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.(adsbygoogle = window.adsbygoogle || []).push({});

[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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# Torque and the Two Conditions for Equilibrium

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