Hooke's Law: Spring Force Negative?

In summary: I always recommend writing out the vector to make it more clear.And, yes, if you want to show the magnitude of a force you can write |F|....although I always recommend writing out the vector to make it more clear.
  • #1
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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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  • #2
Mouadys said:
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). :smile:
 
  • #3
##\vec F=-k(\vec x-\vec x_{eqbm})##
 
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  • #4
berkeman said:
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). :smile:
Thank you for answering.

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)
 
  • #5
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?
 
  • #6
berkeman said:
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.
 
  • #7
Mouadys said:
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... :smile:
 
  • #8
Mouadys said:
It just goes back to x=0
So what direction must the force point for that to happen?
 
  • #9
Dale said:
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.
 
  • #10
Mouadys said:
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.

Mouadys said:
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.
 
  • #11
Dale said:
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 ?
 
  • #12
Mouadys said:
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.
 
  • #13
Now I got it thank you very much for your support.
 
  • #14
You are welcome!
 

What is Hooke's Law?

Hooke's Law is a principle in physics that describes the relationship between the force applied to a spring and the resulting displacement of the spring. It states that the force applied to a spring is directly proportional to the amount of stretch or compression of the spring.

Who discovered Hooke's Law?

Hooke's Law was first discovered by English scientist Robert Hooke in the 17th century. He observed that the stretch of a spring was proportional to the force applied to it, and he published his findings in his book "De Potentia Restitutiva" in 1678.

What is the equation for Hooke's Law?

The equation for Hooke's Law is F = -kx, where F is the force applied to the spring, k is the spring constant, and x is the displacement of the spring from its equilibrium position. This equation is known as the spring force negative equation because the force is always in the opposite direction of the displacement.

What is the significance of the negative sign in Hooke's Law?

The negative sign in Hooke's Law indicates that the force applied to the spring is in the opposite direction of the displacement. This means that when the spring is stretched, the force is pulling it back towards its equilibrium position, and when the spring is compressed, the force is pushing it back towards its equilibrium position.

What are some real-world applications of Hooke's Law?

Hooke's Law is applicable to many real-world situations, such as calculating the force needed to stretch a spring in a car's suspension system, determining the force required to compress a diving board, or measuring the elasticity of materials in engineering and construction. It is also used in the design of various mechanical devices, such as shock absorbers and springs in mattresses and furniture.

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