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Work of a bicep curl Kaplan question

Discussion in 'MCAT Study Question Q&A' started by medschoolworries, Aug 18, 2015.

  1. medschoolworries

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    upload_2015-8-18_11-23-47.png

    I know W=Fd... but why does the bicep not apply the force over the entire height? According to the answer, the height is .01m instead of .1m even though you're lifting it .1m.
     
  2. Neuroplasticity

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    I think since gravity is a conservative force, the work is path independent and this can be solved by work is equal to change in potential energy W=mgh=5*10*0.1=5, or since it is path independent you can make up that it is lifted straight up so W=F*d=(5*10)*0.1=5
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    I think the way they are describing it is by the sum of the torques about the shoulder is 0, kind of like this picture:
    http://freephysicsquestions.com/wp-content/themes/directorypress/thumbs/27.jpg

    So 0=Fbicep*3.5-mg*35 -> Fbicep=500 N
    Then to move the point in the picture above at 2m up 10 cm, the point at 1m does not need to move up 10 cm as well, since the beam can rotate to something like this:
    [​IMG]
    Using similar triangles then, to move the mass up 10 cm, the point at the bicep only needs to move up 1 cm.
     
  3. JKMBC

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    Do you guys understand why, in the solution, "since the mass is located 10x farther from the elbow than the bicep, the bicep must do 10x the force"? Is it because the net torque on the side of the elbow must= 0, and since the weight of the mass is in the downwards direction, the force of the elbow must be in the opposite direction at 10x the force?
     
  4. Neuroplasticity

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    I believe you are correct. Though I think their solution is saying that the net torque of the shoulder is 0.
    Torque = F*dperpendicular
    0=Fbicep*1 - Fmass*10
    Fbicep=10*Fmass
     
  5. Mad Jack

    Mad Jack Critically Caring
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    [​IMG]
     
    Stop hovering to collapse... Click to collapse... Hover to expand... Click to expand...
    GrapesofRath likes this.

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