Solving for Spring Stretch with Two Blocks of Mass m/2

[thonwer]

Homework Statement

A block of mass m, is suspended by a spring with a spring constant k and stretches a length x . If two blocks of mass m/2 are suspended by a spring with the same spring constant k (ii), how much will the spring stretch?

http://imagizer.imageshack.us/v2/800x600q90/12/sidr.png

Homework Equations

Fg=mg

Fe=-kx

The Attempt at a Solution

I think each mass of m/2 stretches the spring x/2, so together would stretch the same as the block of mass m alone.

 

[hilbert2]

Suppose that we have case (i) and the spring and mass are at equilibrium, with the spring being stretched a distance x. The mass is pulling the spring with force mg, but doesn't the support the spring is hanging from have to "pull" the spring from opposite end with the opposite force -mg? (otherwise it would not be a static situation)

In case (ii) the spring is pulled from one end with force mg/2 and from the other end with force -mg/2.

 

[thonwer]

But the question is how much does the spring stretch in (ii), not how is (i) a static situation, you suppose (i) as a static situation of course, but it's not what the problem is about.

 

[hilbert2]

I'm just pointing out that in case (i) the spring is stretched from both ends with force mg and in the case (ii) it's stretched from both ends with only half that force...

 

[thonwer]

Are you supposing that the spring has mass? Because in the problem they say nothing about the mass of the spring, so I suppose it with no-mass. Otherwise I don't understand you very well.

 

[tiny-tim]

hi thonwer!

you should consider the tension (the force) on each side of the spring in both cases

as hilbert2 says …

I'm just pointing out that in case (i) the spring is stretched from both ends with force mg and in the case (ii) it's stretched from both ends with only half that force...

and i don't understand what you are saying here about "static" …

But the question is how much does the spring stretch in (ii), not how is (i) a static situation, you suppose (i) as a static situation of course, but it's not what the problem is about.

… they are both in static equilibrium

(and mass has nothing to do with it)

 

[thonwer]

I still don't get it :( I use Newton's Second Law in (i) Fe+T=mg
in (ii) for each block Fe=mg/2 ?

 

[tiny-tim]

(i) Fe+T=mg
in (ii) for each block Fe=mg/2 ?

i don't understand what you're doing here

what is T?

on which body are those the external forces?

 

[thonwer]

T=tension

The external forces are on the blocks.

I'm a little bit lost I think. Which would be the free body diagram?

 

[tiny-tim]

(i) Fe+T=mg
in (ii) for each block Fe=mg/2 ?

T=tension

The external forces are on the blocks.

Fe (i assume that's the spring force) does not act on the blocks, only the tension (and the weight) acts on the blocks

(the spring is not connected to the blocks: the string, with its tension, is between them)

I'm a little bit lost I think. Which would be the free body diagram?

you should do the free body diagram for the spring …

what external forces are acting on the spring?

 

[thonwer]

What I've done is this:

http://imagizer.imageshack.us/v2/800x600q90/202/7fbl.jpg

I don't get anywhere :S

 

[thonwer]

Ok, I got to the point in which in (i) x=mg/k and in (ii) x'=mg/2k so, the answer is, the string stretches in (ii) x/2 ? or do I have to add the two blocks in (ii) and the string stretches x?

 

[PhanthomJay]

Ok, I got to the point in which in (i) x=mg/k and in (ii) x'=mg/2k so, the answer is, the string stretches in (ii) x/2 ? or do I have to add the two blocks in (ii) and the string stretches x?

you've got to the right point and one of those answers is correct, but how you go it there is a mystery becasuse your free body diagram is not only incorrect, it is not even a free body diagram when you show every internal force in one uncut section and with force directions shown without regard to Newton 3.
In part 1, the force in the spring is mg and it elongation is x. In part 2, the force in the spring is ---?--- and per Hookes law its elongation is----?---? I would guess 50% of beginners would guess wrong. Which group are you in?

 

[thonwer]

In part 1, the force in the spring is mg and it elongation is x. In part 2, the force in the spring is ---?--- and per Hookes law its elongation is----?---? I would guess 50% of beginners would guess wrong.

In part 2, the forces in the spring are the tension and the elastic force, in the blocks are mg/2 and the tension. As the tension is the same in the spring and blocks tension=mg/2. In the spring we have tension=elasticforce => mg/2=kx' => x'=mg/2k. This is half the x of part 1.

 

[tiny-tim]

In part 2, the forces in the spring are the tension and the elastic force, in the blocks are mg/2 and the tension. As the tension is the same in the spring and blocks tension=mg/2. In the spring we have tension=elasticforce => mg/2=kx' => x'=mg/2k. This is half the x of part 1.

thonwer, you keep talking about the "elastic force" (which you also write as F_e)

where have you got this expression from?

your free body diagram, or your force equation, must only use external forces

whatever the "elastic force" is, it isn't external, is it?

do the first spring again, and include the tension in the other string, between the spring and the ceiling ​

 

[thonwer]

With elastic force I mean Hook's Law, F=-kx, it's a translation which I see I've made wrong.

