# Homework Help: Mechanics problem -- 2 masses & spring on a surface

1. Jun 19, 2014

### Satvik Pandey

1. The problem statement, all variables and given/known data

Find the minimum Force required to move M2.
M1=3kg.M2=5kg,$\mu$1=0.4and$\mu$2=0.6.
2. Relevant equations

Net f =Ma

3. The attempt at a solution
For a displacement 'x' force applied on the block M1 by the spring=kx.
From free body diagrams we can get
F=$\mu$1*M1*g+kx.
When spring applies kx on block 1 it will apply kx to block 2 but in opposite direction.The block 2 will move when kx=$\mu$2*m2*g.Combining Two equations we get f= 41.16.
It is not right,somebody told me that I could not do this question in this way because force exerted by the spring is not constant.But maximum force that spring can exert on block is $\mu$2*m2*g and if F exceeds than that then block 2 should start moving.This thing does not have anything to do with the nature of force.

Last edited: Jun 19, 2014
2. Jun 19, 2014

### Matterwave

Can you elaborate on what you're trying to find in this question? You just gave a figure, but you didn't ask a question.

3. Jun 19, 2014

### Satvik Pandey

I am trying to find the minimum force required to move M2.

4. Jun 19, 2014

### Matterwave

What is k?

5. Jun 19, 2014

### Satvik Pandey

k is the force constant of the spring.

6. Jun 19, 2014

### Matterwave

Were you given a numerical number for this?

Your solution actually looks right to me. The only thing I might consider perhaps different is that the first block must move before the second block does. In this case, the first block might change to kinetic friction and so require a little bit less force.

7. Jun 19, 2014

### Satvik Pandey

The value of k is not given in the question.Somebody told me to do this question using Work energy theorem.

8. Jun 19, 2014

### Tanya Sharma

The question asks about the minimum force required . What can you say about the motion of M1 ?

By the way , value of k is not required .

9. Jun 19, 2014

### Matterwave

It seems tenuous that you would use the work energy theorem in this problem. In the end, you need to specify a force on the second block for it to move, and you will recover the same answer as before.

Maybe I'm not seeing something, but from my analysis, when block 1 is moving towards block 2, there are 3 forces on block 1. The pushing force, the force due to the spring, and the (kinetic) frictional force of block 1. When the force due to the spring equals the force required to move the second block, the second block should move. So your answer in your first post should be right, except one should use the coefficient of kinetic friction for block 1 rather than the coefficient of static friction.

Unfortunately, doing this problem the "work energy way", does not tell you anything about the coefficient of kinetic friction either, as that is a purely dissipative force and there's no energy stored anywhere due to that force.

10. Jun 19, 2014

### sankalpmittal

Make an equation:

Work done by force F + work done by friction = Change in Mechanical Energy of the system

From there you can analyse the minimum value of F.

11. Jun 19, 2014

### Satvik Pandey

I did this question in this way.But the answer is wrong. People at other physics sites say that I could not use this way because the force due to spring is not constant.

12. Jun 19, 2014

### Satvik Pandey

By Free body diagram we came to know that there are three forces acting on the block m1.
Block 2 will not move untill the kx =M2 $\mu$2 g.In order to set the block 2 in motion the applied F should overcome frictional force on M1 and force due to spring.The force due to spring
should be equal to M2 $\mu$2 g in order to move block 2.
From free body diagrams we can get
F=μ1*M1*g+kx.(as we only wanted to find minimum F)
By combining the two equation I get wrong answer.

13. Jun 19, 2014

### Tanya Sharma

Ok.

Here lies the problem . F will surely move the block ,but it is not the minimum force .

Just focus on block 1 . Think what should be the minimum velocity of block 1 when block 2 just starts to move .

14. Jun 19, 2014

### dauto

This problem cannot be solved without some information about the initial condition. Doe the motion start from rest? What's the initial position of the block?

15. Jun 19, 2014

### sankalpmittal

Let me be clear. You seem to be finding the equilibrium condition and assuming that force F for this condition is minimum. NO. It's just the equilibrium. Maximum or minimum doesn't apply!

So what you do is set up a constraint as :

Work done by force F + work done by friction = Change in Mechanical Energy of the system

From there you analyse force F.

Also what will be the minimum velocity of bock 1 ? Will that be equal to velocity of center of mass ?

Edit: Force due to spring is of course not constant. It's -kx and is dependent on variable x. Set x such that F is minimum.

Last edited: Jun 19, 2014
16. Jun 20, 2014

### Satvik Pandey

Why this force is not the minimum force?When we apply force(F), (this F must be greater than the frictional force) it moves the block.As the block moves force on it due to spring increases.But at a certain point force exerted by the spring could not increase (certain point means when kx=$\mu$2 m2 g),at this point the block should just start to move.So Min.F should be equal to sum of the frictional forces on block1 and block 2.

17. Jun 20, 2014

### Satvik Pandey

I am wondering why the method that I have used above is wrong.Why the value of F,that I got is wrong???????????

Last edited: Jun 20, 2014
18. Jun 20, 2014

### Tanya Sharma

First you need to answer a simple question and then you will start to understand where the problem lies .

What can be the minimum velocity of block 1 when block 2 just starts to move ?

Last edited: Jun 20, 2014
19. Jun 20, 2014

### Satvik Pandey

In reply to sankalpmittal and Tanya Sharma -
Work done on block=Change in KE.
When 'F' force is applied on the block 1 it is displaced from the initial position.Let 'x' be the displacement at which M2 just moves.
So, Fx-$\mu$1 M1 g x - kx^2/2=M1 v^2/2.
or x(F - $\mu$1 M1 g - $\mu$2 M2 g)=M1 v^2/2.Transposing M1 and 2 to the other side of equation we can easily get v.

20. Jun 20, 2014

### Tanya Sharma

Can we say from the above equation that the higher the value of the force applied the higher the value of 'v' ?

21. Jun 20, 2014

### Satvik Pandey

When we apply F force on block on M1 it willbe displaced from its original position.Let 'x' be the displacement at which block will begin to move.
By using work energy theorem-
Fx-$\mu$1 M1 gx - Kx^2/2=M1v^2/2.
or x(F-$\mu$1 M1 g-$\mu$2 M2 g/2)=M1v^2/2.By transposing M1 and 2 to the other side of equation we can get v but here the value of x is not known.

Last edited: Jun 20, 2014
22. Jun 20, 2014

### Satvik Pandey

Yes,because F is directly proportionate to v.

23. Jun 20, 2014

### Tanya Sharma

And what would be the inverse of the statement I have written.

24. Jun 20, 2014

### Satvik Pandey

If we decrease F then v will be decreased accordingly.

25. Jun 20, 2014

### Tanya Sharma

Good .

Now we need minimum force ,so the velocity 'v' should be minimum .What can be the minimum velocity of block 1? i.e how much small the velocity of block 1 can be ?