
#19
Dec709, 06:18 AM

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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.




#20
Dec709, 07:08 AM

P: 272

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? 



#21
Dec709, 07:34 AM

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P: 40,883

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? 



#22
Dec709, 08:12 AM

P: 272

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.




#23
Dec709, 08:24 AM

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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. 



#24
Dec709, 08:35 AM

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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'. 



#25
Dec709, 08:37 AM

P: 272

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. 



#26
Dec709, 08:41 AM

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P: 40,883

Answer my question about the accelerating car. 



#27
Dec709, 08:43 AM

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



#28
Dec709, 08:55 AM

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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? 



#29
Dec709, 09:28 AM

P: 730

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. 



#30
Dec709, 09:59 AM

P: 272

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. 



#31
Dec709, 10:22 AM

P: 272

I got it. It just struck my mind now! and was so simple.
1.) If I push a wheel then friction helps in rotation by producing torque and my push creates a linear movement (tends to slide the wheel) and together they produce a roll. 2.) If an axle gives torque to wheel, friction provides linear movement and axle provides torque and together they produce a roll. It was simple. But Doc ALl confused me: 



#32
Dec709, 12:13 PM

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Bearing in mind that a couple (torque) is defined as the product of force times distance, I can't recall any textbooks being fussy about where that force came from!




#33
Dec709, 12:19 PM

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If there is a ball on a mat, I can pull the mat towards me. The ball will move towards me AND start to rotate. Nicht war?




#34
Dec709, 12:31 PM

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P: 40,883

The torque is determined in a similar manner. The two 10 N forces you exert produce 0 net torque. The friction force thus produces the net torque. 



#35
Dec709, 01:05 PM

P: 272

OK Doc leave all this(since i've understood) , come to the where we start from , when a wheel rolls up a hill due to torque provided by axle, direction of friction will be up the hill, right??
I think u agreed to me. But in one of my books they have taken it down the hill. Is the book wrong or we are wrong somewhere? 



#36
Dec709, 01:56 PM

P: 279

A body of mass m and moment of ineria J taken about its central axis rolls down an incline plane with friction limited to the surface contact patch. The body is assumed to be rigid.
Taken in the direction of motion on the plane, the sum of forces acting on the rigid body minus the mass times acceleration equals zero. Let F be the force of resolved gravity and f be the friction, then acceleration is given by: [tex]a = \frac{Ff}{m}[/tex] The torque is given by: [tex]T = fr[/tex] Angular acceleration about the central axis: [tex]\alpha = \frac{fr}{J}[/tex] For a body of mass m, the friction and rates of acceleration are functions of moment of inertia J, so bodies of different inertia have different rates of acceleration down the plane. Notice a force acting at the center of mass, or at the axis which includes the center of mass, generates zero torque. Also the bearings in a mechanical axle support a force but not a torque. The brakes on such a machine require a structure to couple braking torque to the frame, bypassing the axle, or the brakes would just go round and round with the wheels. 


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