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A puzzle about the open road

October 10, 2010
tags:

Draw a free body diagram for a car traveling down the road at constant velocity. Many of my kids originally want to say things like “the engine pushes the car forward” but with a question or two, they see that that same engine won’t do much on ice, so it must be the road pushing the car forward. This leads them to this FBD:
FBD for a car traveling at constant velocity

But I have a bit of a problem with this I’ve never been able to fully resolve. Kids also want to draw a backward frictional force in addition to the drag force. And then they get all hung up on how the road can both be pushing the car forward and backward at the same time.

I’ve read about rolling resistance, and some great articles on what makes a car go (this makes a lot of sense, and really digs into a understanding of energy). But I’m still a bit perplexed about whether there should be a frictional force opposite the motion of a car driving down the highway (I think there should), and whether this force would act on all four tires, or just the drive wheels. If this rolling resistance force only acts on the non drive wheels, it would seem to be you could escape this force entirely with all wheel drive, which doesn’t make sense. Also, how big is this force compared to the drag force? My guess is that at highway speeds, rolling resistance is much smaller than the drag force.

So I’d welcome any of your insights to this problem. Can the road exert two frictional forces on the tire in opposite directions?

10 Comments leave one →
  1. October 10, 2010 4:06 am

    Simple answer for the kiddos:
    http://wcydwt.posterous.com/bicycle-wheel-motion-blur

    The part of the wheel that touches the ground is at rest. There is only static friction which keeps the tires from slipping and pushes opposite the direction of the wheel rotation (which is in the direction of the center of mass motion of the wheel). I think this is what most people call traction.

    Of course, the wheel actually deforms when touching the ground so there is rolling friction (which slows down the rotation of the wheel) in addition to traction (which pushes the wheel forward).

    • October 10, 2010 4:14 am

      That’s what I was thinking, but this would say that there are no frictional forces acting opposite the direction of the motion of the car, right?

      And in terms of forces, the traction force is the same force as the rolling friction force that slows down the wheel’s rotation, no?

      And suppose you suddenly took your foot off the gas, greatly reducing the traction force on the car. This would seem to indicate that the only force responsible for changing the momentum of the system of the car would be the drag force, correct?

      • Noah permalink
        October 10, 2010 10:17 pm

        John’s question has bothered me too. In particular his point about all-wheel-drive vehicles. For two wheel drive vehicles it’s pretty clear that friction between the road and the drive wheels provides a forward force while friction between the road and passive wheels provides a backward force. For the all-wheel-drive situation the frictional force from the road must be forward on each wheel when your foot is on the gas (fighting air drag) but when you take your foot off the gas, the direction of friction from the road switches. The alternator and internal friction make the wheels want to stop turning. The forward motion of the car and a backward frictional force from the road induce the wheels to roll without slipping.

        Thinking about friction’s role in walking provokes similar questions. Walking at a steady pace the net frictional force is zero but the direction of friction must be different at different point in our stride. If you insert a banana peel in front of someone’s front foot they fall on their back. If you could somehow insert it under their back foot they’d fall forward.

  2. Theron permalink
    October 11, 2010 1:59 am

    My take is something like this: Rolling friction is an internal force (torque) that works in opposition to the drive torque provided by the engine. The bottoms of the tires only experience the net torque exerted on them by the wheels. Rolling friction only occurs where parts slide past one another (axles, transmission, engine). The reason the car slows down when you take the foot off of the gas is that the net torque at the wheels switches directions, which would switch the direction of the torsional force exerted by the tires on the road from rearward to forward. I would assume that your students are only interested external forces, and should therefore ignore rolling friction.

    • October 11, 2010 2:06 am

      Theron,
      I like this, but I still feel like somethings missing. Let’s let the car be our system. Then N2, rephrased says

      \frac{dp}{dt}=F_{\textrm{net,ext}}

      So if rolling friction is an internal force, it can’t be responsible for changing the momentum of the system right? Then all the loss of momentum is due to the drag force?

      • Theron permalink
        October 11, 2010 10:56 pm

        In my conception, rolling friction is only experienced by the tires when the engine provides less ‘forward’ torque than the rolling friction provides ‘rearward’ torque. The tires only experience the net torque at the end of the axle.

  3. October 11, 2010 11:28 pm

    But aren’t the wheels still part of the system? How can this change the linear momentum of the car the forces are internal to the car?

Trackbacks

  1. In which I find the solution by going back in time « Quantum Progress
  2. A free body diagram challenge: the constant velocity buggy | Quantum Progress

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