Extension in spring with unequal forces acting on its two ends

[@bhishek]

Two masses m1 and m2 are connected by by spring of spring constant k. If two forces F1 and F2 acts on the two masses respectively in mutually opposite direction (i.e. outwards) what would be the extension in the spring and the acceleration of the two masses.

I think that if assuming F1>F2
then the extension in spring should be equal to x= F2/k
and the forces acting on m1 would be F1 in forward and kx(F2) in backward direction.
and on m2 F2 in backward and kx(F2) in forward direction.

Where am I wrong in my assumption?

 

[Doc Al]

I think that if assuming F1>F2
then the extension in spring should be equal to x= F2/k
and the forces acting on m1 would be F1 in forward and kx(F2) in backward direction.
and on m2 F2 in backward and kx(F2) in forward direction.

Where am I wrong in my assumption?

If the spring force equaled F2, what would be the net force on m2? Is that reasonable?

 

[@bhishek]

Thanks

Thank you. I noticed were I was wrong.
In this case the eqn would be
F1-kx = m1*a
kx- F2 = m2*a
Solving we would get
x= (m2*F1+ m1* F2)/(k)(m1+m2)

 

[Doc Al]

Thank you. I noticed were I was wrong.
In this case the eqn would be
F1-kx = m1*a
kx- F2 = m2*a
Solving we would get
x= (m2*F1+ m1* F2)/(k)(m1+m2)

Looks good to me.

Note that you are solving for the special case where there is no relative motion of the masses--the spring extension remains constant.

 

[@bhishek]

Looks good to me.

Note that you are solving for the special case where there is no relative motion of the masses--the spring extension remains constant.

In the same special case with no relative motion if we put m1=m2 and F2=0 (the surface is friction-less, I forgot to mention it earlier) then the extension in the spring comes to F1/(2*k)
Then how could there be extension when there is no friction acting on the masses?

 

[Doc Al]

In the same special case with no relative motion if we put m1=m2 and F2=0 (the surface is friction-less, I forgot to mention it earlier) then the extension in the spring comes to F1/(2*k)

Right. (I has assumed there was no friction since you did not mention any.)

Then how could there be extension when there is no friction acting on the masses?

Why would you need friction?

 

[Chestermiller]

There seems to be something wrong with this problem statement. There appear to be 3 unknowns (the relative displacement x, and the acceleration of each of the two masses a₁ and a₂), but only two equations.

Chet

 

[Doc Al]

There seems to be something wrong with this problem statement. There appear to be 3 unknowns (the relative displacement x, and the acceleration of each of the two masses a₁ and a₂), but only two equations.

We are assuming no relative motion of the masses, so there is a single acceleration. (That assumption should have been made explicit in the problem statement.)

 

[@bhishek]

Right. (I has assumed there was no friction since you did not mention any.)


Why would you need friction?

If the surface is friction less then why should there be any extension in spring. Shouldn't the spring be able to pull the block m2 without any extension?

 

[Doc Al]

If the surface is friction less then why should there be any extension in spring. Shouldn't the spring be able to pull the block m2 without any extension?

A spring cannot pull without extension. (An unstretched spring exerts zero force.)

The force exerted on m1 drags the entire system along. Some of that force is transmitted to m2 via the spring.

 

[@bhishek]

A spring cannot pull without extension. (An unstretched spring exerts zero force.)

The force exerted on m1 drags the entire system along. Some of that force is transmitted to m2 via the spring.

That solves the problem. :D