Clarifying Hooke's Law: Understanding the Formula and When to Use the Negative

[thegreengineer]

Well, I know that Hooke's law establishes that the force applied on a spring is proportional to the displacement. However, I've got a little bit confused about the formula. My textbook manages the formula as the following:

F=-kx

Whereas some websites manage it as this:

F=kx

I still don't know when to use the negative or what it means, help.

 

[ehild]

Well, I know that Hooke's law establishes that the force applied on a spring is proportional to the displacement. However, I've got a little bit confused about the formula. My textbook manages the formula as the following:

F=-kx

Whereas some websites manage it as this:

F=kx

I still don't know when to use the negative or what it means, help.

Check what is the force the formula refers to. If you want to stretch a spring, so as its end displaces by x, you need to exert F(you)= kx force. At the same time, the spring exerts F(spring) = -kx force on your hand.

 

[thegreengineer]

So, the sign has nothing to do with the direction of the displacement? I mean, it doesn't matter whether I'm stretching or contracting a spring?

 

[A.T.]

So, the sign has nothing to do with the direction of the displacement? I mean, it doesn't matter whether I'm stretching or contracting a spring?

The spring force is always opposite to the displacement, so they must have opposite signs. Which one is positive and which negative depends on arbitrary choice of confidantes.

 

[thegreengineer]

So let me understand... you are saying that if both forces are the same however one of them is positive (without considering the minus sign; this is when we speak about the force exerted ON the spring) and the other is negative (considering the minus sign; this is when we speak of the force exerted BY the spring)?

 

[A.T.]

you are saying that if both forces are the same however one of them is positive

I wasn't talking about two forces, just about the force BY the spring vs. the displacement. But yes, the force BY the spring, and the force ON the spring are equal but opposite, according to Newton's 3rd Law.

 

[MrSnoopy]

It is always easier when dealing with most problems complicated or simple ones
is to first think on the basic physics laws.
The Newton's 3rd Law for closed system (i.e. no exchange any matter with its
surroundings and no external forces is acting on the system) states that:
$$\mathrm{d}\mathbf{\mathit F} + \mathrm{d}\mathbf{\mathit F}_{\mathrm{deform}} = 0.$$
For a string, small displacements : force is approximately proportional with displacement (Hooks Law), or $\mathrm{d} \mathbf{\mathit F}_{\mathrm{deform}} = k \mathrm{d}\mathbf{\mathit x}$, where $k$ is known as springs stiffness (physical characteristics of a spring). The acting force in longitudinal string displacement has opposite direction of the force with which string is reacting and thus
$$\mathrm{d}\mathbf{\mathit F} = - k \mathrm{d}\mathbf{\mathit x}.$$

What happens when spring is put into vertical position (z-axis). Now gravitational force is forcing string to compress under its own weight. In this case, equation takes form:
<br /> \sum_i \mathbf{\mathit F}_i = \mathbf{\mathit F}_{\mathrm g}(z) = <br /> ES \frac{\mathrm{d} x}{\mathrm{d} z} = kL \frac{\mathrm{d} x}{\mathrm{d} z}.<br />
where $E$ and $S$ is elasticity module and cross-sectional area, respectively. $L$ represents springs length in undeformed state, $\mathrm{d}x$ spring displacement, $\mathrm{d}z$ infinitesimal change of springs length. In the equation where we sum over all forces. The only force which acts on a spring is gravitational force: $\mathbf{\mathit F}_{\mathrm g}(z) = m\frac{(L-z)}{L} \mathbf{\mathit g}$.

Similar can be done if a spring is mounted on both ends (except additional force must be taken into account).

 

[DocZaius]

I don't know why some websites wrote F = kx. It should be F= -kx since this is a restorative force and k is assumed to always be positive.

Imagine you have a horizontal spring with one end attached to a wall and one end to a mass. Ignore gravity for this one (maybe you can assume the whole setup is resting on a frictionless surface). You denote the position of the mass when the spring is uncompressed or unstretched as x = 0 (I am omitting units here for simplification). What the negative sign in F = -kx tells you is that the force of the spring acting on the mass will be in the opposite direction of the displacement of the mass. So let's say k = 1 and you grab the mass and move it to x = 3, (and thereby stretch the spring), then the force of the spring on the mass will be F = -3. That negative sign tells you that, on your coordinate system, the force points in the negative direction, while your mass is located at a positive position.

Likewise if you had moved your mass to x = -3 (thereby compressing the spring), then the force of the spring on the mass would be 3 (the positive sign indicating the force points in a positive direction.) This is called a restorative force because the force always points in a direction that would restore the mass to its position when the spring is uncompressed or unstretched.

I see there was some confusion about which forces have which sign in this thread's discussion above. Every mention of force that I have made so far was regarding the force of the spring acting on the mass. All the consequences of Newton's 3rd Law still apply if you wish to consider them now.

 

[A.T.]

I don't know why some websites wrote F = kx. It should be F= -kx since this is a restorative force and k is assumed to always be positive.

They either just consider the force magnitude, or they mean the applied not the restorative force.