Friction force has different effect than Applied Force?

[Ocata]

Hi,

I'm wondering what's going on when a friction force of a surface is applied to a moving block as the block slows down to zero and it, the friction force, is the only horizontal force applied. Does the friction force continue at a constant value until the last moment and just drop to 0N or is there a gradual decrease somehow?

I described two scenarios below to possibly clarify what I think is not making sense.Scenario 1:
Suppose I am sitting at a frictionless table and you push a block so that the block is traveling toward me at 10m/s. I am standing on the ground with normal friction. As the block approaches me, I apply force to the block with my hand at a constant force of 5N, so as to slow it down. But I continue applying the force of 5N so that the block slows down to a stop and immediately begins to speed up in the opposite direction, such that I am returning the block to its original speed in the opposite direction.

Scenario 2:
Suppose I push a block that is on a table, with friction, so that the block accelerates to 10m/s. When the block reaches 10m/s, I continue pushing with a force such that the net force on the block is Fnet = 0N and the velocity of the block is traveling at a constant velocity of 10m/s. Suppose I'm applying a force of 5N and the friction of the table is applying -5N on the block. The friction force is constant. Suppose all of a sudden, I immediately stop applying force to the block. The block will slow down by the force of friction from the table just as I slowed down the block with my hand in scenario 1.

The difference is, in scenario 2, once the block reaches a velocity of 0m/s, it will not reverse direction and begin speeding up in the opposite direction.

If the table was actually applying a constant force of -5N on the block, the block should, in theory, reverse direction and begin speeding up in the opposite direction. But instead, the table seems to stop applying a friction force of -5N just as the block reaches 0m/s.

Is the friction force -5N the entire time and just disappears when the block reaches 0m/s? If I plot a Net Force .time graph of scenario 2, will I see the Net Force steadily decrease as the velocity of the block approaches 0m/s or will there be a sharp drop from 5N to 0N (maybe even an instantaneous drop) right when the block reaches 0m/s? What exactly is going on here?

Thank you.

 

[Ocata]

Btw, I revised the first paragraph just in case it did not make sense. Please let me know if my question is not making sense or if I need to clarify further.

 

[ericaclayton]

yes of course!

 

[CWatters]

Friction models are just that.. Models.
Static friction model... The friction force matches the applied force upto a limit beyond which the object will start moving.
Kinetic friction model.. Friction force is usually assumed to be a constant force independent of velocity but that's not always true.

 

[Ocata]

Thank you, I realize that I needed to understand static and kinetic friction a bit better. Now that I went took some time to review and contemplate static and kinetic friction, I can more precisely identify what I'm finding confusing, formulate my question better, and feel more prepared to understand feedback.

Suppose I have a 5kg block on a table with .2 static friction and .1 kinetic friction. When I push the block with a force that exceeds 10N, the block will begin to move. After the block is set in motion, when I apply 5N, the block will have a 0 net force and thus will have a constant velocity.

The first question I have is) when the block is in motion and I apply various levels of force, I can observe the block moving at a constant velocity even though the net force is not zero. For example, as I'm typing this, I am pushing a calculator on my desk. If I push softly, I can observe what seems to be the calculator moving at a slow, but constant velocity. When I increase my force, I increase the speed of the calculator, and when the calculator reaches some desired speed, I can then adjust my applied force such that the calculator maintains a faster, constant velocity. Since the force of friction is 5N for every magnitude of applied force, why can I apply force in such a way that the motion of the calculator (or block) can be observed to have various levels of constant velocity?

My guess is that I may be very subtly adjusting my applied force back and forth to speed up and slow down the calculator. And also, I'm guessing on some (or all) instances, the object may be bouncing off of the applied force (my hand in this case) at rates so fast and or distances so small that they are undetectable to human senses.

So I wonder, why does it seem like an object can have different magnitudes of constant velocity when different magnitudes of force are applied if there is only one applied force that can generate a zero net force?

 

[@navin]

What you are doing is that when you increase the force on your calculator, the forces are unbalanced. So the calculator accelerates. And then, all of a sudden, you decrease the force you are applying to 5N (which is the kinetic friction) and it, in turn, makes the net force on the calculator zero. So, it goes on moving with the same velocity. You can repeat it again and again, but to keep the velocity constant, you need to apply 5N only.
I hope you understood :-)

 

[Ocata]

Yes @navin, thank you, I did understand very well.

 

[CWatters]

Thank you, I realize that I needed to understand static and kinetic friction a bit better. Now that I went took some time to review and contemplate static and kinetic friction, I can more precisely identify what I'm finding confusing, formulate my question better, and feel more prepared to understand feedback.

Suppose I have a 5kg block on a table with .2 static friction and .1 kinetic friction. When I push the block with a force that exceeds 10N, the block will begin to move. After the block is set in motion, when I apply 5N, the block will have a 0 net force and thus will have a constant velocity.

Correct.

The first question I have is) when the block is in motion and I apply various levels of force, I can observe the block moving at a constant velocity even though the net force is not zero.

That's not correct. If the velocity is constant the net force must be zero. Newtons law.

For example, as I'm typing this, I am pushing a calculator on my desk. If I push softly, I can observe what seems to be the calculator moving at a slow, but constant velocity. When I increase my force, I increase the speed of the calculator, and when the calculator reaches some desired speed, I can then adjust my applied force such that the calculator maintains a faster, constant velocity. Since the force of friction is 5N for every magnitude of applied force, why can I apply force in such a way that the motion of the calculator (or block) can be observed to have various levels of constant velocity?

Its possible that friction force isn't constant in this case. You would have to do a more controlled experiment to confirm that a constant friction force is the right model for the rubber feet of a calculator. However once the new velocity has been achieved the net force will be zero regardless.

 

[Ocata]

Thank you CWatters.

The net force will always be zero, so acceleration will always be zero. But the velocity can be different depending how 5N of applied force was reached. Right now, I am just beginning to wonder how changing the way force is applied to the block determines the velocity once 5N of applied force is obtained. I will start a new thread to ask this new question.

 

[CWatters]

I am just beginning to wonder how changing the way force is applied to the block determines the velocity once 5N of applied force is obtained. I will start a new thread to ask this new question.

The short answer is history.

The net force at any instant in time tells you what's happening to the acceleration at that instant but the velocity depends on history (eg the sum or intergral of what happened before that instant).

For example..

F = ma
so
a = F/m

The definition of acceleration gives..
a = dv/dt

dv/dt = F/m

integrate..

v = integral (F/m) + constant

Sorry for lack of proper formatting/notation.