Oct 22, 2014

### Staff: Mentor

Yes. If the total force were <35.4 N, each surface would have <17.2 N, and static friction would be guaranteed to be intact on both surfaces. Otherwise, it's more complicated.

Chet