 

[tiny-tim]

hi thonwer!

With elastic force I mean Hook's Law, F=-kx, it's a translation which I see I've made wrong.

it isn't the translation that's wrong, it's the way you're using it

the F in Hooke's law is the applied force, the external force …

in this case it is the tension in the string

your F and T are the same thing!

as i said, start again with the first spring … what are the external forces on it? ​

 

[thonwer]

I don't understand, I'm sorry. In (i) the spring suffers from the mg of the block and the tension, the block suffers from mg and kx from the spring. Am I right?

In (ii) the spring suffers from tension of both strings, each block suffers from mg/2 and the tension. Is this right? If so, where do I have to put kx ?

 

[thonwer]

Oh, so the F in Hooke's Law is the net force?

 

[PhanthomJay]

In part 1, the force in the spring is mg and it elongation is x. In part 2, the force in the spring is ---?--- and per Hookes law its elongation is----?---? I would guess 50% of beginners would guess wrong.

In part 2, the forces in the spring are the tension and the elastic force, in the blocks are mg/2 and the tension. As the tension iI] the same in the spring and blocks tension=mg/2. In the spring we have tension=elasticforce => mg/2=kx' => x'=mg/2k. This is half the x of part 1.

Yes, correct. But you need to get a better handle on free body diagrams and the difference between internal and external forces. The only forces external to the system are the weights of the blocks. When you look at one of the hanging blocks, you isolate it in a free body diagram to identify the forces acting on it. Its weight acts down and the tension in the cord acts up, and since the block is in equilbrium , you conclude from Newtons first law that the cord tension is mg/2. Note that the tension and weight act on the block , not in the block. (The force in the block is the internal tension mg/2). Same principles apply for the other block.
Now look the free body diagram of the spring. The 2 forces acting on it are the cord tensions mg/2 from each side. The elastic force in the spring is determined by cutting a section thru the spring, and determining from equilibrium that the internal spring force is mg/2.

 

[tiny-tim]

hi thonwer!

Oh, so the F in Hooke's Law is the net force?

i'm not sure what you mean by that, in this context …

let's return to this when you've solved the actual question

In (i) the spring suffers from the mg of the block and the tension, the block suffers from mg and kx from the spring. Am I right?

(btw, we say the spring "feels" or "experiences" or even "enjoys" the forces on it, but not "suffers"! )

a body only experiences forces from whatever it is directly connected to

the block is not connected to the spring (or vice versa) … the string is between, and is connected to both of them

the block experiences its weight (mg) and the tension in the string (T)

the spring does not experience the block's weight (mg): it is connected to the string, not to the block

In (ii) the spring suffers from tension of both strings, each block suffers from mg/2 and the tension. Is this right?

yes

If so, where do I have to put kx ?

kx is only used to find x …

once you know what the applied (external) force is, you divide it by k to find the extension, x

kx is not a force!, it is a formula that converts a force to a distance (for a spring with spring constant k)

as i said, start again with the first spring … what are the external forces on it?

(assume the spring has negligible mass)​

 

[thonwer]

Ok let's see, in (i)T=mg, so, kx=T=mg ?

In(ii) mg/2=T (from the blocks). In the spring only T; as T=mg/2 , kx=T=mg/2 ?

And if that is right, the spring stretches x/2 or x in (ii)?

 

[tiny-tim]

let's stay on (i)

(never mind kx, we can deal with that later)

what are the external forces on the spring? (ie, what is pulling the spring down, and what is stopping it from falling to the floor?)

 

[thonwer]

mg from the block pulls it down and the tension of the string keeps it attached to ceiling.

 

[tiny-tim]

mg from the block pulls it down and the tension of the string keeps it attached to ceiling.

yes! (or rather, the tension of the lower string is mg, and that tension pulls the string down … the block is only connected to the string, so the weight does not act on the spring)

ok, and (assuming the mass of the spring is negligible) how much is the tension of the string that keeps it attached to ceiling?

 

[thonwer]

That tension is equal to the mg (from the block) that pulls it down.

 

[tiny-tim]

That tension is equal to the mg (from the block) that pulls it down.

yes, if we regard the spring and block (and the string between them) as one body, then since they are in equilibrium, the total external force must be zero

since we know that the force down is mg, that means the force up must be mg also

ok, so you know that the first spring has two forces on it, mg up and mg down

and you know that the second spring has two forces on it, mg/2 left and mg/2 right

sooo … ? ​

 

[thonwer]

kx=mg/2 ? is that?

 

[tiny-tim]

kx=mg/2 ? is that?

i'm not sure what you're saying, or why

the first spring has two forces on it, mg up and mg down

the second spring has two forces on it, mg/2 left and mg/2 right

the first spring stretches a length x

so can you explain, in words, how much the second stsring stretches, and why?​

 

[thonwer]

The second spring stretches to the left, and to the right because of the mg/2 forces.
I think, as the spring is pulled from both sides with half the force than in (i), then the second spring stretches x too. But the solution they gave me was that it stretches x/2, and I don't know which is the correct answer.