# Difficulty with the intuition of being able to raise an object by applying a force mg

If the force applied on ab object due to gravity is -mg why is applying a force of mg give upward raise the object if the two forces are equal and opposite in direction. I would think the object would remain still until you apply a force greater than mg.

the object would remain still until you apply a force greater than mg.

Unless i missed something, this is correct.

If you apply upward force mg to cancel the downward weight mg, net force on the object is zero. What does Newton's first law say happens when the net force is zero? An object either remains at rest or it moves with a constant velocity. You forgot the part about moving with a constant velocity.

I didn't watch the video, but move an object up against gravity at constant velocity (an elevator, for instance), you only need to apply a force equal to its weight. However, to GET IT MOVING in the first place, you do in fact need to apply a force greater than its weight for at least a short time.

So teddyayalew you are correct. Just that in simplified examples they might omit this fact, or they probably describe the situation in a steady-state after it has already began moving.

sophiecentaur
Gold Member
2020 Award

Take two equal masses on a (massless) string which goes over a pulley. They are balanced. Ignoring friction, if you give them a small nudge, then one will keep going up and the other will keep going down. The tension in the string on each side will still be equal to the weight of a mass.
At the same time, the rising mass will have increasing gravitational potential energy and the falling one will have decreasing GPE. One is 'doing work' on the other.

However, to GET IT MOVING in the first place, you do in fact need to apply a force greater than its weight for at least a short time.

That's right, and when the object gets to the altitude where you want it, you will probably stop the object by applying a force that is less than its weight for a short time. These two little-bit-more and little-bit-less terms add up to zero, so that the total work that you did against gravity is mgh, the same as if the acts of starting and stopping were simply neglected.

You can also imagine it as a limit problem, allowing a very long time to lift the object, so that the starting and stopping compensations approach zero. Same value of work done, mgh.

arildno
Homework Helper
Gold Member
Dearly Missed

Hi teddy!
If you want a velocity to CHANGE from say 0 to some positive value, you need a tiny time interval in which the object accelerates. In that time interval, the upwards force MUST be greater than mg.

But:
If that force keeps being bigger than mg all the time afterwards, the object will still accelerate, rather than maintain a particular velocity level.
In order to maintain a velocity level,the forces acting upon it must cancel each other out, i.e be equal, but of different signs.

If the force applied on ab object due to gravity is -mg why is applying a force of mg give upward raise the object if the two forces are equal and opposite in direction. I would think the object would remain still until you apply a force greater than mg.

So in conclusion:

If an upward force of mg is "being applied" to an object while you look at it, it could be moving up at a constant velocity, down at a constant velocity, or have a velocity of zero. It cannot be accelerating.

If it was at rest to begin with, you would have to apply more than mg to make it move up. Because you would have accelerated it from rest.