GOAL Apply the equilibrium conditions to the human body. A 50.0-N (11-lb) bowlin
ID: 1433366 • Letter: G
Question
GOAL Apply the equilibrium conditions to the human body. A 50.0-N (11-lb) bowling ball is held in a person's hand with the forearm horizontal, as in Figure (a). The biceps muscle is attached 0.0300 m from the joint, and the ball is 0.350 m from the joint. Find the upward force F exerted by the biceps on the forearm (the ulna) and the downward force R exerted by the humerus on the forearm, acting at the joint. Neglect the weight of the forearm and slight deviation from the vertical of the biceps. STRATEGY The forces acting on the forearm are equivalent to those acting on a bar of length 0.350 m, as shown in Figure (b). Choose the usual x- and ^-coordinates as shown and the axis at O on the left end. (This completes Steps 1 and 2.) Use the conditions of equilibrium to generate equations for the unknowns, and solve. SOLUTION Apply the second condition for equilibrium (Step 3) and solve for the upward force F. Apply the first condition for equilibrium (Step 4) and solve (Step 5) for the downward force R. sigma tou = tou_r + tou_f+ tou_BB= 0 R(0) + F(0.0300 m) - (50.0 N)(0.350 m) = 0 F = 583 N (131 lb) sigmaF_y = F - R - 50.0 N = 0 R = F - 50.0 N = 583 N - 50 N = 533 N (120 lb) LEARN MORE REMARKS The magnitude of the force supplied by the biceps must be about ten times as large as the bowling ball it is supporting! QUESTION Suppose the biceps were surgically reattached three centimeters farther toward the person's hand. If the same bowling ball were again held in the person's hand, how would the force required for the biceps be affected? Explain? (Select all that apply.)Explanation / Answer
let the maximum weight be a
so, again by using the condition that torque about O is zero, we write
Fx 0.02976 = m x 0.347 eqn 1
Also F = m + R eqn 2
Substituting eqn 2 in eqn 1 we get,
( m + R ) x 0.02976 = 0.347m eqn 3
Given maximum value of R = 792 N
Using this in the derived eqn 3,
( m + 792 ) x 0.02976 = 0.347 m
0.02976 m + 23.56992 = 0.347 m
23.56992 = 0.31724 m
m = 74.29680999 (which can be rounded of to 74.3 , in case you are specific about the number of significant figures)
F = m + R = 74.29680999 + 792 = 866.29681 N
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