0
$\begingroup$

enter image description here

I am having trouble understanding the sign/direction of displacement for the equation F=-kx.

I am not going to elaborate on what the specific problem is that goes along with this visual since that's not necessary. Basically the book says that displacement is positive in this problem, but since displacement in the equation $F=-kx$ is referring to the displacement from the unstretched position of the spring to where the spring is now (in the picture, the spring is now at the equilibrium position), shouldn't displacement be negative since displacement in this case is in the negative y-direction. Or is the convention for spring displacement that a stretch means positive displacement and a compression equals negative displacement?

I researched for what the convention of spring displacement is and I can't seem to find unified answers.

$\endgroup$
  • 1
    $\begingroup$ If the book says that the displacement is positive, then it obviously has to be using a convention where the downward direction is positive (and the upward direction is negative). And, no, there is no convention that a stretch has to be positive. You can choose your axes so that a stretch is either positive or negative, and whichever way you choose the negative sign in F=-kx will ensure that the spring force is acting to oppose the stretch or compression, which is consistent with the expected behavior for a spring. $\endgroup$ – Samuel Weir Jul 6 '18 at 20:04
  • $\begingroup$ Possible duplicate of How do I use the signs in the spring force equation $\endgroup$ – sammy gerbil Jul 7 '18 at 9:21
1
$\begingroup$

The force $\vec F$ is the force that the spring exerts on an object external to it (the mass in this case) when there is a displacement $\vec x$ of one end of the spring from its unstretched position.
Let $\hat d$ be the unit vector in the downward direction and you might say that this defines the positive x-direction.

-

$\vec F = -k\vec x \Rightarrow F\hat d = -k x \hat d \Rightarrow F = - kx$ where F is the component of the force in the $\hat d$ direction and $x$ is the component of the displacement in the $\hat d$ direction.

If $x$ is positive (displacement downwards with the spring being stretched) then $F$ is negative ie the force on the mass is upwards.
If $x$ is negative (displacement upwards with the spring being compressed) then $F$ is positive ie the force on the mass is downwards.
So the equation $F=-kx$ correctly predicts the direction of the force for a given displacement of the end of the spring.

Now suppose that you decided that you wanted to use the unit vector $\vec u$ which points upwards with $\hat u = -\hat d$

$\vec F = -k\vec x \Rightarrow F\hat u = -k x \hat u \Rightarrow F = - kx$ where F is the component of the force in the $\hat u$ direction and $x$ is the component of the displacement in the $\hat u$ direction.

If $x$ is positive (displacement upwards with the spring being compressed) then $F$ is negative ie the force on the mass is downwards.
If $x$ is negative (displacement downwards with the spring being stretched) then $F$ is positive ie the force on the mass is upwards.
So again the direction of the force that the spring exerts on the mass is correctly related to the displacement of the end of the spring.

In the example that you have given the author of the book that you are reading has chosen down as the positive direction $(\hat d)$.

—-

Update as OP defined $x$ incorrectly in the original question.

Using down as positive $\hat d$ and $x_0 \hat d$ as the displacement of the end of the spring from its position when unstretched to the equilibrium position.
At the equilibrium position $-kx_0 \hat d + mg \hat d =0 \Rightarrow kx_0 = mg$.

Now move the end of the spring by a displacement $\vec x = x \hat x$ from the equilibrium position.

The net force on the mass is now $-k(x_0+x) \hat d + mg \hat d = -kx \hat d \Rightarrow F_{\rm net} = -kx$ which is the equation you were asking about.

The equation correctly predicts the net force on the mass as being upwards if $x$ is positive ie below the equilibrium position and the net force on the mass as being downwards if $x$ is negative ie above the equilibrium position.

$\endgroup$
  • $\begingroup$ Wait I realized that x is the displacement from the equilibrium position so why is x=0 at the unstretched position shouldn't x=0 at the equilibrium position. $\endgroup$ – 54284User Jul 6 '18 at 22:02
  • $\begingroup$ In your question you stated that the displacement was from the unstretched position. $\endgroup$ – Farcher Jul 6 '18 at 22:13
  • $\begingroup$ I realized I was wrong because x actually means displacement from equilibrium position $\endgroup$ – 54284User Jul 6 '18 at 22:18
  • $\begingroup$ @54284User I have updated my answer. $\endgroup$ – Farcher Jul 6 '18 at 22:45
  • $\begingroup$ +1 The important lesson here is that it is always better to begin with vector equation and deduce the scalar equation from it. $\endgroup$ – Deep Jul 7 '18 at 10:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.