The wheel on old Horse-Cart problem

  1. There is a site that explains the old problem of why the horse can pull the cart and move even though Newton's third law says every action has an equal and opposite reaction. It does it well except on this page when it summaries everything.

    I don't quite agree with how it deals with the frictional force.

    For a wheel, when it spins on normal road, it exerts a force backwards and the ground via friction exerts a force on the wheel forwards. This is what makes the wheel hence cart move forward on a normal road. If there is no friction than the wheel will just spin on the spot. Imagine spinning the wheel in air, it will send air molecules backwards but the force from the air molecules is not directional so do not allow the wheel to travel forwards. The fact that the earth doesn’t move wrt the cart means that for an observer on earth, the cart moves forward.

    On that website, it shows the force by wagon on ground (which is exerted via the wheel from the diagram) is to the right. And the force by ground on wagon (via wheel) is to the left. This would only be correct if the wheel was locked and can’t turn. But that would be dumb and wasn’t specified on the website. So if the wheels are allowed to turn than the website got their facts wrong. Am I correct?

    However, I see that it wants to show that by introducing friction, the net force on the cart is reduced. So it might be hard to encoprate what I suggest because on the one hand they want to show the ground reduces the motion of the cart but in reality for a wheel the ground exerts the forward force. Maybe they could show that the net force is exerted by the ground to the cart via the wheel. How would you show everything correctly and systematically?
    Last edited: Jan 9, 2007
  2. jcsd
  3. russ_watters

    Staff: Mentor

    Yes, it is an oversimplification.
  4. rcgldr

    rcgldr 7,691
    Homework Helper

    No, the wheel exerts a forwards force on the ground, and the ground opposes this force with a backwards force. The diagram in the link is correct.

    The horse's feet produce a backwards force against the ground, and the ground responds with a forwards force. If the cart is going at a constant speed, then the horse's force is canceling out the drag force of the wheel.
  5. Jeff, how can that be? The wheel is spinning clockwise when the cart is facing right as shown in the picture. This means that anything that sticks directly below the wheel, the wheel will make it move to the left. Hence the wheel will exert a left force that is in line with its clockwise motion to the ground. It's just like how people walk. The leg moves backwards or clockwise if you like.
  6. Doc Al

    Staff: Mentor

    The diagram and chart on that website are perfectly correct. (Note that it describes the initial acceleration of the horse/cart from rest.)

    Don't confuse a passively turning wagon wheel with a driving wheel like on an accelerating car. Think of it this way: The wheel begins spinning clockwise, so there must be a force exerting a clockwise torque on the wheel. That force is friction, acting to the left on the wheel. (And the reaction force is the wheel exerting a rightward force on the ground.)

    On an accelerating car, the power wheels are forced to turn clockwise via the engine/crankshaft assembly. That clockwise motion is resisted by the ground, resulting in a rightward friction force on the tires that propels the car forward. In the cart/horse example it's the horse that is driving the cart forward, not the friction on the wheel. (Of course, it's friction on the horse that propels the horse.)

    The example of walking is similar to the accelerating car. You force your leg to move and it pushes off the ground, propeling you forward.
  7. russ_watters

    Staff: Mentor

    It doesn't go into anywhere near that level of detail. It doesn't mention acceleration, torque, differentiate between static and dynamic friction, etc. I'm not sure I would assume it really takes them into account. The arrows are different lengths, but it is tough to glean much useful from that. And if it does take into account the moment of inertia and torque of the wheel, why isn't there a curved arrow for them?

    The specifics of the diagram are so specific to the assumptions of the case that it is tough to make that diagram a generalization or figure out what exacly the specific case they are talking about it. For example, for a reasonably loaded cart on a reasonably well-designed wheel, on a reasonably smooth road the moment of inertia of the wheel and static friction and rolling resistance are all so insignificantly small, why bother mentioning it? And if you do mention it, why shouldn't you also mention wind resistance or even simply differentiate between the static cases and dynamic (equilibrium or accelerating) ones?

    I don't like this example at all - there are too many possible perturbations.
    Last edited: Jan 10, 2007
  8. Doc Al

    Staff: Mentor

    Are you looking at the same diagram that pivoxa15 linked to in his first post? This is just meant to be a simple depiction of some of the 3rd-law pairs involved when the horse begins to move the cart. The direction of the friction force (from the ground, at least) on the wheel is correctly displayed in the diagram. Or were you responding to something else?
  9. Doc Al

    Staff: Mentor

    It's just to illustrate the principle of action-reaction pairs in Newton's 3rd law. Static friction might be small, but it is necessary in order to start the wheel turning.

    You are thinking too much like an engineer; what's needed here is to think like a high school physics teacher. :wink:
  10. BobG

    BobG 2,351
    Science Advisor
    Homework Helper

    I agree with Russ. The diagram is an oversimplification.

