Direction of Friction: Rotating a Wheel Up a Hill

[R Power]

Hi guys
Consider a wheel rolling up the hill by some external force. What will be the direction of friction?
I know it's a basic question but i got confused last night with friction in rotation and rolling.

 

[Monocles]

The direction of friction would have to be down the hill in order for the wheel to roll in the correct direction.

 

[SystemTheory]

The friction generates torque to cause rotation of the wheel. When you apply the right hand rule your fingers wrap in the direction of the force:

http://hyperphysics.phy-astr.gsu.edu/hbase/tord.html

 

[R Power]

"The direction of friction would have to be down the hill in order for the wheel to roll in the correct direction."

if i give it a force then it's ok i understand u or myself
but consider an axle of a car , it gives a wheel a torque , so in this case torque has already been provided so will friction act up the hill?

 

[SystemTheory]

My original logic in this post was incorrect, so it is deleted (12/06/09)!

A diagram is included with a comment below.

 

[R Power]

but the reaction force is friction. Isn't it?

 

[Doc Al]

but consider an axle of a car , it gives a wheel a torque , so in this case torque has already been provided so will friction act up the hill?

Yes. The friction of the road will act up the hill thus propelling the car up the hill.

 

[R Power]

in this case why will car propell up the hill?
If i apply external force to a wheel , friction will act downwars providing the wheel with torque, which will make it rotate and the linear force i provided combining with the rotation will create a "roll".
But in above case torque is provided b axle it's ok but what about linear component of force required too create a roll. If u say it's friction I say that frition will just give an anti torque i.e just trying to stopp the rotation of wheel. What will give the wheel a linear component??Bcoz linear component together with rotation produces 'roll'.

 

[Doc Al]

Friction from the road provides a linear force as well as a torque.

 

[R Power]

Why will friction provide a linear force. Actually friction provides a linear force a radius distance apart from center of rotation which creates a torque(a torque is always a couple so in this case one force is at the top of wheel due to my push and other and opp is due to friction).
What propells it forward?

 

[russ_watters]

Both. The forces come in pairs, otherwise the wheel would freely accelerate when you apply a force to it.

 

[R Power]

Why would wheel freely acc when i apply force to it? When i would give it force , the force will be on all points on it and also at the top. Now at bottom friction will give an opp force , the couple creates a torque and this would produce rotation but since a linear force was present at all other points too(my force) it would move forward along with rotation and this is called rolling.

 

[russ_watters]

Why would wheel freely acc when i apply force to it?

F=ma, so if you apply a single force to an object, it will accelerate.

When i would give it force , the force will be on all points on it and also at the top. Now at bottom friction will give an opp force , the couple creates a torque and this would produce rotation but since a linear force was present at all other points too(my force) it would move forward along with rotation and this is called rolling.

Right - the engine provides a torque in one direction, the friction with the ground provides a torque in the other direction. In constant speed motion, forces come in pairs (at least), that sum to zero.

 

[SystemTheory]

Sketch attached. Friction is upward and resolved weight of a cylinder/wheel is downward along the plane angle.

If no torque acts independently at the axis of rotation, the wheel is rolling downhill, the friction causes a clockwise torque about the axis applying the right hand rule.

Suppose a second counter-clockwise torque (like at a car axle) acts independently about the central axis. The magnitude of the upward friction force must grow in reaction to this torque to either (a) slow the downward motion; (b) hold the wheel in place on the ramp; or (c) move the wheel up the ramp. I'm fairly certain this description will make sense now.

 

[R Power]

system theory
consider the case of a wheel going up the hill.
Though i understand the case of wheel rolling down, direction of friction clearly.
What i am asking is : When wheel rolls down firction produces torque and thus rotation and there is a downward motion due to gravity which together with rotation creates a "roll".That's clear .OK.
But when wheel goes up especially wheel on an axle , torque is provided by axle , friction just counter acts to oppose that torque in uphill direction, then what provides linear motion to wheel to roll instead of rotate.
In previous case gravity provided linear motion and friction provided rotation and togehter they both created rolling motion.
But in later case where does linear motion come from?
ReAD CAREFULLY!

 

[Doc Al]

What i am asking is : When wheel rolls down firction produces torque and thus rotation and there is a downward motion due to gravity which together with rotation creates a "roll".That's clear .OK.

The translational acceleration is determined by both gravity and friction.

But when wheel goes up especially wheel on an axle , torque is provided by axle , friction just counter acts to oppose that torque in uphill direction, then what provides linear motion to wheel to roll instead of rotate.

Friction and gravity once again determine the translational acceleration.

In previous case gravity provided linear motion and friction provided rotation and togehter they both created rolling motion.
But in later case where does linear motion come from?

In both cases friction helps determine the linear motion. You seem to be stuck on the idea that friction provides a torque but not a linear force.

