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 ? = 54.5? with the horizontal. Assume the forearm is uniform, has a length of script l = 0.336 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.

Explanation / Answer

Consider the elbow point (where the humerus is) as the Torque Point Consider CCW as positive Consider CW as negative Sum of the torque about the torque point is zero Tsin(54.5)(.336) - 1.6(9.8)(.168) = 0 T = 9.63 N Now use the sum of the forces in both the x and y direction to determine the reaction forces at the humerus: Sum of forces in x-direction = 0 -Tcos(54.5) + Rx = 0 Rx = 5.59 N Sum of forces in y-direction = 0 Tsin(54.5) - (1.6)(9.8) + Ry = 0 Ry = 7.84 N BOL

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