# Direction of Friction!

by R Power
Tags: direction, friction
 Sci Advisor PF Gold P: 10,214 For any of this to work, you need a frame. So why not assume that the wheel is massless and that the frame has the mass. Then we need not consider Moment of Inertia - just forces and mass. No one seems to have read / understood / responded to my statement that it can all be reduced to levers. The wheel thing just adds confusion. And you can even forget friction if you drive the car / lever on a rack rather than a friction surface. All the basics will remain the same - the essence shines through.
 P: 279 I think the study of bodies rolling downhill is a better approach in terms of common sense. Add a counter-torque acting at the central axis in the equations above, and when F - f = 0 the acceleration a = 0 and rotational acceleration alpha = 0, as when a brake is applied. The moment of inertia J cannot be eliminated from problems of motion because it absorbs rotational kinetic energy, which takes energy away from the translational rotational energy. Place a body on a frictionless plane and the initial potential energy U = mgh, or mass times standard gravity times height. The only way to account for the slower exit velocity of rolling bodies is to store energy in the rotational inertia J. I know inertia complicates the example a bit, but I think R Power is trying to visualize the common sense interpretation of rolling motion in bodies and vehicles, so it is worth the effort to make some case studies based on a traditional dynamic solution framework. Your frame of reference may have merit, but I'm not sure J can be eliminated except under the assumption of constant velocity, and in downhill motion, that is not the case.
 Sci Advisor PF Gold P: 10,214 We really need to sort out the model we are discussing. Is the wheel a wheel on a massless frame or is it a massless wheel on a massive frame? That would be a much better model to study as it is realistic. If we are talking of cars, the wheel MI is negligible, surely. If there is no friction, the wheel will not rotate in any case and the car will go nowhere - just slipping down the hill, perhaps. This thread just goes in circles because no one has decided what we're talking about. There has been some serious nonsense written as a consequence of several misunderstandings, I fear.
HW Helper
P: 6,795
 Quote by R Power Consider a wheel rolling up the hill by some external force. What will be the direction of friction?
Insufficient information. Is the external force greater than m g sin(θ), where θ is the angle of the hill? You need to know the direction of the net force.

Net force = (external force) - (m g sin(θ))

If the net force is up hill, then the friction force is down hill, if the net force is down hill, then the friction force is up hill. If the net force is zero, then there is no friction force. The direction of the velocity doesn't matter, only the direction of the acceleration.

This ignores rolling resistance which can apply an opposing force and opposing torque at the same time to decelrate a rolling object free of any other forces.
 P: 272 If i give a force to a wheel uphill where net force is in uphill direction. That's ok! But if an axle provides torque to a wheel up hill as if a car moving uphill, then what will bw direction of friction? I think friction will act uphill i.e in the direction of movement of car.
 Sci Advisor PF Gold P: 10,214 Of course it will. It's the only force in an uphill direction to make the car go that way. It's a REACTION force - that means it acts against the force causing it. If the (driven) wheel pushes down the hill the car is pushed uphill because of friction. When g dominates and the car runs downhill, the friction force is uphill and slows the car down. This is all so elementary. Just think about Newton's laws and apply them here, rigorously. Yet again, I don't think the situation has been described unambiguously and understood by both sides of the argument. Are we talking driven vehicles or trailers? The difference is significant. There is no point continuing until that's cleared up.
 P: 272 I just wanted to confirm this only in the beginning and even I don't know where this thread has gone. In one of my books direction of friction on wheels accelerating up is taken downwards, so I thought I may be wrong in my mind and just wanted to confirm. That was all. And yeah! we were comparing driven vehicles and trailers, !
PF Gold
P: 10,214
 Quote by R Power I just wanted to confirm this only in the beginning and even I don't know where this thread has gone. In one of my books direction of friction on wheels accelerating up is taken downwards, so I thought I may be wrong in my mind and just wanted to confirm. That was all. And yeah! we were comparing driven vehicles and trailers, !
"Wheels accelerating up"??? Driven up or driving up?
That is the whole point. The answer is different for each case.
 P: 272 Driven up by axle. No trailer , just like drive wheels of a rear wheel drive car.
 Sci Advisor PF Gold P: 10,214 Ah well, that appears to be wrong. The force ON the ground is downwards; it has to be for the reaction (friction) force to push the car uphill.
P: 279
 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!
This sketch shows the rear arm of a shaft powered motorcycle. For simplicity let all torques and forces act on the rear wheel. If the resolved weight for F acts to the right (downhill) a braking torque must be applied (reaction friction force f acts uphill) such that velocity is constant. Thus the motorcycle is either at rest or at constant velocity with F = f.

I hope this picture is "worth a thousand words" as far as the confusion on this thread.
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