Which force does friction counteract on an inclined plane?

[sharpnova]

This is not a specific problem. It pertains to all the problems I'm working on and only requires that I have one concept explained.

If an object is being pulled up a plane by a force parallel to the plane, does the friction work against the parallel component of the gravity or the force pulling the object up?

I have the parallel component of the gravity as weight_object * cos(angle), the pulling force in the exact opposite direction, and the friction as u * weight_object * cos(angle)

The problem is I don't see which way to apply the friction. If the object was sliding down the plane, then friction would work against the gravity force. But shouldn't the friction also work against the act of pulling the object up the plane? I don't know if that means I ignore the friction, apply it in one direction only, or do something else to calculate the "net" friction force.

 

[BvU]

The answer to your first questions is: it depends !
Corny, I'm sorry. But friction opposes motion, so if the gravity wins from the pulling force, friction is with the pulling force. And if the pulling force is greater than the downward gravity component, then the friction is with the gravity, i.e. downwards.
Note that the friction coefficient times the normal component expression calculates a maximum friction force. In the range where | gravity minus pulling force | is less than the maximum friction force, the magnitude of the actual friction force is less than this maximum value (namely such that the acceleration is zero , so that gravity minus pulling force and friction combined add up to zero)

 

[stockzahn]

If there is a relative motion between two bodies for each body the direction of the friction force can be applied against the direction of the motion:

object sliding down → friction force at the object pointing "up" & friction force at the plane pointing "down"
object pulled up → friction force at the object pointing "down" & friction force at the plane pointing "up"

 

[sharpnova]

Thanks that cleared everything up. Is there a way to select a best answer or anything like that here?

 

[BvU]

Your thanks is adequate reward ! We're all colleagues and not competitors. And both answers agree.

 

[BvU]

Let's not turn PF into a mutual admiration society , but: thanks !

PS I like the PF culture in the sense that after a while you can distinguish the relative authorities a little bit. And that's good. Any ranking would be an insult: high rank is no insurance against goofs, whatever the voters may claim.

 

[sharpnova]

Sounds good to me :D

I've never been a fan of stackexchange's voting system. It has made the place somewhat toxic in my opinion.

 

[haruspex]

friction opposes motion

Sorry, BvU, but I have to intercede here.
It's better to say that friction opposes relative motion of surfaces in contact. The shorter form confuses many students when they have to deal with e.g. a car driving uphill.

 

[sharpnova]

Could you briefly tell me a little bit about what a car driving uphill would be like, friction wise. I'm trying to visualize it and it seems with the wheels rotating, the friction would be eliminated from the equation. But I'm sure that's not the case.

 

[haruspex]

Could you briefly tell me a little bit about what a car driving uphill would be like, friction wise. I'm trying to visualize it and it seems with the wheels rotating, the friction would be eliminated from the equation. But I'm sure that's not the case.

Could the car get up the hill if there were no friction?

 

[sharpnova]

No I don't think so. The wheels would just spin in place. So how would the friction be accounted for there? Since I think the wheels have no relative motion to the ground?

 

[haruspex]

The wheels would just spin in place. ... I think the wheels have no relative motion to the ground?

You don't see a contradiction between those two statements? Friction opposes relative motion of the surfaces. It may or may not be able to oppose it so completely that it prevents it.

 

[sharpnova]

I think the friction is why the wheel grips the ground.

Is it just that if the incline were steep enough then the friction wouldn't be enough and then the car would stay in place or ultimately slide back?

 

[haruspex]

I think the friction is why the wheel grips the ground.

Is it just that if the incline were steep enough then the friction wouldn't be enough and then the car would stay in place or ultimately slide back?

That's true, but is not the point I'm making. Assume the car is managing to get up the hill without the tyres' slipping. Which way does the force of friction act?

 

[sharpnova]

I think it's acting... in the same direction as the car's motion? Against the road-parallel component of the car's weight

 

[haruspex]

I think it's acting... in the same direction as the car's motion? Against the road-parallel component of the car's weight

Quite so. The trap is that some students learn "friction opposes motion", so think it must act down the slope, opposing the motion of the car. The correct view is that it acts to oppose the relative motion of tyre surface against road.
Of course, we've been assuming driving wheels here. The situation is different for non-driving wheels.

 

[sharpnova]

Yep for a wagon or a plane taking off or landing the friction would operate like a classical block-on-an-incline situation.

 

[haruspex]

Yep for a wagon or a plane taking off or landing the friction would operate like a classical block-on-an-incline situation.

Not so fast.
Suppose a rear-wheel-drive car is going up hill but slowing down. What is happening to the front wheels?

 

[sharpnova]

Not so fast.
Suppose a rear-wheel-drive car is going up hill but slowing down. What is happening to the front wheels?

Being pushed up by the rear wheels driving and retarded by friction?

 

[haruspex]

Being pushed up by the rear wheels driving and retarded by friction?

The front wheels have no direct knowledge of the rear wheels. They just 'feel' the axle, the road, and gravity.
"Retarded" is a bit ambiguous here. Are you referring to linear motion or rotational motion? Which way does the friction act, and why?

 

[sharpnova]

I'm not sure what you thought I was saying about knowledge. I was just saying that the locomotion of the powered wheels would generate a force in the direction of the car's travel that would be a part of what was "happening to the front wheels"

As for friction I was referring to the linear motion. I'm not sure how to deal with the rotational motion. Probably the most elusive part of the thought experiment to me

 

[haruspex]

I was just saying that the locomotion of the powered wheels would generate a force in the direction of the car's travel that would be a part of what was "happening to the front wheels"

Yes, but that's too indirect. The rear wheels do not apply any forces directly to the front wheels. What forces are acting on the front wheels?

As for friction I was referring to the linear motion.

Remember that the friction here is acting at the bottom of the wheels. If the frictional force is acting on the rolling wheel in its direction of motion, that tends to accelerate its linear motion. But the torque it applies about the axle opposes the wheel's rotation, slowing that down. This apparent paradox is resolved by the fact that there is also a force at the axle.

As this is not part of your homework assignment, I'll stop quizzing you on this and just lay it out.
Suppose there's no friction on the front wheels, somehow. Although the car slows down up the hill, there's nothing tending to slow the rotation of the front wheels. The car could come to a stop and the front wheels would just carry on spinning. In the real world, with friction, the front wheels do spin more slowly and stop. The friction opposes the slipping of the tyres on the road, so acts uphill, opposing the rotation of the wheels. Because of the rotational inertia of the wheels, this means the front wheels tend to pull the car up the hill. The reaction from the axle pulls the wheels down the hill, and that is what slows their linear motion.

Complicated, huh?