ReAD CAREFULLY!

Good idea!

 

[R Power]

In both cases friction helps determine the linear motion. You seem to be stuck on the idea that friction provides a torque but not a linear force.

But friction provides linear force only at the bottom part where wheel touches the ground and that linear force is used for producing torque. Isn't it? Or that linear force also pushes the wheel forward?

 

[Doc Al]

But friction provides linear force only at the bottom part where wheel touches the ground and that linear force is used for producing torque. Isn't it?

Since friction is applied at a distance from the center, it contributes to the torque on the wheel. But it also contributes to the linear acceleration of the wheel.

Or that linear force also pushes the wheel forward?

Of course it does. When you use Newton's 2nd law to find the linear acceleration, you must consider all the forces regardless of where they act.

 

[sophiecentaur]

Would we all be having such a terrible problem is we replaced the wheel / road / drive shaft system by a lever and just considered the first mm of movement? That's all that's happening, really. It's jut that the wheel provides you with a set of levers one of which is always in contact with the road.

 

[R Power]

Since friction is applied at a distance from the center, it contributes to the torque on the wheel. But it also contributes to the linear acceleration of the wheel.

Let me say I give a force of 10 Newton uniformly over whole of a wheel. Now every point on the wheel will have an acceleration coresponding to that force or I can say every point will experience a force of 10 N. The bottom point touching the ground will also experience a force and it will apply same force to ground. Now a reactive force or friction will act on the same point in opp direction. Also, there existed same force on the top most point. So a couple will be formed of 10N each force and the wheel will start to rotate. So now 10 N or whole of friction force is used in creating couple, where does linear component come from?

If u still say friction will give linear component, then relook that force at top point was 10N, so we need a 10N at bottom for rotation to start, so friction provided that 10N which was it's instataneous max. value. So, all 10N used in rotation , where does linear force come from?

 

[Doc Al]

Let me say I give a force of 10 Newton uniformly over whole of a wheel. Now every point on the wheel will have an acceleration coresponding to that force or I can say every point will experience a force of 10 N. The bottom point touching the ground will also experience a force and it will apply same force to ground. Now a reactive force or friction will act on the same point in opp direction. Also, there existed same force on the top most point. So a couple will be formed of 10N each force and the wheel will start to rotate. So now 10 N or whole of friction force is used in creating couple, where does linear component come from?

I don't know what you mean by applying a 10 N force 'uniformly over the wheel'. Perhaps you mean that you are applying a net force of 10 N on the wheel in such a way as to exert no torque? And you also have another force at the top of the wheel? Please try describing the situation again.

If u still say friction will give linear component, then relook that force at top point was 10N, so we need a 10N at bottom for rotation to start, so friction provided that 10N which was it's instataneous max. value. So, all 10N used in rotation , where does linear force come from?

Again, I don't quite understand what you're saying here. Realize that all forces contribute to the linear acceleration of an object. This is just Newton's 2nd law: ΣF = ma. ΣF stands for the net force, including everything.

Think of this: Imagine you are driving a car on level ground. You step on the gas to accelerate. What external force acts on the car to accelerate it?

 

[R Power]

I didn't have any other force on the top of wheel. I simply applied a force of 10 N to the wheel,so each point on the wheel will experience a force of 10 N. So the topmost point will also experience a force of 10 N. Now bottom point will also experience 10 N force and exert it onto ground bcoz it is in contact with ground. So ground will exert a equal and oppp force of 10 N on it as friction. The 10N force at top and frictional 10N at bottom will produce torque. Friction can't provide any linear force then since all 10N friction is used in creating couple with the 10 N force at toppoint.

 

[sophiecentaur]

Apart from accelerating the wheel (but let's assume it is massless), itself, the force "around" it is irrelevant. What is important is the forces where the wheel is in contact with something else. There is a couple, from the periphery of the shaft - say two forces - one at the top and one at the bottom of the hub. There is another couple, consisting of a forward force from the road, due to friction and a reaction force (backwards) against the centre of the shaft, because of the reaction due to the mass of the car.

I suppose there must be another couple involved, due to the weight of the car which is well forward of the driveshaft, which stops the car from 'wheelying'. (In a front wheel drive, the force is from the rear wheels). This is greater than the couple turning the wheel, so the car goes forward without tilting up. I am finding it difficult to draw a diagram which shows this extra couple and its effect. Perhaps you have to consider it as a reactive couple, appearing against the couple applied to the wheel.
I guess that when a wheely starts, the forward acceleration is briefly less, as the force against the ground is reduced.

