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Mechanics question: Roller passing over a step

  1. Feb 18, 2017 #1
    1. The problem statement, all variables and given/known data
    A gardener pulls a roller of mass 85 kg over a step. The roller has a radius of 0.25 m. The handle is attached to an axle through the centre of the roller at an angle of 45 degrees to the horizontal.

    Here's a link to the diagram below, it's figure 1 and 2 on question 18.

    https://books.google.co.uk/books?id...ing a roller of mass 85kg over a step&f=false


    1) Determine the magnitude of the force that will just move the roller when the force is applied as shown in figure 1.

    2) Determine the force which would be required if the handle was pulled horizontally.





    2. Relevant equations

    Moment of a force = force x perpendicular distance

    3. The attempt at a solution

    I don't know where to begin. I'm guessing you have to take the moments about a point but the question doesn't make sense to me. I would have thought the force would have depended on the steepness of the step which isn't given.

    Any help would be appreciated, thanks.
     
  2. jcsd
  3. Feb 18, 2017 #2

    BvU

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    First step is to find the point around which the rotation takes place. Second step is a moment balance around that point. a fraction more force and it goes, a fraction less and it doesn't. So that is the force they are asking for.

    Big help is to make a free-body diagram of all the forces acting. For the moment balance, you don't need all of them, fortunately.
     
  4. Feb 18, 2017 #3
    The way i would solve it is the following:
    You can consider a point A ,which is where the roller touches the step, as your rotation axis and you can assume that positive torque is counter-clockwise. So now you have to ask to ΣΤ>=0 (T=torque) and you will find the lowest value of F so as the roller to move.

    PS: Dont forget to analyze the W in compoments or use any other methods of finding torques if you dont know this one.
     
  5. Feb 18, 2017 #4

    BvU

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    Giving it away eh ? Let mtig ponder over his exercise and learn from it !
     
  6. Feb 18, 2017 #5
    You are right, i thought to help him with this one and let him practise with same type of excersices after.
     
  7. Feb 18, 2017 #6
    Thank you for the replies. I understand what the question is asking now I think.

    anticlockwise moment = F x 0.25

    clockwise moment = 0.25 x 85 x 9.81 x cos 45

    F = 85 x 9.81 x cos 45

    So F = 590N

    I am still trying to understand where the moment of the forces are coming from. It seems like the force from the person is providing an anti-clockwise moment. But then the contact between the step and the roller is also providing an anticlockwise moment. Or is the contact from the step because of the result of the force applied by the person so they are the same moment?

    I'm also struggling to see how the component of weight is providing a clockwise moment?

    Thanks
     
  8. Feb 18, 2017 #7

    BvU

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    Think again. Wrt what point are you setting up the moment balance ?
     
  9. Feb 19, 2017 #8
    So I set up the moment balance where the edge of the roller is in contact with the step.

    The force from the centre of the roller applied by the person would be my anticlockwise moment.

    I am unsure where or how a component of the weight of the roller provides the opposite moment of force.
     
  10. Feb 19, 2017 #9

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    Thing to do is make the drawing ...
     
  11. Feb 19, 2017 #10
    Yes I've done one now and I understand it now, thank you for the help. What got me I think was that on moment questions I am used to a straight bar as oppose to something round.

    Thanks.
     
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