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 ?=17? with the horizontal, as shown. (Figure 1) (Figure 2) Use g=9.8m/s2 throughout the problem.

Find the tension T in the deltoid muscle.

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

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

Answer 1

A:

Take the moments about the attachment point to the scapula:

T*Sin(17)*.15 - 3.6*9.8* 0.33 = 0

T*Sin(17)*.15 = 3.6*9.8*0.33

T = 3.6*9.8*.33/(.15*Sin(17)) = 265N

B:

Take the sum of the forces in the "y" direction:

3.6*9.8 - 265*Sin(17) + F_sy = 0

F_sy = 265*Sin(17) - 3.6*9.8 = 42.3 N (downward)

C:

Take the sum of the forces in the "x" direction:

T*Cos(17) - F_sx = 0

F_sx = T*Cos(17) = 265*Cos(17) = 253.4 N away from the body.

Answer 2

Use the formula ?=Fdsin(?). I'll take positive as clockwise. Some people take positive as anticlockwise - it doesn't really matter providing you are consistent. Also I'll use metres rather than cm.

? = (-50.0x0.20xsin(60?)) + (+50.0x0.20xsin(0)) + (+50.0x0.20xsin(30?)) + (+50.0x0.20xsin(30?+60?))
= -8.66 + 0 + 5 + 10
= 6.34 Nm (positive, so direction is clockwise)