Action Reaction Spring Scale Question

[Ineedhelpwithphysics]

Homework Statement

Two 100-N weights are attached to a spring scale as
shown. Does the scale read 0, 100 N, or 200 N, or does it
give some other reading?

Relevant Equations

No equations just newtons third law

Here is the image the answer is say 100 but why. Why isn't it 0. Is it because of the pulley but even with the pulleys the tensions in both ropes are 100 N.

 

[ergospherical]

The spring scale displays the tension within it! Which is the same as the tension within the rope, which is $100 \ \mathrm{N}$.

 

[Ineedhelpwithphysics]

What i don't get? Why not 0

 

[Ineedhelpwithphysics]

Both ropes have 100

 

[ergospherical]

What i don't get? Why not 0

If you pull on both ends of a spring scale with your hands, you'd expect it to be non-zero, right?

 

[Ineedhelpwithphysics]

Yeah i guess it would no zero but wouldn't it be zero if i pull with equal and opposite ends?

 

[ergospherical]

The overall force on the spring is zero, sure, but the tension in it isn't.

 

[Ineedhelpwithphysics]

okay so why isn't it 200 since boht ropes have a tesion of 100

 

[ergospherical]

You can think of the tension in the spring as the force that each little element of the spring applies to the adjacent little elements of spring.

If you think about the little element of spring right at the end of the spring, then it's experiencing a force of $100 \ \mathrm{N}$ from the rope and a tension force $T$ from the rest of the spring. So, the tension must also be $100 \ \mathrm{N}$.

 

[berkeman]

okay so why isn't it 200 since boht ropes have a tesion of 100

The question is made a bit confusing on purpose, to make you think about the setup and mentally go through different equivalent scenarios to figure it out.

You can see that if the left pulley and weight were replaced by a fixed support, both ropes would have 100N tension and the overall differential force on the scale would be 100N, right? Since the ropes and pulleys and the scale are not moving and accelerating, the situation is no different from one weight on the right side and a fixed support/tie at the left.

Now your next question would be what if I eliminated both weights? Well, if you kept the tension of 100N to a fixed support on the right and had a fixed support on the left, you would still have 100N in both ropes and a total net spring force of 100N separation.

The confusing part of this question (done on purpose by the author/instructor) is to show two 100N weights, but the tension in the non-moving system is still 100N differential overall. Hope that helps.

 

[PeroK]

okay so why isn't it 200 since boht ropes have a tesion of 100

The first thing to reason is that the spring is no different from the string itself. If you replaced the spring with more string you would have a common tension througout.

If the tension in the spring is $200N$, then the tension in the string must also be $200N$.

Now, if the tension in the string is $200N$, then look at the forces on either of the $10kg$ masses. The forces would be unbalanced, and the masses would be accelerating upwards.

So: the force exerted by the spring on the string to the right must be $100N$; and the force exerted by the spring on the string to the left must be $100N$. That's clear.

The way I look at it, that defines what we mean by the tension in the string. It's an elastic force that applies equally in both directions.

You could, I guess, define the tension in the string as twice this. But, then the tension would be twice the force that the spring exerts at either end. And, in all our diagrams and calculations we would have $F = \frac T 2$, where the force exerted by a spring or string with tension $T$ would be $T/2$.

Isn't it just simpler and better to define the tension as equal to the force?