Mechanics problem -- 2 masses & spring on a surface

[Satvik Pandey]

Homework Statement

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

Homework Equations

Net f =Ma

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.

 

[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.

 

[Satvik Pandey]

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.

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

 

[Matterwave]

What is k?

 

[Satvik Pandey]

What is k?

k is the force constant of the spring.

 

[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.

 

[Satvik Pandey]

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.

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

 

[Tanya Sharma]

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.

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 .

 

[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.

 

[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.

Is it a slow loading or a sudden push ?

 

[Satvik Pandey]

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.

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.

 

[Satvik Pandey]

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 .

By Free body diagram we came to know that there are three forces acting on the block m1.
Block 2 will not move until 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.

 

[Tanya Sharma]

By Free body diagram we came to know that there are three forces acting on the block m1.
Block 2 will not move until 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.

Ok.

From free body diagrams we can get
F=μ1*M1*g+kx.(as we only wanted to find minimum F)

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 .

 

[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?

 

[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.

 

[Satvik Pandey]

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 .

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.

 

[Satvik Pandey]

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.

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

 

[Tanya Sharma]

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.

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 ?

 

[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.

 

[Tanya Sharma]

So, Fx-$\mu$1 M1 g x - kx^2/2=M1 v^2/2.

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

 

[Satvik Pandey]

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 ?

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.

 

[Satvik Pandey]

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

Yes,because F is directly proportionate to v.

 

[Tanya Sharma]

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

 

[Satvik Pandey]

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

If we decrease F then v will be decreased accordingly.

 

[Tanya Sharma]

If we decrease F then v will be decreased accordingly.

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 ?

 

[Satvik Pandey]

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 ?

The minimum velocity of block 1 can be 0.
x(F -$\mu$1M1g-$\mu$2M2g/2)=0
or F =g($\mu$1M1+$\mu$2M2/2).
YES I got the answer.But I still have confusion that why my first approach to the solution(using Newton's Law) was wrong.

 

[Tanya Sharma]

The minimum velocity of block 1 can be 0.
x(F -$\mu$1M1g-$\mu$2M2g/2)=0
or F =g($\mu$1M1+$\mu$2M2/2).
YES I got the answer.

Very good .

But I still have confusion that why my first approach to the solution(using Newton's Law) was wrong.

Because that force will be more than the minimum force you have calculated .In other words the block 1 will be moving when block 2 just starts to move .

But we wanted a minimum amount of force .This condition was captured using the Work Energy theorem by putting the velocity of block 1 zero .

 

[sankalpmittal]

Good at least you followed my advice. I told you work energy constraint was the key. xD

You said : But I still have confusion that why my first approach to the solution(using Newton's Law) was wrong.

I again repeat : Equilibrium does not mean that force is minimum/maximum. So that concept doesn't apply. I don't know what was that value you calculated, it wasn't equal means that condition for v to be zero wouldn't have met.

 

[Satvik Pandey]

Very good .
Because that force will be more than the minimum force you have calculated .In other words the block 1 will be moving when block 2 just starts to move .

But we wanted a minimum amount of force .This condition was captured using the Work Energy theorem by putting the velocity of block 1 zero .

Thank you very much for clearing my doubts.

 

[Satvik Pandey]

Good at least you followed my advice. I told you work energy constraint was the key. xD

You said : But I still have confusion that why my first approach to the solution(using Newton's Law) was wrong.

I again repeat : Equilibrium does not mean that force is minimum/maximum. So that concept doesn't apply. I don't know what was that value you calculated, it wasn't equal means that condition for v to be zero wouldn't have met.

Thank you.

 

[Tanya Sharma]

Thanks you very much for clearing my doubts.

I am glad you were able to solve the problem

 

[ehild]

The minimum velocity of block 1 can be 0.
x(F -$\mu$1M1g-$\mu$2M2g/2)=0
or F =g($\mu$1M1+$\mu$2M2/2).
YES I got the answer.But I still have confusion that why my first approach to the solution(using Newton's Law) was wrong.

Your approach assumed that both masses are moving. Only in that case are the forces of friction μmg. If one of the mass is in rest, friction is static. Can be less than μmg.

The masses in the problem can move in a peculiar way. Initially both are in rest, so the force F has to overcome μ₁m₁g and the elastic force of the spring. But the other mass is in rest, till the elastic force overcomes the frictional force acting to it. At the same time, the first mass can get into rest, and the elastic force alone drives the second mass forward. So the average applied force can be less than μ₁m₁g+μ₂m₂g.

ehild