You are able to hold out your arm in an outstretched horizontal position because

Question

You are able to hold out your arm in an outstretched horizontal position because of the action of the deltoid muscle. Assume the humerus bone has a mass M1=3.6kg, length L=0.66m and its center of mass is a distance L1=0.33m from the scapula. (For this problem ignore the rest of the arm.) The deltoid muscle attaches to the humerus a distance L2=0.15m from the scapula. The deltoid muscle makes an angle of (theta)=17degrees with the horizontal, as shown. (Figure 1) (Figure 2) Use g=9.8m/s2 throughout the problem.

Figure 1:

Figure 2:

Part A: Find the tension T in the deltoid muscle.

Part B: Using the conditions for static equilibrium, find the magnitude of the vertical component of the force Fy exerted by the scapula on the humerus (where the humerus attaches to the rest of the body).

Part C: Now find the magnitude of the horizontal component of the force Fx exerted by the scapula on the humerus.

You are able to hold out your arm in an outstretched horizontal position because of the action of the deltoid muscle. Assume the humerus bone has a mass M1=3.6kg, length L=0.66m and its center of mass is a distance L1=0.33m from the scapula. (For this problem ignore the rest of the arm.) The deltoid muscle attaches to the humerus a distance L2=0.15m from the scapula. The deltoid muscle makes an angle of (theta)=17degrees with the horizontal, as shown. (Figure 1) (Figure 2) Use g=9.8m/s2 throughout the problem. Figure 1: Part A: Find the tension T in the deltoid muscle. Part B: Using the conditions for static equilibrium, find the magnitude of the vertical component of the force Fy exerted by the scapula on the humerus (where the humerus attaches to the rest of the body). Part C: Now find the magnitude of the horizontal component of the force Fx exerted by the scapula on the humerus. Figure 2:

Explanation / Answer

Ty = T sin theta, Tx = T cos theta

Torqueclockwise = Torqueanticlockwise

(Ty) L2 = (M1 g) L1

(T sin theta) L2 = (M1 g) L1

T = M1 g L1 / ((sin theta) L2) = (3.6 * 9.8 * 0.33 ) / ((sin 17) 0.15)

T = 265 N

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Using the second condition of static equilibrium

sigma Fnet - y = 0

Fy + Ty - (M1 g) = 0

Fy = (M1 g) - T sin theta

Fy = (3.6 * 9.8) - 265 * sin 17

= -42 N [negative indicates direction is downward]

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using the second condition of static equilibrium

sigma Fnet - x = 0

Fx = T cos theta

Fx = 265 * cos 17

= 254 N [direction of horizontal force is i right side]

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