# Tension & friction prblem (limiting case)

1. Sep 28, 2004

### shashank010288

I am finding it very difficult to find frictional force & tension in this problem, since the system remains at rest. It requires something more than FBD & force eqations.

Q. Two blocks 10 kg(u=o.f) & 6kg(u=0.2) ( u:co-ef of friction) lie on a horizontal table, connected by string. (inextensible.) The 10 kg block is connected by a string passing over a pully to a 2kg block which hangs freely.
similarly, the 6kg block is connected to a 4kg block by string pully, which hangs freely. find the force of friction on the 2 blocks & tension in the string joining them.

U get 2 eqations
T+ F(on 6)= 4g

T=F(on 10) + 2g

2 eq. & 3 unknowns. How to solve?

Last edited: Sep 28, 2004
2. Sep 28, 2004

### Pyrrhus

Remember $$F_{f} = \mu N$$

3. Sep 28, 2004

### Leong

10288,
What is the coefficient of friction of the 10 kg block ?

4. Sep 28, 2004

### Gokul43201

Staff Emeritus
Where does it say the system remains at rest ??

Even if it were at rest, it would not require anything more. But clearly, there is no reason to assume the system is at rest, based on the given information.

I see 2 equations, but only 1 unknown : T. And that is a problem, since clearly, the values of F(fr) are known for both blocks. However, there is one more variable that you have to include : the unknown acceleration, 'a'.

5. Sep 29, 2004

### shashank010288

I am extremely sorry.
The coeff of friction for 10 kg block is 0.5. It is wrongly printed as 0.f

& For gokul, whatever u r saying is surely wrong

6. Sep 29, 2004

### Gokul43201

Staff Emeritus
Only because I didn't know what "0.f" meant !

If that number was low enough, it would allow for the blocks to slide, with some acceleration, 'a'.

7. Sep 29, 2004

### arildno

I believe shaskank thinks he's been given the coefficients of maximal static friction, rather than the coefficients of kinetic friction..

8. Sep 29, 2004

### Gokul43201

Staff Emeritus
Yes, that does make it kinda complex. I'll need to sit down and think about it a little more...

9. Sep 30, 2004

### aekanshchumber

I think unless blocks are hanging or accelerated there will no tension in the string you see what i understand from your question is that two blocks are resting on a horizontal table connected with a string and pulley, which is useless, there is lying there. there is no force on the block which will produse tension in the string.
If i am mistaken at any point please tell otherwise this que. could not be solved.
thanks

10. Sep 30, 2004

### Pyrrhus

Read the problem, again. Even if it was at rest, it will be tension, because there's a block hanging freely, being attracted by gravity, so the Friction of both blocks must exert a force to neutralize the tension.

11. Sep 30, 2004

### aekanshchumber

Thanks, i got it.
As the blocks are at rest, let there be tension T in the string, it neutralize the force of friction as it tries to bring them closer but friction resists it. Divide the components of the tension vector horizontal one will be equal to the force of friction.
i think it will do it.

12. Sep 30, 2004

### shashank010288

PLz calculate its value by finding the equations.
I think its unsolvable as it involves very complex mechanism of friction.
Think hard & u will realize that its very very conceptual problem.

13. Sep 30, 2004

### Pyrrhus

Actually shashank read what Gokul43201 and arildno said. At rest or moving it won't be unsolveable, only if what arildno said holds true, it will be complex (i doubt unsolveable, but i will have to try)

14. Oct 2, 2004

### Staff: Mentor

This is an interesting problem since you don't even know the direction of the frictional forces, much less their magnitudes. (All you know up front is the maximum static friction that the blocks can support.)

But if you assume that you wish to find the minimum tension in the middle string, it's easy to solve. (Imagine that you gently release the hanging weights---that will give you the minimum tension.)

First imagine that there is no string connecting the 10kg and 6kg blocks. The static friction of the 10kg block (maximum of 5g in standard units) is more than enough to support the hanging 2kg mass. But the 6kg block (maximum friction of 1.2g) would not be able to support the hanging 4kg mass. The middle string needs to supply a minimum tension of 2.8g. In that case the friction force on the 6kg mass is at maximum (1.2g directed towards the 10kg mass), but the friction force on the 10kg mass is only 0.8g directly away from the 6kg mass.

15. Oct 2, 2004

### arildno

A very neat solution, Doc Al!

Last edited: Oct 2, 2004
16. Oct 10, 2004

### shashank010288

I appreciate DOC AL's work, but that's quite easy to get.
But, my question is,
Why cant the system give a single solution?

17. Oct 11, 2004

### Ganesh

I agree with Shashank.
Once the system is arranged and all the required data is given, only a single solution must exist in the field of classical physics.