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Finding angle required for equilibrium on a slope

  1. Feb 1, 2015 #1
    1. The problem statement, all variables and given/known data
    A heavy uniform cylindrical drum is placed, with its axis horizontal, on a slope inclined at an angle α
    to the horizontal. It is prevented from sliding or rolling down the slope by a triangular wedge. The weight of the wedge is negligible compared with the weight of the drum. The angle at the point of wedge is β, and the coefficient of friction between the wedge and the slope is μ, where μ>tan(α). Show that, however smooth the surface of the drum, the wedge will keep it in equilibrium provided that β is between α and arctan(μ).
    2. Relevant equation.
    Force of friction <= coefficient of friction * magnitude of normal contact force
    F=ma
    Moment of a force=force * perpendicular distance to pivot
    3. The attempt at a solution
    Resolving along and perpendicular to the plane of the slope and taking moments about the center of mass of the drum gives equations involving: the weight of the drum; the contact forces between the wedge and the drum and between the wedge and the slope; and the angles α and β. However I cannot find a way to manipulate them in order to derive the required inequality.
     
    Last edited: Feb 1, 2015
  2. jcsd
  3. Feb 1, 2015 #2

    PeroK

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    When you drew your diagram, did you see why β must be at least α?
     
  4. Feb 1, 2015 #3
    Yes, β must be at least α so that the wedge face is above the horizontal.

    With regards to the upper bound on β, something of the form tan(β)<μ must be shown, but I still can't find a way of eliminating the various forces in the ratio sinβ/cosβ
     
  5. Feb 1, 2015 #4

    PeroK

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    If the cylinder is in equilibrium, how many forces are acting on it? Have you drawn a free-body diagram?
     
  6. Feb 1, 2015 #5
    Yes, I have drawn a free-body diagram.

    There are 5 forces: the weight of the cylinder; the normal contact force and parallel friction force from where the wedge touches the cylinder; the normal contact force and parallel friction force from where the cylinder touches the slope. Taking moments about the centre shows that the two frictional forces must be equal in magnitude for the cylinder to be in equilibrium.
     
  7. Feb 1, 2015 #6

    PeroK

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    What did you get for the two normal forces? Magnitude and direction?
     
  8. Feb 1, 2015 #7

    PeroK

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    Okay, I just checked this out. There is a neat solution, where you don't actually have to calculate those forces.

    How far have you got? It's difficult to help if you don't show what you've done.

    Hint: concentrate on the force between the cylinder an the wedge.

    PS I'd assume the cylinder is smooth and ignore those frictional forces on the cylinder. I'm not sure they'd be relevant in any case.
     
  9. Feb 1, 2015 #8
    Okay, I think I have gotten a solution along the lines of your hint, though I don't believe we need to neglect friction.

    Considering the forces on the wedge: we have the normal contact force from the cylinder, R; the force of friction, F (parallel to the wedge face and acting in a direction into the slope); a reaction force from the slope, G; and a force of friction from the slope (acting up the slope).

    Resolving perpendicularly to the slope face gives G=Rcos(β)+Fsin(β). Also, F/R<=1 (since the maximum of F is some coefficient of friction * R).

    We want the wedge to be in equilibrium, so we must have:
    Rsinβ<=μ(Rcos(β)+Fsin(β))+Fcos(β)<=μR(cos(β)+sin(β))+Fcos(β)

    sin(β)<=μ(sin(β)+sin(β))+(F/R)cos(β)<=μcos(β)+μsin(β)+cos(β)=(μ+1)cos(β)+μsin(β)

    Thus,

    tan(β)<=(μ+1)/(1-μ)

    (μ+1)/(1-μ)>μ, so tan(β)<μ is a sufficient condition (provided arctan(μ)>α)
     
    Last edited: Feb 1, 2015
  10. Feb 1, 2015 #9

    PeroK

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    I'm not sure how you managed to factor F out of your equation! Beta must depend on F.

    note that the normal force between the wedge and slope is increased by any friction between the wedge and cylinder and the tangential force is decreased. So, with friction on the cylinder beta can be greater than for a smooth cylinder.

    For a smooth cylinder you should get ##tan\beta = \mu##

    Your equation can't be correct!
     
  11. Feb 1, 2015 #10

    PeroK

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    If you simplify your equation correctly then you do get a neat solution involving R and F, which reduces to the above when F = 0.
     
  12. Feb 1, 2015 #11
    In the line Rsinβ<=μ(Rcos(β)+Fsin(β))+Fcos(β)<=μR(cos(β)+sin(β))+Fcos(β), I used the fact that F/R<=1 (which implies F<=R) as follows:

    Rsinβ<=μ(Rcos(β)+Fsin(β))+Fcos(β)=μRcos(β)+μFsin(β)+Fcos(β)<=μRcos(β)+μRsin(β)+Fcos(β)=μR(cos(β)+sin(β))+Fcos(β)

    Yes, I agree that any friction between the wedge and cylinder will increase β. (μ+1)/(1-μ)>μ (and tan(β)<=(μ+1)/(1-μ)) suggest this.
     
  13. Feb 1, 2015 #12

    PeroK

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    You should get:

    ##tan\beta = \frac{\mu R + F}{R - \mu F}##
     
  14. Feb 1, 2015 #13

    haruspex

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    There can only be a frictional force between wedge and cylinder if there is also one between slope and cylinder. (Take moments about cylinder's centre.) If there is friction at both places, the magnitude of that friction is indeterminate. Both zero will always be a solution, since no friction is required to hold the cylinder in a vee. As noted, you're looking for the least upper bound on beta, which is when F = 0.
    Coefficients of friction can be much greater than 1.
     
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