Understanding Signs of Net Force in Mass-Spring Systems

[chattymatty]

Hello!

I am having trouble with a simple concept...trying to figure out the signs of net force of a mass on a spring.

I understand that for a mass on the spring, there is a downward force of gravity (mg) and upwards restoring force of the (-kx). How do we equate these, if we regard the positive direction in the same direction as gravity? How would we equate these, if we, in contrast, regard the positive direction in the same direction as the restoring force?

I believe that the final conclusion will be mg = kx. But where are the signs? How did the signs cancel out?

Thank you!
Chatty Matty

 

[Doc Al]

At the equilibrium position, the net force on the mass is zero.

The thing to realize is that regardless of sign convention the spring force acts up (magnitude kx, where x is the distance below the unstretched position) and gravity acts down. What might be throwing you off is the minus sign in Hooke's law: F = -kx. The minus sign just tells you the direction of the restoring force with respect to displacement x. (If x is down, F is up.)

Using up as positive, the spring force is +kx and gravity is -mg. Add them up to get: kx - mg = 0.

Using down as positive, the spring force is -kx and gravity is +mg. Add them: -kx + mg = 0.

 

[chattymatty]

Hi Doc Al,

When considering the forces, how come we did not consider the minus sign in Hooke's law? So when we consider the spring force (whether we take up or down being positive), is the spring force just "kx"? Depending on the direction we take as positive/negative will dictate then the a +kx or -kx, and not from Hooke's law?

Thanks!

 

[Doc Al]

Hooke's law tells you the actual direction of the force. (Not by blinding plugging into the equation, but by understanding it.) The force is always opposite to the direction of the stretch (or compression) with respect to the unstretched position. If you pull the spring in one direction, the restoring force points in the opposite direction.

The magnitude of that force is always just kx (where x is taken as positive); the sign that you use depends on the coordinate system and sign convention that you are using.