Consider the following. A standard man holds the upper arm vertical and the lowe

Question

Consider the following. A standard man holds the upper arm vertical and the lower arm horizontal with an object of mass M = 5 kg resting on the hand, as illustrated in the figure above. The mass of the lower arm and hand is one-half the mass of the entire arm. As shown in the figure below, there are four forces acting on the lower arm: the external force Fext, exerted by the bones and ligaments of the upper arm at the elbow (fulcrum); the tension T, exerted by the biceps; a force w due to the weight of the holding object; and the weight wF of the lower arm. The points along the lower arm at which the forces act are I1 = 3 cm, I2 = 14 cm, and I3 = 44 cm, as shown in the figure below. (Assume the mass of the lower arm is 2.3 kg.) Arrangement of the biceps muscle (1) and the triceps muscle (2) in the upper arm. The biceps is connected to the scapula (6) and the radius (3), which runs parallel to the ulna (4) in the lower arm. The muscle actions of the biceps and triceps rotate the lower arm, relative to the humerus (5), at the elbow. (a) Calculate the vertical component of the force Fext. (Enter the magnitude only.) N (b) Calculate the vertical component of the tension T. (Enter the magnitude only.) N

Explanation / Answer

Here ,

for the vertical component of tension force T ,

as the mass is in equilibrium,

let us balance the moment of force about joint ,

5 * 9.8 * 0.44 + 2.3 * 9.8 * 0.14 - 0.03 * T = 0

solving for T

T = 823.9 N

the tension in the muscle is 823.9 N

for the verical force at the joint ,

balancing the forces in vertical direction

Fext + T - 5 * 9.8 - 2.3 * 9.8 = 0

Fext = -752 N

the external force is -752 N

hence ,

a) force Fext is 752 N

b) vertical tension in the force T is 823.9 N

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