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Force P that will start the block to move

  1. Aug 21, 2009 #1
    1. The problem statement, all variables and given/known data.

    Find the force P and theta that will start the block to move.
    The illustration can be found at the attachments.

    Given:
    W= 200 N
    Radius of the circle 400 mm
    length of the block 100 mm

    The unknowns are the P and theta.

    2. Relevant equations.

    Summation of Fx=0
    Summation of Fy=0

    Those are the equations that I see that can be used.

    3. The attempt at a solution.

    Note that there is no normal force between the cylinder and the surface since the system will start to move.

    There are three forces acting on the system: the weight (W), force P, and normal force between the block and the cylinder (Nb). The horizontal angle of the Nb can be acquired by using right triangle where hypotenuse equals the radius and and one leg equals radius minus length of the block. By using sine function, the angle can be acquired. There are three unknowns left: P, theta, and Nb. There are only 2 equations that I can see: the summation of horizontal component and the summation of vertical component. What should I do?

    Thank you.
     

    Attached Files:

    Last edited: Aug 21, 2009
  2. jcsd
  3. Aug 21, 2009 #2

    Doc Al

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    I assume the problem is to find the minimum force P (and the angle it makes) that will start the cylinder going over the step.

    Hint: Consider torques.
     
  4. Aug 21, 2009 #3
    yes that is it. I think you cannot use moments in here. How should I start using torques?
     
  5. Aug 21, 2009 #4

    Doc Al

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    Moments and torques are the same thing. Why can't you use them? (You've got to.)

    Consider the torques about the edge of the step. What torque does the weight produce? What torque does force P produce? (At what angle must P act to produce the most torque?)
     
  6. Aug 21, 2009 #5
    so the axis is the point of application of the block and the cylinder? okay i try it.
     
  7. Aug 21, 2009 #6
    I still got no result. My answer becomes 0=0. At that point, Nb has no torque and the W and Py has the same distance which is around 264. and Px has a distance of 300 from the point. When I solved it, I cannot get a result. Why?
     
  8. Aug 21, 2009 #7

    Doc Al

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    I'm not quite sure what you're doing. Forget Px and Py, just consider P. What angle must it make to maximize the torque it produces? (What angle must the P vector make with the line from the corner of the step to the center of the cylinder?) Then find the minimum value of P such that its torque is just enough to overcome the torque due to the weight of the cylinder.
     
  9. Aug 21, 2009 #8
    Do you mean step as the corner of the block? I got Nb about 50. I think I'm wrong. Do you mean that torque of P equals torque of W?
     
  10. Aug 21, 2009 #9

    Doc Al

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    Yes. The cylinder will pivot about the corner of the block/step. Calculate torques about that point.
    Yes.
     
  11. Aug 21, 2009 #10
    My problem now is the angle. I cannot combine sin theta and cosine theta. The equation is like: sin theta = 1-cos theta. Can I substitute square root of 1 - cos square theta to sin theta to get the angle. What other methods can I use?
     
  12. Aug 21, 2009 #11

    Doc Al

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    I don't understand what you are doing. Answer the two questions I asked in post #7.
     
  13. Aug 21, 2009 #12
    Now I have P, I cannot get the angle. Can you give me another hint?
     
    Last edited: Aug 21, 2009
  14. Aug 21, 2009 #13

    Doc Al

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    I'm still waiting for you to answer my questions from post #7. (In order to find P, you must have the angle it makes with the radius.)
     
  15. Aug 21, 2009 #14
    I have P but no angle. My equations are Pcos theta = 0.66R and Psin theta = 200 - 0.75R. I cannot really get it. There are two unknowns now but I still cannot get it. The P I get is about 130. When I substitute I cannot get R or theta.
     
    Last edited: Aug 21, 2009
  16. Aug 21, 2009 #15

    Doc Al

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    I'm going to quote myself:
    Please answer the two questions I ask. (It's really just one question, restated.)

    When you answer those questions, you'll have the angle. Then you can find P using one simple torque equation.
     
  17. Aug 23, 2009 #16
    Up to this day, it is still negative. However, I will try to figure it out. I have 1 week remaining. Ok. Thank you.
     
  18. Aug 23, 2009 #17

    Doc Al

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    :confused:

    Let me ask you this. If you wanted to turn a wrench with maximum torque, which direction would you push or pull? (See: http://hyperphysics.phy-astr.gsu.edu/hbase/torq.html" [Broken])
     
    Last edited by a moderator: May 4, 2017
  19. Aug 24, 2009 #18
    I think I know now. The force P should be perpendicular to the step. If the angle of the normal between block and cylinder is 30, then the theta will be 60. Am I right?
     
    Last edited: Aug 24, 2009
  20. Aug 24, 2009 #19

    Doc Al

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    It should be perpendicular to the line from the edge of the step to the center of the cylinder.
    I'm not sure how you are defining your angles.
     
  21. Aug 24, 2009 #20
    The normal is also directed to the center of cylinder.
     
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