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Homework Help: Torque Calculation. Short- Am I correct?

  1. Apr 26, 2012 #1
    1. The problem statement, all variables and given/known data

    What is the torque on the steering wheel?

    2. Relevant equations

    I don't know how to draw on here.

    Steering wheel is circular.
    It has a Diameter of 300mm
    A force of 20N acts downwards toward the 3 oclock position
    A force of 10N acts Upwards toward the 9 oclock position

    3. The attempt at a solution
    Torque=Force X Distance


    Torque=10 X (0.3)

    This is my first torque calculation and I am working unadvised. I would like to know if I am doing it correctly by using the diameter instead of the radius. Also If I have made any mistakes could somebody please explain what I have done wrong! Thank you in Advance!
  2. jcsd
  3. Apr 26, 2012 #2
    Hello there !

    When you are talking about Torque you should ALWAYS mention your relevant point ( ie Torque about a point A on the rim of the steering wheel or at its centre etc ).Don't forget that Torque is a vector !!!

    Torque about the centre of the wheel :

    T= 20 x 0.15 + 10*0.15 = 30*0.15 = 4.5 KGm !!!

    Both forces are trying to rotate the wheel to the same direction so you should NOT subtract them !
    Last edited: Apr 26, 2012
  4. Apr 26, 2012 #3
    Ok so I was completely off the mark on that one!

    You have used a radius perpendicular to the application point of the torque.
    And you have resolved the torque for both forces and added them...
  5. Apr 26, 2012 #4
    Exactly . if you are confused about the Forces ( ie when they look that they are in the opposite directions ) you should try "rotating" one of the forces until you reach the other one so that you can decide whether you should subtract or add them.
  6. Apr 26, 2012 #5
    Thank you for the tip! You are very helpful!
    What about if the point of application of the force is away from where you want to calculate the force? For example a spanner moving a nut... Would it have no effect or would I use the entire length of the spanner up to the centre of the nut?
  7. Apr 26, 2012 #6
    If you grabbed the spanner from its end then you would use its entire length , if you held it from its middle then you would use L/2 etc . The distance that you will need is the distance from the axis of rotation ( so in the wheel it was the radius , and in this case of the spanner it is its whole length because the axis of rotation is passing perpendicular to the nut and passing through it ).
  8. Apr 26, 2012 #7
    That's why steering wheels in public buses are huge because the wheels of the bus are also huge . The bigger the distance from the axis of rotation the easier it is to rotate anything. Try , opening a door by pushing it from it's fixed end...LOL . Or try steering a lorry with a 20 cm steering wheel . You got the point i hope :P
  9. Apr 26, 2012 #8
    Thanks for your help!
    I am pretty sure I understand it now!
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