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Angles and rotational equilibrium

  1. Dec 26, 2011 #1
    1. The problem statement, all variables and given/known data
    This is not a homework question, I was just curious.

    I'm trying to find the angle θ that an unbalanced arm produces in the following case:


    2. Relevant equations

    3. The attempt at a solution
    r(98 N)(cos θ)=r(49 N)(cos θ)

    However, the cos θ cancels, so how does one solve?
  2. jcsd
  3. Dec 26, 2011 #2
    There is one solution for θ you haven't accounted for.
    If you look at your equation and think of different values of θ, all except one value would make that equation false.

    Mathematically, you are not allowed to divide by zero, so if you consider this one solution, cosθ will not cancel out.

    Also, what if r = 0? Then θ could be anything and it would still be in equilibrium.

    My math Professor had told us a number of times to be careful when cancelling out variables in equations, because what if a variable = 0?

    Instead, you should do it like this:
    r(F1)cosθ = r(F2)cosθ
    r(F1)cosθ - r(F2)cosθ = 0
    rcosθ(F1 - F2) = 0

    now you have r = 0, cosθ = 0 or θ = n∏/2
    Last edited: Dec 26, 2011
  4. Dec 26, 2011 #3
    Thank you for the answer, I will remember not to cancel haphazardly. However, I don't understand how this could be correct.

    This suggests that θ=90° no matter the balance of forces, but this is not the case; everyday experience tells us otherwise.
  5. Dec 26, 2011 #4
    Hmm, can you give an example? Cause I think in the case of gravity and assuming there is no friction or outside forces (besides the pivot point holding the system up), the system would be stable at 90°.
  6. Dec 26, 2011 #5
    A trivial example would be something like this:


    The force exerted as "effort" is less than that of the load, but it remains stable, although it is not at 90° (never mind that it is not possible here due to the ground).
  7. Dec 26, 2011 #6
    Well in your lever example, it looks like the effort applied to the end is always at a 90° angle, and so the torque is always constant. The load on the other hand has a gravitational force that always acts downwards, and so only the component that is perpendicular to the stick contributes to the torque. The parallel components are most likely canceled out by static friction, otherwise the load would start sliding.

    The first example on the other hand, we are assuming that forces on both sides are due to gravity. Since one of the gravitational forces is greater than the other, their perpendicular components will always be unequal, so there will always be a nonzero net torque except when the plank is perfectly vertical.
  8. Dec 26, 2011 #7
    Okay, I guess that wasn't the best example.

    What about this one?

  9. Dec 27, 2011 #8
    There's also the mass of the board to take under consideration.
    Let M be the mass of the board, R be the distance between the board's center of gravity and the pivot point:
    r1mcosθ = r2mcosθ + RMcosθ
    r1mcosθ - r2mcosθ - RMcosθ = 0
    cosθ(r1m - r2m - RM) = 0
    Clearly θ is not 90 degrees, so cosθ is no 0 and we can divide both sides by it, but that's not to say that 90 degrees isn't a solution, it is, but there are other solutions that make this equation true.
    m(r1 - r2) = RM
    R = m(r1 - r2)/M
    We can do a little bit of estimation to see how far the CM of the board is from the pivot point.
    The shoe on the right looks a little closer to the pivot point than the shoe on the left, so let's say the difference, r1 - r2 = 3cm. Each shoe looks to be about 1kg, and the board looks like it's 2.5kg.
    R = 1kg(3cm)/2.5kg = 1.2cm
    As you can see R is quite small, but the fact that the board is relatively heavy compared to a shoe makes the torque the board contributes to be significant.

    Also notice how the board is not pivoting about a fixed point, but a cylinder. Assuming that there is enough static friction to keep the board from sliding, the location of the pivot point is going to vary depending on where the board makes contact with the cylinder (think of a circle and a tangent line). So naturally the board is going to want to reorientate itself so that its center of mass is a bit further away from the pivot point. So maybe the "stable angle" has to do with how that angle affects R.

    What I find strange is why the board is tilted with the shoe closer to the edge up. I think it would be the other way around, but maybe the board isn't perfectly centered with the cylinder to begin with.
  10. Dec 27, 2011 #9
    Ah, I understand now. I think my mistake was not considering the mass of the board for the first case, which is why the answer that it would go to 90° does not work in non-ideal real-life situations.

    Thank you very much, greatly appreciated.
  11. Dec 27, 2011 #10
    Assuming the board is uniform in density the mass makes no difference since all the torques cancel. In an ideal situation on an infinitely small pivot point the board will just tip completely over (or to 90 degrees if it is fixed to the rotation point by a nail or some such), the reason this does not happen with the board on the drums(?) is because the point of contact moves as the board rotates which changes the lengths of the lever arms until the torque is zero.
  12. Dec 27, 2011 #11
    But this occurs on a stationary pivot as well. Try balancing a pencil on your finger where one side is heavier than the other; it reaches an equilibrium.
  13. Dec 27, 2011 #12
    Because as the pencil rotates counterclockwise the left arm shortens and the right one lengthens or the reverse for clockwise rotations. Changing the lengths of the lever arms changes the torques.
  14. Dec 27, 2011 #13
    I don't think that's the only factor.

    What originally prompted this question was playing with a toy car. I held it by the wheel in the middle (making it the fulcrum), and it tipped towards the heavier side. Changing the weight on either side changed the balance accordingly. It did not go to a vertical position when it was unbalanced.

    Here's a picture:

  15. Jan 2, 2012 #14
    A model car is a little different, this is why you get this seemingly odd result, the car will rotate until the center of mass is under the fulcrum. To test this do your experiment again but when the car balances draw an imaginary line directly down through the fulcrum to the ground, the center of mass should lie on that line and so you should be able to balance your car on your finger (or a paddle pop stick depending on the size) at that point.
  16. Jan 2, 2012 #15
    Oh, I see. That makes a lot of sense, thanks.
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