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Wheelbarrow over a step

  1. Nov 26, 2013 #1
    1. The problem statement, all variables and given/known data
    A person pushing horizontally a uniformly-loaded 45.2 kg wheelbarrow is attempting to get it over a horizontal step. The maximum horizontal force that person could apply is F = 410 N. What is the maximum height of the step, as a fraction of the wheel's radius, the person could get the wheelbarrow over? Assume that the height of the step is h=nR.

    2. Relevant equations
    τ=Fr
    F = mg



    3. The attempt at a solution
    Calculating the wheelbarrow at the edge of step.

    hence
    Fr = mg

    and I'm stuck here of how to apporach on to the next step
     
  2. jcsd
  3. Nov 26, 2013 #2

    CWatters

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    This isn't a question about friction. Try drawing a diagram for a step height equal to about 1/2 the radius of the wheel (n=1/2). What forces act on the wheel?
     
  4. Nov 26, 2013 #3

    haruspex

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    To draw CWatters' diagram, you need to choose some point in the process of getting the wheel up the step. It should be obvious at what point the force needed is maximised; if not, put in an unknown for, say, the angle that the radius from wheel centre to point of contact with step makes to the horizontal.
     
  5. Nov 27, 2013 #4
    diagram.png

    This is the image I have came out with.
    From this picture I can conclude that
    τ = Frsinθ

    The known values are:
    F = 410

    I know that Force of gravity is pushing down and the torque for wheel pushing upward. Hence,
    mgR/2 = FRsinθ

    (mg)/(2F) = sinθ

    Is the approach correct? because I could not find the fraction for it
     
  6. Nov 27, 2013 #5
    I don't think you have to worry about torque here. Just find the normal force upward as it depends on how high the step is and set it equal to the weight.
     
  7. Nov 27, 2013 #6

    haruspex

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    F is applied horizontally.
    Which angle is theta? Is it what I suggested?
    How did you get mgR/2?
     
  8. Nov 27, 2013 #7
    Theta is the angle between the radius and the point of contact with the step. However, I am confused with "the angle that the radius from wheel centre to point of contact with step makes to the horizontal."

    I got mgR/2 from the thought of finding the center of mass
     
  9. Nov 27, 2013 #8

    CWatters

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    Can I suggest a slightly different diagram. I would also look at the torques about the point where the wheel meets the step.
     

    Attached Files:

  10. Nov 27, 2013 #9
    Haruspex,

    Wheel.png

    Diagram.png

    I have decided to remove my thought on torque and drew the body diagram again.

    sin θ = (R-h)/R
    θ = 1-h/R
    θ = 1-n

    Is this what you are suggesting?
     
  11. Nov 27, 2013 #10
    So according to your diagram
    tanθ=Mg/2F

    The torque where the points meet. The formula for torque is
    τ=Frsinθ
     
  12. Nov 27, 2013 #11

    haruspex

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    I think you mean sin θ = 1 - n.
    The new diagram is better, but you have the normal wrong. When a point contacts a smooth surface, the normal is the normal to the surface, in this case the wheel. Correct that, then write out the horizontal and vertical statics equations.
     
  13. Nov 27, 2013 #12

    CWatters

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    I was thinking that to rotate about the point of contact with the step there must be a net clockwise torque.

    Clockwise you have F(rSinθ)

    Anticlockwise you have 0.5mg(rcosθ)

    The brackets are just to highlight the distance term.

    The factor of 0.5 comes from the fact that roughly half the weight is supported by the wheel on one end and half by the man on the handles at the other. I guess it depends what the expression "uniformly distributed" means but I reckon this is a reasonable approximation.

    So I think the condition to climb the step is..

    FrSinθ - 0.5mgrcosθ > 0

    Edit: I'm happy to be convinced that another approach is better !
     
  14. Nov 27, 2013 #13

    haruspex

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    It says the force is horizontal. I think you have to take that literally or there's not enough information to answer the question. Of course, when pushing a wheelbarrow on the flat the person has to take some of the vertical load, but typically the handles are long enough that the wheel takes most of it. When trying to get up the step, there's no longer a need for the person to apply any vertical force. It can be purely horizontal now.
     
  15. Nov 27, 2013 #14
    So after changing the direction of the normal force. Here are the force vectors I get:

    F = Fn
    Fg = ma

    τnet = 0

    τnet = -Frsinθ + mgR/2sinθ


    Fr = mgR/2sinθ
    F = mg/2sinθ

    2F/mg = sinθ

    2F/mg = 1-n
    Am I missing something? Or did I do write an equation wrong.
     
    Last edited: Nov 27, 2013
  16. Nov 27, 2013 #15

    haruspex

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    You seem to have made the normal force (Fn?) horizontal now. It should be radial to the wheel. F is horizontal, mg is vertical. There could also in principle be a frictional force from the step tangential to the wheel. You can show there is not, but you should not assume it.
     
  17. Nov 27, 2013 #16
    if Fn = normal force.
    So, if it is tangential to the wheel, the best way to make it work would be the torque. There is no frictional force in this case.

    τ=Frsinθ

    F = Fn in such case.

    is this what you are suggesting?
     
  18. Nov 27, 2013 #17

    haruspex

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    No, not tangential, radial. If there is a frictional force, that will be tangential.
    If you take moments about the point of contact with the step, as CWatters suggested, you won't need to worry about this normal and frictional pair of forces. Neither will have any moment about that point. What torques do mg and F have about that point?
     
  19. Nov 27, 2013 #18

    Iw2 would be the radial.

    Iw2 = F

    τnet = -Frsinθ + mgR/2sin(90-θ)

    I guess these are the formulas
     
    Last edited: Nov 27, 2013
  20. Nov 27, 2013 #19

    haruspex

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    That's a formula for (twice) rotational kinetic energy!
    Close, but lose the /2. That assumes the wheelbarrow's centre of mass is half way between the wheel and the handles. They're designed to have the centre of mass much further forward (when loaded), as close as possible to being directly over the wheel without risking tipping forwards. So you should take the weight as acting down through the centre of the wheel.
    And r = R, I hope.
    At equilibrium, what is the value of τnet?
     
  21. Nov 28, 2013 #20
    τnet = 0;

    FRsinθ = mgRsin(90-θ)

    F/(mg)=cotθ

    (mg)/F = tanθ

    Solve for θ

    since sinθ = 1-n.

    This way we could solve for n. I guess this is the way
     
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