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Force - spring on an arc

  1. Mar 21, 2010 #1
    1. The problem statement, all variables and given/known data

    As test of strength, a diabolical trainer sets up the following apparatus. The trainee must maintain a variable pulling force which is always tangent to a nearly frictionless, semicircular surface . By slowly varying the force, a block with mass 23.0 kg is moved (at a very slow constant speed), and the spring to which it is attached is stretched from position 1 to position 2 (through an angle of 36 degrees). The spring has negligible mass and force constant 450 N/m. The end of the spring moves in an arc of radius a~=~ 49.0 cm.

    http://session.masteringphysics.com/problemAsset/1000054849/9/YF-07-41.jpg

    What is the necessary size of the trainee's variable pulling force at position 2?

    2. Relevant equations

    F=-kx

    3. The attempt at a solution

    I found the length of the arc first

    36/180 = .2

    C = 2(pi)R(1/2)
    C = 0.49(pi)

    Arc = .2(.49*pi)
    Arc = 0.307876 m

    then i multiplied that by the spring force constant

    F = 0.3079(450)

    F = 138.5 N

    I think this is wrong cause i am not taking the mass into consideration but i wasnt sure where or how to include the mass of the block.

    Any suggestions?

    Thank you
     
  2. jcsd
  3. Mar 21, 2010 #2

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    Suggestions

    1. Work with symbols and plug in the numbers at the very end. It will be easier for you to see what's going on.
    2. Draw a free body diagram of the block at position 2, put in all the forces and find the net force.
    3. Use Newton's Second Law. What is the acceleration in this case?
     
  4. Mar 21, 2010 #3
    2. Ok drew my free body diagram. I have weight going down, the normal force perpendicular to the surface of the arc, the pulling force in the same place as in the picture, and the force of the spring in the opposite direction of the pulling force.

    2. I believe the net force would be ZERO because it is being pulled at a constant speed and is not accelerating therefore giving it a net force of zero.

    3. wouldn't the acceleration be zero because its being pulled at a constant speed?
     
  5. Mar 21, 2010 #4
    Would i find the sum of the forces in the x and y directions and then find the resultant force and that will be my trainee's variable pulling force at position 2?
     
  6. Mar 21, 2010 #5
    I got it. Thank you :)
     
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