In an emergency situation, a person with a broken forearm ties a strap from his hand to clip on his shoulder as in the figure below. His 1.60-kg forearm remains in a horizontal position and the strap makes an angle of 0 = 53.5° with the horizontal. Assume the forearm is uniform, has a length of e = 0.324 m, assume the biceps muscle is relaxed, and ignore the mass and length of th hand. (a) Find the tension in the strap. N (b) Find the components of the reaction force exerted by the humerus on the forearm. (Assume the positive x-direction is to the right and the positive y-direction is upward.) Ry = N Ry =

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In an emergency situation, a person with a broken forearm ties a strap from his hand to clip on his shoulder as in the figure below. His 1.60-kg forearm remains in a horizontal position and the
strap makes an angle of 0 = 53.5° with the horizontal. Assume the forearm is uniform, has a length of e = 0.324 m, assume the biceps muscle is relaxed, and ignore the mass and length of the
hand.
R
(a) Find the tension in the strap.
(b) Find the components of the reaction force exerted by the humerus on the forearm. (Assume the positive x-direction is to the right and the positive y-direction is upward.)
Ry =
N
R, =
N
Transcribed Image Text:In an emergency situation, a person with a broken forearm ties a strap from his hand to clip on his shoulder as in the figure below. His 1.60-kg forearm remains in a horizontal position and the strap makes an angle of 0 = 53.5° with the horizontal. Assume the forearm is uniform, has a length of e = 0.324 m, assume the biceps muscle is relaxed, and ignore the mass and length of the hand. R (a) Find the tension in the strap. (b) Find the components of the reaction force exerted by the humerus on the forearm. (Assume the positive x-direction is to the right and the positive y-direction is upward.) Ry = N R, = N
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