 

[sophiecentaur]

RPOWER
If you consider driving the wheel with a roller at the top (which may simplify matters helpfully) then you have one couple, consisting of the roller force and the friction force separated by the diameter of the wheel.
There will also be a couple due to the mass of the car, acting at the axis of the wheel and the reaction force of the road, separated by the radius.
Because of the rigid body, front wheels / the weight of the car (mentioned above) the car can't rotate or deform so the force acts against the mass of the car - pushing it forwards.
The argument would apply if you consider a lever, too.
I must say that, apart from having made me think very usefully, about the problem I don't see where this is getting us.
The system clearly works and all that remains is to reconcile some intuitive thoughts with a bit of real mechanics and to choose the appropriate forces and distances to consider. No one has 'got it wrong'.

 

[R Power]

sophie,
let us take case of a wheel and i push it with some force , friction creates a couple with force i applied and causes rotation. Now since i provided a linear force (not torque), it will create a roll bcoz rotation+linear motion = rolling motion. Here friction creates rotation and my push on wheel gives it linear motion so together they create a roll. Note that friction doesn't provide any linear movement to wheel here , it only provides it torque.
Do u agree with me? First tell this then we 'll talk of axles.

 

[Doc Al]

I didn't have any other force on the top of wheel. I simply applied a force of 10 N to the wheel,so each point on the wheel will experience a force of 10 N.

I don't know what you mean. Did you apply a force of 10 N spread out over the entire wheel? That's not the same as every point on the wheel experiencing a force of 10 N. (That would mean you applied an infinite net force to the wheel.) Extremely confusing!

So the topmost point will also experience a force of 10 N.

See above.

Now bottom point will also experience 10 N force and exert it onto ground bcoz it is in contact with ground. So ground will exert a equal and oppp force of 10 N on it as friction.

No. Just because you (supposedly) are exerting 10 N of force on the bottom of the wheel does not mean that the wheel exerts 10 N of force on the ground.

The 10N force at top and frictional 10N at bottom will produce torque. Friction can't provide any linear force then since all 10N friction is used in creating couple with the 10 N force at toppoint.

Again, you seem to be creating some strange arrangement of forces. In any case, you (supposedly) exerted 10 N of force at the top and bottom of the wheel. Those two forces exert a net linear force but no torque. Friction exerts both a linear force and a torque. (A force doesn't get 'used up' in creating a torque!)

Answer my question about the accelerating car.

 

[Doc Al]

Note that friction doesn't provide any linear movement to wheel here , it only provides it torque.

Wrong. All forces contribute to the net force on the object, which is used to determine the linear acceleration.

Friction does two things: It contributes a linear force and a torque.

 

[sophiecentaur]

sophie,
let us take case of a wheel and i push it with some force , friction creates a couple with force i applied and causes rotation. Now since i provided a linear force (not torque), it will create a roll bcoz rotation+linear motion = rolling motion. Here friction creates rotation and my push on wheel gives it linear motion so together they create a roll. Note that friction doesn't provide any linear movement to wheel here , it only provides it torque.
Do u agree with me? First tell this then we 'll talk of axles.

If your wheel is massless, it will rotate as it rolls with 'zero' friction force and accelerate at an infinite rate. Already, you seem to have introduced a model which I can't get along with. If the wheel has mass, then this is actually adding to the complexity of the next stage of putting in on a car with an axle.
Your friction, in the above case, is actually in the other direction from in the case of a driven wheel.
And, of course, all this would apply equally to a 'rolling road' on which there is no forward motion of a car. Every motion would be rotary.
I'm still not sure what you are arguing about. Have you spotted a flaw in reality?

 

[dE_logics]

If something is pushing the cart, then under ideal conditions there will be no friction when considering the cart and the hill only.

Now depending on what's pushing the cart, the direction of friction force will be opposite to the direction of motion of the cart (though friction we get the normal reaction (relative to the cart) which the cart needs to move). Now here I've assumed that that the normal reaction is given by the hill itself...it can be other things too like a propulsion system, in which case the hill will not give the normal reaction and the reaction will even not be through friction.

 

[R Power]

first of all let me tell you how i applied force. I simply pushed the wheel. U r confused about top point and bottom point experiencing 10N force i think.
See, a body is made of a number of particles(say infinite), when a body accelerates under effect of a force , each particle of body acc by same amount, that means each particle of body experiences same force. This is what i meant.
So the bottom point or particle at bottom or group of particles at bottom and top will also experience a force of 10N.
This is what I meant.
Now reconsider my model and tell me if friction can provide any linear force?
Bcoz friction is produced due to particles of body at bottom hitting ground due to force i applied to wheel. So friction force equal to 10N and opp in direction is produced. This friction force along with force on particles at top creates a couple.
Now particle(s) at top experienced a force of 10N and same is friction's max instataneous value in opp direction. This creates torque. So there is no friction force left to provide linear motion.
Also in this case friction is in direction opp to wheel's linear movement so why it will provide wheel a linear movement . It will just give torque.
But Doc All said it will provide linear movement.