In an emergency situation, a person with a broken forearm ties a strap from his

Question

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 ? = 53.5? with the horizontal. Assume the forearm is uniform, has a length of = 0.342 m, assume the biceps muscle is relaxed, and ignore the mass and length of the hand. (a) Find the tension in the strap. N (b) Find the components of the reaction force exerted by the humerus on the forearm. Rx = N Ry = N

Explanation / Answer

we have, a. T sin(53.5) + Ry = Mg Tcos(53.5) = Rx Tsin(53.5) x 0.342 = 1.6 x 9.8 x (0.342/2) Tension = T = 9.753 N b. Ry = 7.84 N Rx = 5.8 N

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