# B How can kinetic friction force be constant if....?

1. Jun 21, 2016

### BlueQuark

As we all know, for the most part, the kinetic friction force is, for the most part, constant. After moving my cup across my table, this thought crossed my mind. If I move my cup across the table with a constant speed, then the force I'm applying must be equal to the kinetic friction force.

Let's say I apply a stronger force to get it at a constant speed that is faster than the other speed. I was pushing a force higher than the kinetic friction force. In that case, the cup should be accelerating instead of remaining in equilibrium.

I suspect I've fallen for a common misconception of sorts. Can someone explain this? Thanks!

2. Jun 21, 2016

### Staff: Mentor

It looks to me like you analyzed the logic right but still choose to accept your initial wrong assumption instead of the correct conclusion you drew later. Why?: Yes, if you apply a force greater than the kinetic friction, the cup will accelerate.

3. Jun 21, 2016

### BlueQuark

This is what is confusing me. I can demonstrably push a higher force than the kinetic friction force and get it to move at a constant speed. Does the kinetic friction force increase with speed?

4. Jun 21, 2016

### Tallus Bryne

If you push with a force greater than kinetic friction, the object speeds up. If you want to "get it to move at constant speed" no matter if it is a higher or lower speed than you started with, you then must stop the acceleration by going back to a state of equilibrium where again the push/pull is equal to the kinetic friction.

5. Jun 21, 2016

### Staff: Mentor

How are you measuring that?
No, with the caveat that real life can be a bit more complex than that simplistic equation. But over a low range of speeds, with hard, dry surfaces, it should hold reasonably well.

6. Jun 23, 2016

### Dr. Manoj

If you are applying force there is an acceleration. If there is increasing acceleration, you speed will not be constant. If you need constant speed maintain constant acceleration. Else if cup isn't moving even you maintained constant acceleration, maybe your force isn't crossing limiting static frictional force. Let me know if I'm wrong or correct. Thank you.

7. Jun 23, 2016

### nasu

Careful, Dr Manoj.
For constant speed you don't need to maintain constant acceleration. Unless that constant is zero.
Even if there is constant (and not increasing) acceleration the speed won't be constant.
You may have some problem with the proper meaning of the terms.

8. Jun 23, 2016

### Drakkith

Staff Emeritus
Not necessarily true. If the force you are applying is exactly countered by another force then there is no net force and thus no acceleration.

The change in the acceleration is irrelevant. Any acceleration means that the speed is not constant.

Constant speed means zero acceleration which means zero net force.

9. Jun 23, 2016

### Nidum

Your hand is acting as a velocity source . Within limits your hand can move an object at a set speed independent of the reaction force from the object being moved .

Last edited: Jun 26, 2016
10. Jun 23, 2016

Staff Emeritus
"Doctor" Manoj, it is dishonest and deceitful to claim a qualification that you have not earned - like a doctorate. Additionally, it is insulting to those of us who spent years earning one. Finally, you bobbled this question - to claim to be an expert (i.e. calling yourself "doctor") and then providing wrong answers is worse than useless.

You might want to demonstrate this, then. I think if you actually did these measurements you would find that either you are reducing the force after some time, or the speed is not constant.

11. Jun 25, 2016

### pixel

Maybe he's a medical doctor, or a doctor of law, or has a Ph.D. in English, etc.

12. Jun 25, 2016

### SammyS

Staff Emeritus
@pixel ,

Look up his profile. a 17 yr old student

13. Jun 29, 2016

### Tazerfish

It probably feels like you apply more force than the initial kinetic friction because you move the cup faster.
You have to do more work in that case.
The amount of power is $P=F v$ so it might feel like you apply more force because it is harder to apply the same force.