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Torque question please help

  1. Mar 30, 2008 #1
    The question:
    A plank, of length L = 3.4 m and mass M = 22.0 kg, rests on the ground and on a frictionless roller at the top of a wall of height h = 1.60 m (see the figure). The center of gravity of the plank is at its center. The plank remains in equilibrium for any value of θ >= 59° but slips if θ < 59°.
    a.Calculate the magnitude in newtons of the force exerted by the roller on the plank when θ = 59°.
    b.Calculate the magnitude in newtons of the normal force exerted by ground on the plank when θ = 59°.
    c.Calculate the magnitude in newtons of the friction force between the ground and the plank when θ = 59°.

    image: http://smg.photobucket.com/albums/v231/er1smesp00n/?action=view&current=physics-1.gif

    so I was just solving it as if it was a normal plank against wall style question

    Sum of Forces in X direction=Force of roller - frictional force=0
    Sum of Forces in Y Direction=Normal(upward force from ground)-mass(gravity)=0
    Sum of Torques = Force of roller(height)-mass(gravity)cos(59°)(1.7)

    I know these formulas are wrong since the Normal force does not equal mass*gravity like it should according to the equations above. How do you solve this problem.
  2. jcsd
  3. Mar 30, 2008 #2
    First you notice that the roller does not apply a force purely in the x direction. You tell me the next step ...
  4. Mar 30, 2008 #3
    so i assum then that the roller has a force F(roller) along the y axis in the same direction as the Normal force

    Sum of Forces in y direction=Normal+Force or Roller along Y - mg=0
    Sum of Forces in x direction should remain the same as before Froller=friction
    Sum of Torque= Force of roller along x(height)-Normal(0)-mg(cos59°)(1.7)+Force of roller along y(1.6)/sin59°=0

    but what answer do they want for the force by the roller on the plank?

    and I can't figure out the math I could use... aren't there too many variables now?
  5. Mar 30, 2008 #4
    What direction is the force vector from the roller? Keep in mind that it is a frictionless roller (hint). It might help you to draw the diagram, showing the vectors as well as their x- and y-components.

    There should really be only one unknown for that roller force (hint, again).
  6. Mar 30, 2008 #5
    I don't know the direction of the force roller
    I thought it was only in the x direction because it was frictionless
    then I thought perhaps theres an upwards force also because the plank is not merely resting against it but resting on it
    is it something to do with the fact it's a roller?
    is there only an upward force?
  7. Mar 30, 2008 #6


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    Welcome to PF!

    Hi Natbird! Welcome to PF! :smile:

    As belliott4488 said, draw a diagram … and draw it to show the area round the top of the wall, very large, not like the tiny .gif picture, and with a nice round roller!

    Now can you see what direction the force from the roller is in? :smile:
  8. Mar 30, 2008 #7
    I think what belliott and tinytim are asking you is, is there a relationship between the critical angle and the direction of the roller force? In other words, can you take two unknowns (Froller-y and Froller-x, the legs of a right triangle) and reduce them to a single unknown hypotenuse and the sine or cosine of a known angle?
  9. Mar 30, 2008 #8
  10. Mar 30, 2008 #9


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    Hi Natbird! :smile:

    Nice diagram … I like the colours!

    (But what is that curvy force? The roller is frictionless, and this is a static problem, so there's no turning … that force shouldn't be there. :frown:)

    But, although you've drawn lots of forces, you haven't drawn the force between the roller and the plank.

    Which direction do you think you should draw it (and remember, it has to go through both the roller and the plank …)? :smile:
  11. Mar 30, 2008 #10
    Natbird - sorry to be dancing around the answer, here, but it will really do you good to find the answer yourself, so I hope you're not too frustrated.

    Look at the point of contact between the roller and the plank. The roller is frictionless, so can exert no "side to side" forces, right? Kind of like a frictionless surface that can exert only a normal force.

    What would that look like from a cylindrical roller??
  12. Mar 30, 2008 #11
  13. Mar 30, 2008 #12
    I'm confused how is a cylinder roller different in this case then? it it's Normal force mgcos59*?
  14. Mar 30, 2008 #13


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    Yes, your red arrow is the force from the roller! :smile:

    But your reasoning is weird.

    Forget components … the force isn't there because the components somehow create it … the force is there, well, because it's there!

    You may need to find the components later - they may be vertical and horizontal, or normal, or …

    Wait and see!

    Let's see … you now have the direction of that force … so the next step is to find its magnitude.

    The system is in equilibrium (not moving), so take moments about some point that will help you find it. Which point do you think that should be? :smile:
  15. Mar 30, 2008 #14
    On the most recent FBD you posted, the big red arrow (with the two yellow arrows) is what we've all been hinting at. Before, you had two unknowns (the 2 yellows) and now you have one unknown (red arrow's magnitude) and one known (angle). You should have all you need now. Three equations with three unknowns (normal, friction, and roller forces), after you take tinytim's advice and fix your torque equation.
  16. Mar 30, 2008 #15
    taking the axis at the ground below the roller
    sum of x direction=Fsin59*-friction=0
    sum of y direction=Fcos59*+N-mg=0
    sum of torque=Fsin59*(h)-N(h)/sin59*-mgcos59*(1.7)=0

    solve for F by setting the N= to mg-Fcos59*

    are those formulas correct??
  17. Mar 30, 2008 #16
    Really close! Everything except the torque eq. looks good to me.

    First, what is the point around which you're defining the torque? It's important to pick this point and then to define your r vector relative to that point. It can be anywhere, since there's no rotation and therefore no net torque around any point, so pick a sensible point. Keep in mind that if the r vector to any one of the forces happens to be zero, then you can ignore that force since it exerts no torque around that point.

    Anyway, think through r X F for each force that's not through your selected point, and then make sure you've got the sines and cosines right ... you're almost there!
  18. Mar 30, 2008 #17
    If you choose your axis in a different place you may find that your torque equation simplifies. You eliminated the roller from the torque equation because, since this force acts 0 m from your chosen axis, it exerts no torque. Try to choose an axis that eliminates two terms rather than one.
  19. Mar 30, 2008 #18
    I think the sum of the torques is actually equal to the equation

    since if I take the axis from the point I did the plank would rotate counterclockwise
  20. Mar 30, 2008 #19
    if I take the orgin as the spot where the plank touches the ground would that be easier

    sum of torques=Fsin59*+Fcos59*(h)/sin59*-mgcos59*(1.7)=0
  21. Mar 30, 2008 #20
    This still doesn't look right to me ... try taking it one step at at time:

    Where is the axis of rotation? Just state it for the record.
    What forces exert torques about that axis?
    For each of the forces you just named, what is the r vector, and what is the angle between this r and the force vector?
    For each force, which direction is the torque that it exerts?

    Once you've answered all those, you can sum them up with opposite signs for oppositely directed torques (this should happen automatically if you do the cross products carefully, using the right hand rule correctly) and set the sum equal to zero.
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