Understanding Friction - Topic

Bean`

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Mar 3 2015 03:42pm

does anyone know of a good way to understand friction on incline and decline planes when speeding up/slowing down/constant velocity? im getting so terribly confused solving problems when i dont understand which way friction is ever pointing. i keep getting my signs messed up which is leading me to the wrong answer. i have tried googling and youtubing but everything i find is more complicated than where i am at in class right now.

all i need to know is how speeding up, slowing down, and constant velocity affects friction when moving up a ramp and understanding the same for moving down a ramp.

This post was edited by Bean` on Mar 3 2015 03:42pm

carteblanche

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Mar 3 2015 07:20pm

iirc friction is always in the opposite direction of motion. so if you're moving up an incline, friction acts down the incline. if you're sliding down, then friction is going up.

constant velocity just means the net force (gravity, friction, etc) is zero.

Bloo_Guardian

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Mar 3 2015 10:23pm

Let "F" be the force in the direction of motion and "f" be the force of friction. Assume the direction of motion is the positive direction.

If you are going up a ramp or down a ramp, these will be the same.
Speeding up: F > f so total net force is positive
Slowing down: F < f so total net force is negative
Constant velocity: F = f so total net force is 0.

Amjohnston81

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Mar 3 2015 10:51pm

They are always opposite of each other in direction, this opposite signs.

TheStealthTarget

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Mar 3 2015 10:54pm

Quote (carteblanche @ Mar 3 2015 08:20pm)

Quote (Bloo_Guardian @ Mar 3 2015 11:23pm)

first, let me say i agree 100%... the next thing to do it just look at the terms... as friction increased, the velocity should decrease, so the equations should help speak to this as well.

Fioravanti

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Mar 4 2015 08:29am

Quote (Bloo_Guardian @ Mar 3 2015 11:23pm)

If F = f then the object in question would not be in motion.

Tclasp

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Mar 4 2015 09:32am

Quote (xAoHxDiesel @ Mar 4 2015 09:29am)

If F = f then the object in question would not be in motion.

If F=f then the object is not accelerating. It could still be moving at constant velocity

betoggc

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Mar 5 2015 06:59pm

There are much more in physics to worry about than friction

Well, when you get a ramp, you have to understand that you will always have the gravity acting there.

So, when you are sliding down the ramp, the gravity will help your "box" to keep moving. When you are sliding up the ramp, the gravity will act against your movement.

Well, let's see the first case (box already is sliding down the ramp).

P - the force of gravity on the box (P = m.g)
N - the force of the ramp on the box
Fr - Friction (always against the movement)

To keep the velocity constant, you have to have an equilibrium of all forces which are acting on the box. All we have to do is to sum all forces that are acting on the box and the result must be zero.
But to do that sum, we have to hide the force "P" for a moment and put it in two components: Px and Py. We have to do that to sum forces.

Now look what we have now:

Look, the force P still exists, but I just put two other forces that represents the force P together. Look this picture:

Well, now we have to abide two circunstances together in order to have the box sliding down with velocity constant:
N = Py
Fr = Px

It's a must to have both conditions above in order to have the box with velocity constant. So, ...

N = Py = P.cos(a)
Fr = Px = P.sin(a)

The formula of friction is Fr = μ . N

So,
μ . N = P. sin (a)
μ . P.cos(a) = P.sin(a)
μ . cos(a) = sin(a)
μ = sin(a) / cos(a)
μ = tan(a)

Well, this is what I can help. Feel free to PM me if you want the other case, which the box is sliding up the ramp.

I hope I could help. Please someone correct me If I'd commit some mistakes.

edit: changed some typo

This post was edited by betoggc on Mar 5 2015 07:00pm

Bean`

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Mar 5 2015 08:56pm

Quote (betoggc @ Mar 5 2015 08:59pm)

There are much more in physics to worry about than friction :P

Well, when you get a ramp, you have to understand that you will always have the gravity acting there.

So, when you are sliding down the ramp, the gravity will help your "box" to keep moving. When you are sliding up the ramp, the gravity will act against your movement.

Well, let's see the first case (box already is sliding down the ramp).

http://i.imgur.com/zMOdEEF.jpg

P - the force of gravity on the box (P = m.g)
N - the force of the ramp on the box
Fr - Friction (always against the movement)

To keep the velocity constant, you have to have an equilibrium of all forces which are acting on the box. All we have to do is to sum all forces that are acting on the box and the result must be zero.
But to do that sum, we have to hide the force "P" for a moment and put it in two components: Px and Py. We have to do that to sum forces.

Now look what we have now:

http://i.imgur.com/LRJst26.jpg

Look, the force P still exists, but I just put two other forces that represents the force P together. Look this picture:

http://i.imgur.com/qmQm80P.png

Well, now we have to abide two circunstances together in order to have the box sliding down with velocity constant:
N = Py
Fr = Px

It's a must to have both conditions above in order to have the box with velocity constant. So, ...

N = Py = P.cos(a)
Fr = Px = P.sin(a)

The formula of friction is Fr = μ . N

So,
μ . N = P. sin (a)
μ . P.cos(a) = P.sin(a)
μ . cos(a) = sin(a)
μ = sin(a) / cos(a)
μ = tan(a)

Well, this is what I can help. Feel free to PM me if you want the other case, which the box is sliding up the ramp.

I hope I could help. Please someone correct me If I'd commit some mistakes.

edit: changed some typo

Thanks man, but what I dont understand now is which way is friction when slowing down moving up a ramp and direction of friction speeding up down a ramp. options are no friction, down the ramp, up the ramp

also when you are you summing your forces and have mg*sin and moving up the ramp, you make it negative right? since the x component of the weight moving up the ramp is negative? also what about friction?

in other words newtons 2nd law sum of all forces = m*a, where sum of all forces in x direction is Wx+fx ; if moving up a ramp say for constant velocity just for example, it would be -Wx+fx right? what about constant velocity moving down ramp? Wx-fx? i want to make sure i am getting my signs correct.

Amjohnston81

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#10

Mar 5 2015 10:37pm

As you are going up the ramp, even slowing down, the friction is opposite of motion, your motion is still up the ramp, but its just at a slower velocity. Speeding down a ramp, friction would be facing towards the peak of the ramp.

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