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Mechanical advantage: lever arm

  1. May 19, 2016 #1
    1. The problem statement, all variables and given/known data
    In order to lift the 30 kg mass, a Force F is applied to the massless meter stick shown below. How much weight must the fulcrum bear if the mass is to be lifted?

    2. Relevant equations
    R/E = l/L

    3. The attempt at a solution
    So I began with the formula above and this is my set up:

    300/F = 0.25/0.75. I got 900 N for F. However, the problem asks about the fulcrum. Logically I would add the 900 N with the 300 N of the resistance force by the mass, I get 1200 N. Since the answer is 400 N, it appears that I have to divide the Force lever arm distance to determine the force on the fulcrum? I think this is the right reasoning here but would appreciate any insight in to this problem.
     

    Attached Files:

  2. jcsd
  3. May 19, 2016 #2

    Doc Al

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    Better check that formula. Another way to check: clockwise torques must equal counter-clockwise torques.
     
  4. May 19, 2016 #3
    If I approach it using torques, it seems counterintuitive to set the center of mass at the fulcrum since the board is massless. The only other alternative is the set the c.m. at the 30 kg block.
     
  5. May 19, 2016 #4

    Doc Al

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    You're not setting the center of mass of anything. You are just choosing a convenient axis of rotation for computing torques. The fulcrum is the natural choice. (But you can use any point.)

    Since the board is assumed massless, its weight is irrelevant.
     
  6. May 19, 2016 #5

    SteamKing

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    The whole point of using a lever is to reduce the effort it takes to lift the 30 kg mass in the first place.

    Simply picking up the mass directly would require only applying 300 N to it, which implies that the "answer" of 400 N is wrong. Same for 900 N and 1200 N. Both of those forces greatly exceed the amount of force required to simply lift the 30 kg mass with your bare hands, and avoid having to set up a lever in the first place.
     
  7. May 19, 2016 #6
    Ok. You all are absolutely right. I set up the mechanical advantage equation wrong. It should be:

    R/E=L/l
    So, 300 N/E = 0.75/.025

    This was originally how I set it up and got 100 N, which makes more sense than 400 N, as you said, since the purpose of M.A. is to reduce the force needed to lift an object. However, the correct answer is 400 N, which to me, is wrong.

    By the way, how would I set this up using torques?
     
  8. May 19, 2016 #7

    Doc Al

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    The value of 100 N for the applied force F is correct. And that should make more sense than your earlier value of 900 N. You can apply the same logic you tried before to get the force at the fulcrum.

    The mechanical advantage equation is derived using torques.
     
  9. May 19, 2016 #8

    SteamKing

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    I spoke too soon in my previous post.

    Looking at the problem statement, the question wants to know what the force on the fulcrum is, rather than the value of F. But you still need to calculate F so that you can find the force on the fulcrum. The force on the fulcrum must equal the sum of the weight of the 30 kg block and the applied force F on the other end of the lever.
    Torque = force × distance

    Find a convenient point about which to calculate the torque (here, the fulcrum is just as convenient as any other point) and write your expressions for the torque each force creates about the reference point. By picking the fulcrum as your reference, whatever force is located there does not create any torque. When calculating torques, make sure you distinguish between clockwise and counterclockwise rotation.
     
  10. May 19, 2016 #9
    Ok! Yes, I looked at the M.A. formula and I see how it is derived from torque. If I cross-multiply the equation I get:

    Rl = EL

    This makes much more sense. Thank you!
     
  11. May 19, 2016 #10
    Ok, since this is a massless board and we now know that force applied is 100 N, the weight of the box is 300 N. Thus, the force on the fulcrum is the sum of these two forces.
     
  12. May 19, 2016 #11

    SteamKing

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    Correct.
     
  13. May 19, 2016 #12

    Doc Al

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    Good.

    Just for fun, try solving this a different way: Calculate torques about the end of the board (where F is applied).
     
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