    If the cart is loaded, what bears the weight of the load? Somehow, the force of the load has to get transfered to the ground.

    That's transfered by the bottom of the axle pushing on the bottom of the hub, which also pushes the bottom of the wheel into the ground. That's creating a lot of friction to turn the wheel, especially if weight of the load bows the axle. With no ground friction, the wheel wouldn't turn. The resistance of the ground is what allows the wheel to overcome that friction and go ahead and turn anyway.

    The better the lubricant, i.e. the better its ability to reduce friction even with a heavy load pushing against it, the less force it takes to keep the cart moving for a given load.
  11. russ_watters

    Staff: Mentor

    Perhaps, but you still used a lot of words and mentioned some forces that didn't appear on the page. If it has arrows for friction between the ground and the wheel (it doesn't say wheel), why doesn't it also mention the friction between the wheel and the axle separately?

    What I would prefer to use here is a horse pulling a plow. There are no torques and the dynamic friction (as opposed to the horse's internal physiology) is what limits the speed. No need then to wonder if they really took those other things into account. Even better, all the forces are equal to each other. All you have is:


    Yeah, maybe I'm nitpicking, but I do that. No biggie, I think we all understand it even if we disagree on its utility.
    Last edited: Jan 10, 2007
  12. Doc Al

    Staff: Mentor

    True, "wheel" was my word. What the page mentioned was the force between ground and wagon--but the ground only interacts (horizontally) with the wagon via friction on the wheel. I indentified that force as friction acting on the wheel because the OP's confusion had to do with the relationship between the direction of the wheel's turning and the direction of the friction force.

    No need to mention friction between wheel and axle (take the simple case and assume there is none, if you like), as that is internal to the "wagon". To start the wheel turning, you must have some static friction from the ground acting on the wheel in the direction specified. (Of course the amount of friction needed depends on various unspecified factors, including axle friction.)

    As you say, no biggie.
  13. I think I am slowly sinking in what's in this post.

    I was thinking of a driving wheel on a car but the cart is about horse pulling the cart to the right and the ground producing friction which turns the wheel clockwise. If the horse continues to pull the cart than the friction will gurantee the turning of the wheel hence the successive right movement of the cart. The site later claimed that the wheel was to reduce friction which now makes sense. Friction is much less turning the wheel than if pulling without wheels because of reduced contact with the ground.

    As you said, the horse is like a car and moves forward by exerting a force to the ground and the ground propels the horse forward via friction - just as shown on the diagram.
    Last edited: Jan 11, 2007
  14. I don't quite get what the diagram tries to "explain", but here's how I see the situation:

    The horse is pushing the cart. It just does it from the front of the cart by pushing its chest against the harness.
    Assuming the axle is ideally frictionless (or at least negligible), then the only friction involved is that which keeps the horse's hooves from slipping backwards.
    The load on the cart supported by the wheels is not much different from a load balanced on top of a stick standing up on the ground, except that when you tip it over, another stick magically comes down in front of the first stick, and there are an infinite number of magic sticks, so the centre of gravity of the load does not drop.

    In order to get going, the horse initially has to take a step *back* so that his centre of gravity is too far forward and he topples forward in a bit of an arc (unless he's on some ideal ice, and he falls down). Fortunately, he can move his feet forward (but not too far forward), extend them to gain altitude, and almost regain his balance. He keeps toppling forward and rising, but in a fairly smooth fashion.

    Once he overcomes the inertia of the cart and load he does not have to work so hard, so his steps can be placed farther forward each time, and there is less "toppling" because his centre of gravity is more nearly above his hooves.

    Since we assume the cart axle is frictionless, then we can dispense with it altogether and treat it as a bigger horse with legs that do not bend (much).

    The critical thing (in my model) is the notion that the horse is continually toppling forward with his centre of gravity in front of his feet, and that his first step has to be backwards with half of his feet. His hooves do not so much claw at the ground to pull his body forward, like some sort of severed hand in a horror movie (or the drive wheels of a tractor).

    If the axle friction does exist, then the horse has to compensate by allowing himself to be a bit more off-balance than before, and he has to raise himself from a lower point at each step than before. That's more work.
  15. rcgldr

    rcgldr 7,691
    Homework Helper

    The horses center of gravity is always in front of his rear feet, so there's no issue with the horse being able to generate a backwards force on the ground from it's hind legs. However the horse's center of gravity is always behind it's front legs. Since the horse has 4 legs, not just 2, balance isn't an issue. Leaning forwards can help generate more force, but for mild forces it's not required.

    If the cart only has two wheels, then normally the center of gravity is set just a bit in front of the contact point of the wheel and ground so that there's is a bit of downforce on the harness on the horse. I'm not sure if there would be an advantage to have the cg behind the axle in the case of a racing cart. With 4 wheels, (or 3), the stability problem is eliminated.
Know someone interested in this topic? Share this thead via email, Google+, Twitter, or Facebook

Have something to add?