AW 45.0 N (10 lb) weight is held in a person's hand with the forearm horizontal,...

 

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AW 45.0 N (10 lb) weight is held in a person's hand with the forearm horizontal, as in Figure 8.11. The biceps muscle is attached d = 0.0301 m from the joint, and the weight 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. Solution Apply the second condition for equilibrium (step 3). Apply the first condition for equilibrium (step 4). Biceps Ulna F = Strategy The forces acting on the forearm are equivalent to those acting on a bar of length 0.350 m, as shown in Figure 8.11b. Choose the usual x- and y-coordinates as shown and the axis O on the left end. (This completes Steps 1 and 2.) Use the conditions of equilibrium to generate equations for the unknowns, and solve. -1-1 (b) Figure 8.11 (a) A weight held with the forearm horizontal. (b) The mechanical model for the system. R= K ET = TR+TP+TBB = 0 R(0)+ F(0.0301 m) - (45.0 N)(0.350 m) = 0 F = N lb kd Fy=F+ R - 45.0 N = 0 R = F - 45.0 N = 523 N - 45.0 N R = N lb W AW 45.0 N (10 lb) weight is held in a person's hand with the forearm horizontal, as in Figure 8.11. The biceps muscle is attached d = 0.0301 m from the joint, and the weight 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. Solution Apply the second condition for equilibrium (step 3). Apply the first condition for equilibrium (step 4). Biceps Ulna F = Strategy The forces acting on the forearm are equivalent to those acting on a bar of length 0.350 m, as shown in Figure 8.11b. Choose the usual x- and y-coordinates as shown and the axis O on the left end. (This completes Steps 1 and 2.) Use the conditions of equilibrium to generate equations for the unknowns, and solve. -1-1 (b) Figure 8.11 (a) A weight held with the forearm horizontal. (b) The mechanical model for the system. R= K ET = TR+TP+TBB = 0 R(0)+ F(0.0301 m) - (45.0 N)(0.350 m) = 0 F = N lb kd Fy=F+ R - 45.0 N = 0 R = F - 45.0 N = 523 N - 45.0 N R = N lb W

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In the image were presented with a physical scenario typical of a biomechanics problem involving the static equilibrium of a forearm holding a weight ... View the full answer