The ratio of compression of springs from three different setups

[Kaushik]

Homework Statement

Three springs are given. Their spring constants are k, 2k and 3k respectively. Their compressions are $x_1 , x_2 , x_3$ respectively. (Find the figure attached)

Relevant Equations

attached below.

I got the compression for the first one.
Second one, i am a bit confused. But i got $x_2 = \frac{F}{2k}$
In the third one, what will the compression be?

I have to get the ratio $x_2 : x_3$ . Then i will be able to get my answer.

Thanks.

 

[kuruman]

The figure is not very informative for spring 2. Is the mass at the other end oscillating or is it kept from moving? Your answer assumes that the mass is not moving.

To answer your question about 3. Look at 1. Is the force shown in the picture the only force acting on the spring? Doesn't the wall on the left also exert a force on the spring? How large is that force? Other than the spring constant, what's the difference between picture 1 and picture 2?

 

[Kaushik]

The figure is not very informative for spring 2. Is the mass at the other end oscillating or is it kept from moving? Your answer assumes that the mass is not moving.

To answer your question about 3. Look at 1. Is the force shown in the picture the only force acting on the spring? Doesn't the wall on the left also exert a force on the spring? How large is that force? Other than the spring constant, what's the difference between picture 1 and picture 2?

There are no other information given about the spring 2.
The answer given is $6:3:2$ .
What do you think? ( i don't know if i can ask this question to you. ) Is the mass free to move or not?

So in the third one i can't consider $x_3 = \frac{2F}{3k}$ as the force exerted by the spring is on opposite direction? But what i though was ( while writing that equation ) , the compression however would remain the same. Is my intuition wrong?

In the first one, the wall does exert force on the spring. But that force cancels out the force with which the spring tries to pull the wall.

Waiting for you reply.

Thanks.

 

[kuruman]

The answer given is 6:3:2.

I am not sure how to interpret this. If it's a ratio can you simplify it so that it looks like $m : n$ where $m$ and $n$ are integers?

The expressions for $x_1$ and $x_2$ in the figure are correct. The expression for $x_3$ is incorrect. If a spring is compressed by $x$, it exerts a force of magnitude $kx$ at each one of its ends in a direction away from it. The forces shown in the drawing for spring 3 are needed to keep the spring compressed and at rest. I suspect the same applies to the drawings for the other two springs except that all the external forces needed to keep compressed and at rest are not shown.

 

[Kaushik]

I am not sure how to interpret this. If it's a ratio can you simplify it so that it looks like $m : n$ where $m$ and $n$ are integers?

The expressions for $x_1$ and $x_2$ in the figure are correct. The expression for $x_3$ is incorrect. If a spring is compressed by $x$, it exerts a force of magnitude $kx$ at each one of its ends in a direction away from it. The forces shown in the drawing for spring 3 are needed to keep the spring compressed and at rest. I suspect the same applies to the drawings for the other two springs except that all the external forces needed to keep compressed and at rest are not shown.

So what will the compression be in spring 3. I tried, but could not arrive at any result.

Won't the spring 3 be compressed by $\frac{F}{3k}$ from the right and $\frac{F}{3k}$ from the left?

 

[kuruman]

There is no compression from the left and the right, there is only compression from the relaxed position. If the spring has relaxed length $l$ and it is compressed by amount $x$ so that its new length is $l-x$, then the force that the spring exerts at each of its ends has magnitude $kx$. Now if you know that the spring is at fixed compression and is not accelerating, a force of what magnitude $F$ needs to be applied at each end for that to be the case? Answer: $F=kx$.

 

[Kaushik]

There is no compression from the left and the right, there is only compression from the relaxed position. If the spring has relaxed length $l$ and it is compressed by amount $x$ so that its new length is $l-x$, then the force that the spring exerts at each of its ends has magnitude $kx$. Now if you know that the spring is at fixed compression and is not accelerating, a force of what magnitude $F$ needs to be applied at each end for that to be the case? Answer: $F=kx$.

So in the second case, is the $Force = F_{normal}$ from the mass to prevent it from acceleration?
Or are there some external force that prevents it from accelerating?

