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Torque and the Two Conditions for Equilibrium

  1. Apr 8, 2006 #1
    A cook holds a 2 kg cartoon of milk at arm's length. What Force [tex] F_{B} [/tex] must be excerted by the biceps muscle? (Ignore the weight of the forearm) See figure attached. [tex] \theta = 75 degrees [/tex] Please help explain what I'm doing wrong. Correct answer is 312 N.

    [tex] \Sigma\eta = F(sin 75)(.08 m) - (2 kg)(9.8 m/s^2)(.25 m) = 67 N [/tex]

    [tex] \Sigma F_{y} = F_{y} + (67 N)(sin 75) - (2kg)(9.8 m/s^2) = 0 [/tex]
    [tex] F_{y} = -45.4 N [/tex]

    [tex] \Sigma F_{x} = (67 N)(cos 75) [/tex]
    [tex] F_{x} = 17 N [/tex]

    [tex] F_{B} = \sqrt{-45.4^2 + 17^2} = 48. 5 N [/tex]

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    Last edited: Apr 8, 2006
  2. jcsd
  3. Apr 9, 2006 #2
    I don't really understand your work. You are solving for some variable F, which i don't know where it came from.

    Draw a FBD if you haven't already. Your diagram should have the two forces that are given in the given diagram (Fb, Fm), and a force at the elbow. Treat the elbow as a pin joint. From there you should be able to see how you can solve for Fb.
  4. Apr 9, 2006 #3


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    You need to take the moments around the elbow joint, so your moment arm for the carton is incorrect. Also look carefully at the diagram where the angle [itex]\theta[/itex] is defined. You only need to consider rotational equilibrium in order to solve this problem, that is the sum of the moments of the two forces around the elbow joint must be zero is all that need to be considered.
  5. Apr 9, 2006 #4

    [tex] \Sigma\eta = F_{B} - F_{g} [/tex]
    [tex] F_{B}(cos 75)(.08 m) = (2 kg)(9.8 m/s^2)(.33 m) [/tex]
    [tex] F_{B} = \frac{(2 kg)(9.8 m/s^2)(.33 m)}{(cos 75)(.08 m)} = 312 N [/tex]
  6. Apr 10, 2006 #5


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    Its a pleasure.
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