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Question

<p><img src="https://s3.amazonaws.com/answer-board-image/b34facfa-d44c-4a0c-977f-ce83098e17ed.jpg" alt="image from custom entry tool" /><br /><br /><br />The drawing shows a lower leg being exercised. It has a F = 44 N weight attached to the foot and is extended at an angle &#952; with respect to the vertical. Consider a rotational axis at the knee. <br /><br /><br />(a) When &#952; = 90.0&#176;, find the magnitude of the torque that the weight creates.<br />_____N&#183;m<br /><br />(b) At what angle &#952; does the magnitude of the torque equal 12 N&#183;m?<br />_____&#176;</p>
<p>&#160;</p>
<p>I need someone to show me how to work this problem and give me the correct answer</p>

Explanation / Answer

Given that force F = 44.0 N The torque is given by = r F sin 90     = (0.55 m)(44 N)       =24.2   N.m b)
Torque is           = rF sin        = sin-1 (/rF)           = sin-1 ((12. N.m)/(0.55 m)(44.0 N))           = 29.7269 o The torque is given by = r F sin 90     = (0.55 m)(44 N)       =24.2   N.m b)
Torque is           = rF sin        = sin-1 (/rF)           = sin-1 ((12. N.m)/(0.55 m)(44.0 N))           = 29.7269 o

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