Physics inertia question

[The_Hamster_King]

@petew83

Say I wanted to start the glass moving, get it up to a steady speed, and then stop it in the exact same way as I started it so that the entire process is exactly reflective across the time axis visually (imagine the hands are invisible and you only see the glass). How would the distribution of forces play out? Is it extremely complicated?

If friction between the glass and the table is negligible, the forces would be symmetrical.

But if the friction is high enough to have an effect, it changes things. Friction resists you when you’re speeding the glass up, but helps you when you’re slowing the glass down. As a result the required forces are asymmetrical.

[Cheshire_Human]

@Mangetout

I’ve got a feeling that the OP wants us to ignore friction altogether, but is perhaps under the impression that a constant force will only accelerate an object to a given velocity, after which the force would cause no further increase in velocity.

Actually, the problem, here, is that the OP is asking a question based on a totally faulty view of the physics involved. The OP is making some very basic, but false, assumptions, in the question itself. Unless it’s determined what those false assumptions are, there’s no way to properly answer this question. Trying to guess what the OP meant isn’t going to work.

OP, why don’t you try to rephrase it, so we can get an idea of what you believe is happening. Then we can address what’s wrong with your view of the situation you are trying to describe. Then we can maybe answer your actual question.

[CookingWithGas]

@sh1bu1

Imagine that you put a spring scale between you and the box. The scale will register more 'weight" when you are accelerating the box. Once you reach the speed you desire and stop accelerating the box (i.e., a constant speed) the weight registered by the spring scale would be less. This means you are not applying a constant force throughout.

Oh, I get that, that wasn’t my point. I have two sticking points. One is

@Chronos

…if the applied force is ever enough to overcome the friction, it’ll always be enough to overcome it, and the object will accelerate indefinitely.

This implies that if you keep the scale reading the same, you’ll keep accelerating, which is counterintuitive. I will be the first to admit that intuition is often wrong, but would like to confirm this point so I understand it better. Similarly for

@Simplicio

If you push something with a constant force, friction or no friction, it will never reach a constant speed, it will accelerate until you reduce the force on it. If you push something with a constant force, friction or no friction, it will never reach a constant speed, it will accelerate until you reduce the force on it.

I know that f=ma so that constant force applied to a mass will result in constant acceleration. This is easy to understand in a rocket-in-space situation. But doesn’t dynamic frictional force increase somehow in proportion to the velocity the object is moving?

And further

@zut

This is one of those situations where your physical intuition fails. When you push a box across the floor, you’re really applying whatever force is necessary to exactly counterbalance the friction and maintain your walking speed. If you run instead of walk, you’re still (for the most part) applying whatever force is necessary to exactly counterbalance the friction and maintain your running speed, and it’s essentially the same force you apply at a slower speed. It might feel like you’re applying a much greater force, but you’re not.

I interpret that to imply that once I reach my target speed, the scale will show the same force whether I am pushing the box at 1 km/sec or 5 km/sec. That’s the part that is counterintuitive. I think I’ll go get my luggage scale and try this…

What are the formulas involved here?

[Cheshire_Human]

Missed the edit window, and I didn’t refresh the page since I left to move snow. It seems the discussion with the OP is already going in the right direction. Ignore my previous post.

[Chronos]

The formula for dry kinetic friction is F = mu[sub]k[/sub]*N, where F is the magnitude of the frictional force, mu[sub]k[/sub] is the coefficient of kinetic friction (a constant that depends on the surfaces involved; typical values are in the vicinity of 1 or less), and N is the magnitude of the normal force, the “pushing force” that the two surfaces are exerting on each other (which will typically be equal to the weight of the object, if it’s sitting on a level, non-accelerating surface with no other vertical forces). At least, that’s the simple version: The value of mu[sub]k[/sub] will in reality vary somewhat depending on speed and the normal force and other factors, but it’s pretty close to constant, and it won’t in general vary in the way you expect, anyway.

[petew83]

@Cheesesteak

#1) You apply 15 units of force to the glass, the glass generates 5 units of frictional force, so you accelerate the glass with 10 net units of force. When you decelerate the glass, in order to have the same rate of deceleration, you need to apply 10 net units of force to slow it down. That means applying 5 units of force in the opposite direction, so you have 10 net units to slow it down.

#2) You apply 10 units of force, the glass generates 5, you get a net 5 units of force accelerating the glass. To decelerate, you want 5 net units of force in the opposite direction, which you get by letting go of the glass, and letting friction do all the work for you.

#3) You apply 7 units of force, the glass generates 5, getting a net 2 units of force accelerating the glass. To decelerate, you want 2 units of force going the opposite direction, so you should reduce your force to 3 units, but keeping the same direction, to get the net of 2.

This is perfect, thank you (#1 is exactly what I described). What is the name for the ‘acceleration time period’ and the ‘deceleration time period’ in this example?