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Wheel calculations (Power, torque, etc.)

  1. Apr 29, 2015 #1
    Hi there, I've forgotten how to solve these problems, eg:

    If I've got two shafts, two wheels per shaft meaning four wheels in total, each wheel is 60 mm in diameter, and the power consumed by the system in providing a constant horizontal velocity of 0.15 m/s is 200W, what is the torque in each wheel? The wheels are rubber coated, and the wheels are acting upon a steel surface (take coefficient of friction to be 0.7).

    This is how I've solved the problem:
    Power = Torque x angular velocity
    w = v / r = 0.15 / 0.030 = 5 rad/s

    Torque = P / w = 200 / 5 = 40 Nm
    Now since there are 4 wheels (instead of just 1 wheel) AND two shafts with two wheels in each shaft, how do I interpret this torque of 40 Nm? Is the force per wheel just force = Power / velocity or is this force divided amonst the 4 wheels?

    With the friction, since the system is moving at a constant speed of 0.15 m/s, will the thrust force equal the frictional forces? The force here doesn't equal the force calculated via the torque of 40 Nm, is this because of a transmission efficiency?

    Thanks in advance, really annoying how I still haven't grasped these basic concepts (I'm studying Mechatronics engineering at uni)
  2. jcsd
  3. Apr 29, 2015 #2
    Yes, the force is divided amonst the 4 wheels. Yes, the thrust force equals the frictional forces. The thrust force equals the force calculated using a torque of 10 Nm at each wheel.
  4. Apr 29, 2015 #3


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    If this is a real system, it may be worth pointing out that, when going round a curve, the speeds of the wheels need to be different, to avoid slipping and power loss. In practical systems, it is common practice to use a (three) differential(s)
  5. Apr 30, 2015 #4


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    At constant speed the sum of all forces equals zero. That comes from Newtons F = m*a. So yes the magnitude of the thrust should equal friction (if there are no other losses).


    How are you calculating the frictional force? It's not just F = μ*N. That would be the maximum force before the wheels start to slip. The actual frictional force will be lower (and hard to calculate - unless you calculate it from the thrust).
  6. Apr 30, 2015 #5


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    This is only true if the wheels are all going at the same speed and in a practical situation you can't be sure of that.
    The "friction" forces in this problem will not involve any loss of power - it could be a rack and pinion system if there is no slippage.
    That value F is the limiting friction force and the actual value could be anything less than μ*N, I think.
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