How force exerted by spring is always opposite to the direction of displacement in Hooke's law

Suppose a spring lying on a horizontal table, displaced from its equilibrium length by an external agent. The external agent is removed, the spring will head back to its equilibrium length. Here, the direction of spring force and displacement will be same.

But according to Hooke's law,

$$\mathbf{F}=-k\Delta\mathbf{x}$$

The minus sign tells us that the force exerted by spring is always opposite to the direction of displacement.

How is this? Please explain the reason for the minus sign.

Thanks.

• You are not writing the hooke's law correctly, it is supposed to be represented in the vector form as $\vec{F}=-k\vec{x}$. And the Wikipedia article on the hooke's law spells it out for you as to how to interpret the hooke's law. – user350331 Jan 15 '17 at 7:48

Imagine a spring which has a force $\vec F_{\rm sy}$ applied on it by you and this produces an extension $\vec x$.
You then have $F_{\rm sy}=k\vec x$

However it is usual to be interested in the force the spring exerts on you $F_{\rm ys}$.
Using Newton's third law $\vec F_{\rm sy}=-\vec F_{\rm ys }$ so $\vec F_{\rm ys}= - k \vec x$.

Introducing a unit vector in the positive x-direction $\hat i$ and let the magnitude of the forces $F_{\rm ys}$ and $F_{\rm sy}$ be $F$.
$\vec F_{\rm ys}= - k \vec x$ becomes $F \hat i = - kx \hat i \rightarrow F=-kx$ in terms of components in the positive x direction.

• this has not any relation to my question. I have asked a specific question that Hooke's law states displacement and force is always opposite in direction. This is true only when body is moving from mean to extreme position. But when moving from extreme to mean this does not seems true. – Mudassir Malik Jan 15 '17 at 8:43

What you say is correct. But, it's not what Hook's law tells you.

Hook's law says, the force exerted on a mass, permanently or temporarily attached to the spring, is proportional to the difference between the instantaneous length of the spring and the equilibrium length. In the direction which points toward the equilibrium.

When the external agent is removed, the displacement of the spring from it's equilibrium hasn't changed. The force is still pointing to the equilibrium state. And, everything is fine.

Occasioned by your comment @Fracher's answer :'But when moving from extreme to mean this does not seems true', i think that you mixed up the vectors 'displacement' and 'velocity' which can have opposite directions.

So i will try to explain the situation as simple as i can (and it is):

Hooke's law for a spring is often stated under the convention that $F$ is the restoring (reaction) force exerted by the spring on whatever is pulling its free end.

In that case, the equation becomes: $F = − k X$

since the direction of the restoring force is opposite to that of the displacement.' (as stated in wikipedia).

This is illustrated in the following diagram :(from karantonibg.blogspot.gr) The external agent is removed, the spring will head back to its equilibrium length. Here, the direction of spring force and displacement will be same.

No!

$x$ is not direction of change. It is just direction.

• If the spring is stretched to the left, then $x$ points leftwards.

• Force points rightwards, because it tries to go back to original length.

• That force makes the end move back towards the right.

• When the end has moved a bit, the remaining stretching is still towards the left, but it is becoming smaller.

• When it has moved half the way, the remaining stretching is still towards the left, but it is getting smaller towards the right.

• When the end is almost back at original position, the stretching is still a tiny bit towards the left, but almost disappeared.

$x$ shows the direction of stretching; not the direction in which the stretching changes. (That would rather be some kind of "velocity"). And the force will always appear to pull back to original size, so always opposite to the stretching.