 

[kuruman]

One more time, whenever you have a spring that is compressed from relaxed length $l$ to new length $l-x$ the spring exerts a reaction force of magnitude $kx$ at each end against whatever it is that compresses it. For example, if you compress the spring between your two fingers, each finger will exert force $kx$ on the spring to reduce its length and each of your fingers will experience force $kx$ pushing it away from the spring. The external forces on the spring are those exerted on it by your fingers. Because they are equal in magnitude and opposite in direction (both fingers are pushing in) the net force on the spring is zero and the spring does not accelerate. This picture is shown in drawing 3 of your picture. The forces labeled "F" are the external forces.

The other two drawings show only one external force F pushing to the left. Does this mean that it is the only force acting on the spring or are we to assume that the spring is not accelerating because there is another external force F at the left end pushing to the right? Well, drawing 1 has a vertical line on the left side of the spring which could be interpreted to be a wall that cannot move. Drawing 2 shows just a mass with nothing to keep it from moving and that is why I asked about it originally. To repeat myself, I suspect this is a simple problem in which all three springs are not accelerating but only drawing 3 is a complete force diagram showing all the forces acting on the spring.

 

[Kaushik]

To repeat myself, I suspect this is a simple problem in which all three springs are not accelerating but only drawing 3 is a complete force diagram showing all the forces acting on the spring.

Thanks for the detailed explanation.

According the answer given, we can assume that there are other external forces that are acting on it which keeps it from accelerating.

, if you compress the spring between your two fingers, each finger will exert force kxkxkx on the spring to reduce its length and each of your fingers will experience force kxkxkx pushing it away from the spring.

So the compression is only $\frac{F}{3k}$ that is caused due to the one of the force $F$ that acts on the spring from either side. The other force is just to keep the system at equilibrium. Am I right?

 

[kuruman]

So the compression is only $\frac{F}{3k}$ that is caused due to the one of the force $F$ that acts on the spring from either side. The other force is just to keep the system at equilibrium. Am I right?

Yes, $x_3=\dfrac{F}{3k}$ and no, you cannot say that one force compresses the spring and the other one keeps it from accelerating. Say you believe that the force on your right compresses the spring and the force on your left keeps it from accelerating. If you looked at the compressed spring from behind the screen, the role of each force will be changed as your left and right are reversed. Does that make sense? Both forces together compress the spring and both forces together keep it from accelerating. You cannot separate their roles.

 

[Kaushik]

Yes, $x_3=\dfrac{F}{3k}$ and no, you cannot say that one force compresses the spring and the other one keeps it from accelerating. Say you believe that the force on your right compresses the spring and the force on your left keeps it from accelerating. If you looked at the compressed spring from behind the screen, the role of each force will be change as your left and right are reversed. Does that make sense? Both forces together compress the spring and both forces together keep it from accelerating. You cannot separate their roles.

So both compress the spring by $\frac{x_3}{2}$ each?

 

[jbriggs444]

So both compress the spring by $\frac{x_3}{2}$ each?

"You cannot separate their roles" means that you cannot separate their roles. To ask about how much compression results from one force alone is as sensible as asking about the sound of one hand clapping.

 

[kuruman]

So both compress the spring by $\frac{x_3}{2}$ each?

No, if the spring is compressed by $x_3$ both forces $F$ together compress the spring by $x_3$. If you wanted the spring to be compressed by $x_3/2$, then both forces would have to be $F/2$. Stop trying to consider the effect of only one force on the spring, it serves no useful purpose. If force $F$ on the right is there but the other force $F$ on the left is not there, the spring will not be compressed by $x_3/2$; it will accelerate according to $F=ma$. Just remember that if a spring's length changes from its unstretched state by $x$ then a force $kx$ appears at each end of the spring. This force is away from the spring if it is compressed and towards the spring if it is stretched.

 

[Kaushik]

the sound of one hand clapping.

Nice Analogy. Thanks a lot.

 

[kuruman]

To ask about how much compression results from one force alone is as sensible as asking about the sound of one hand clapping.

When people mention to me the sound of one hand clapping, I demonstrate what it sounds like by slapping my cheek with one hand. This is analogous to drawing 2 in which only one force $F$ is acting on the spring and the mass accelerates with constant acceleration. If we know the acceleration, we can find the compression.

To @Kaushik : The solution of this might clarify things for you. If the spring is compressed by $x$, it must exert force $kx$ on the mass which is free to move under the influence of this force. By Newton's 2nd law,
$$F_{net}=ma = kx~\rightarrow~x=\frac{ma}{k}$$

 

[jbriggs444]

When people mention to me the sound of one hand clapping, I demonstrate what it sounds like by slapping my cheek with one hand.

Yes, there is some hyperbole in the epithet.