Spring force negative?

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Main Question or Discussion Point

Hello,
Following Hooke's law, the force applied by a string on an object attached to one of its ends is F = -kx
But here is my question : if we consider the equilibrium coordinate x=0 of a horizontal string, and the string is stretched until its end reaches a coordinate x1>0. By applying hooke's law we find F = -k*x1 but since x1>0 so that means F<0, but as I know the magnitude of a force cannot be negative...so am I missing something or should I use |F| for magnitude or what ?

Thank you

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berkeman
Mentor
Hello,
Following Hooke's law, the force applied by a string on an object attached to one of its ends is F = -kx
But here is my question : if we consider the equilibrium coordinate x=0 of a horizontal string, and the string is stretched until its end reaches a coordinate x1>0. By applying hooke's law we find F = -k*x1 but since x1>0 so that means F<0, but as I know the magnitude of a force cannot be negative...so am I missing something or should I use |F| for magnitude or what ?

Thank you
Welcome to the PF.

Of course it can. It all depends on coordinates. Have you learned about vectors yet? (Vectors have a magnitude and a direction).

robphy
Homework Helper
Gold Member
$\vec F=-k(\vec x-\vec x_{eqbm})$

Welcome to the PF.

Of course it can. It all depends on coordinates. Have you learned about vectors yet? (Vectors have a magnitude and a direction).

In fact, yes I have learned about vectors but I personally have never seen a negative magnitude, and since as I have studied the magnitude of a vector is ||F|| = sqrt(Fx^2 + Fy^2 ....) so doesn't this basically mean the magnitude of F is |Fx| ? ( Fx = -kx)

berkeman
Mentor
Just think about a mass initially at x=0 with a spring that goes off to the left. It is on a frictionless horizontal surface.

You pull the mass out to the right at x=1cm and let it go. What happens next?

Just think about a mass initially at x=0 with a spring that goes off to the left. It is on a frictionless horizontal surface.

You pull the mass out to the right at x=1cm and let it go. What happens next?
It just goes back to x=0 (and negative values depending on k) under the effect of a force directed from right to left.

berkeman
Mentor
It just goes back to x=0 (and negative values depending on k)
If it rests on a frictionless surface, it oscillates back and forth to +/- 1cm. What are the forces on the mass as it oscillates? Include the magnitude of the force in the x direction...

Dale
Mentor
It just goes back to x=0
So what direction must the force point for that to happen?

So what direction must the force point for that to happen?
It goes from the right to the left, and I get that spring force has an opposite direction of the stretching/compression.
What worries me is getting a force with a negative value, which is something I have never seen, as I always hear that the magnitude of a force is always positive.

Dale
Mentor
It goes from the right to the left, and I get that spring force has an opposite direction of the stretching/compression.
That is what the negative sign means. The negative of a vector simply points in the opposite direction.

What worries me is getting a force with a negative value, which is something I have never seen, as I always hear that the magnitude of a force is always positive.
That is true. The magnitude of x is the same as the magnitude of -x and both are positive.

That is what the negative sign means. The negative of a vector simply points in the opposite direction.

That is true. The magnitude of x is the same as the magnitude of -x and both are positive.
Then how would I explain a stretch of x=1m to result in a positive force magnitude ?
Let's take K=2N/m
F = -2*1=-2N
It gives a negative value
Whilst with x=-1 F=2N
So am I supposed to use |F| to write magnitude or what ?

Dale
Mentor
Then how would I explain a stretch of x=1m to result in a positive force magnitude ?
Let's take K=2N/m
F = -2*1=-2N
It gives a negative value
Whilst with x=-1 F=2N
So am I supposed to use |F| to write magnitude or what ?
Remember that force is a vector. So a force of -2 N is a force of magnitude 2 N pointing in the -x direction. Properly it should be written as F = (-2,0,0) N to make it clear that it is a vector, but in this case the 0 components are supposed to be understood from the context.

And, yes, if you want to show the magnitude of a force you can write |F|. In this case that would be |(-2,0,0)| = 2.

Now I got it thank you very much for your support.

Dale
Mentor
You are welcome!