Restoring Force vs Applied Force for Hooke's Law

According to Hooke's Law, $$F=-kx$$ where $$F$$ is the restoring force, $$k$$ is the spring constant and $$x$$ is the length of extension/compression.

When an applied force compresses a spring, a restoring force will act in the opposite direction.
When a spring is compressed and is in equilibrium (not extending/compressing), the restoring force should be equal to the applied force since the resultant force is 0 and the acceleration is 0. (Thus the spring is not extending/compressing)

However, the act of compressing a spring will require the applied force to be greater than the restoring force even at an infinitesimally small time interval, am I wrong?

Is it more correct to say that $$F=kx$$ where $$F$$ is the applied force or $$F=-kx$$ where $$F$$ is the restoring force? I think that $$F=-kx$$ the better option, but I'm pretty confused here. Does it matter at all?

• Compress the spring such that the end of the spring moves at a constant speed, and the two forces will at all times be equal to each other (except for right at the beginning and right at the end, but these times can be made arbitrarily small, theoretically). – march Dec 17 '18 at 16:50
• If the applied force was not equal to the springs restoring force at any time there would be acceleration. The spring can be compressed and have no net force on it – Triatticus Dec 17 '18 at 16:51
• If the spring is compressed at an accelerating rate, the applied force and restoring force are not equal. Would x be proportional to the applied force or the restoring force? – helpme Dec 17 '18 at 18:19
• Newton's 3rd law requires that the force and opposing force be equal at all times. – David White Dec 17 '18 at 20:35
• The 3rd law partner forces act on different bodies – helpme Dec 17 '18 at 20:39

Would x be proportional to the applied force or the restoring force?

The equation $$F=-kx$$ is for the spring (restoring) force. If we apply a force equal to this ($$F_{app}=kx)$$ then the spring edge where the force is applied will not accelerate (it will be stationary or moving at a constant velocity depending on what the velocity is when we set our force to start being equal to $$kx$$). Note that this is a very specific case

You are right, in order to stretch/compress a spring at rest we need to initially apply a force different than $$kx$$, but this is true for other scenarios too. For example, to lift a box we need to first apply a force larger than its weight to start it's velocity upwards. Then we can reduce our force to be equal to the weight to lift the box at a constant speed, or we can keep our force constant and the box will accelerate upwards. The same analogy can be made for the spring at each position $$x$$, since the force depends on $$x$$.

So overall, the spring (restoring) force is always given by $$F=-kx$$ where $$x$$ is the displacement from equilibrium. The applied force is really anything you want it to be, and depending on what it is the spring will respond accordingly based on the initial conditions. For example, we can hang a mass from a spring so that the mass remains stationary, or we can stretch and release that same mass so that oscillations occur. Notice how in either scenario the applied force is a constant $$mg$$, but the spring behavior is different due to the initial conditions of the system

I think that for your question, the sign of the force is not something that matters. What really matters is that an unbalance between the applied force and the elastic restoring force is effectively needed in order for the applied force to accelerate the spring.

If you think of this problem as static then the it takes infinite time for the applied force to compress the spring. This is the only case in which you can consider the applied and the restoring elastic force equal.

When we say $$F=kx$$, we are essentially saying that based on $$kx$$ of the spring, we can determine the applied force. The same goes for an accelerating object. The $$ma$$ of the object will be equivalent to the applied force.

In response, the spring exerts a restoring force back to its equilibrium position whether it is stretched or compressed. Since we are taking the direction of applied force to be positive, the direction of the restoring force will thus be negative, given by $$F=-kx$$.