The magnitude of the vertical component of the force Fy exerted by the scapula on the humerus is -42 N.
Let Fy be the vertical component of the force exerted by the scapula on the humerus
using the second condition of static equilibrium, the net force acting on the system is zero.
∑ Fnet.y = 0
Fy + Ty - (M₁g) = 0
Fy = M₁G - TsinΘ
substituting the numerical values ( image *attached*) in the equation and solving for Fy.
Fy = (3.6 kg)(9.8 m/s²) - (265.47 N) sin 17°
= -42.3 N
= -42N
here, the negative sign indicates that the direction of vertical force is in the downward direction.
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(complete question)
You are able to hold out your arm at an outset
the horizontal position because of the action of the deltoid muscle. Assume the humerus bone has a mass of M1 = 3.6Kg, a length of L=0.66m and its center of mass is a distance of L1=0.33m from the scapula. (For this problem ignore the rest of the arm.) The deltoid muscle attaches to the humerus at a distance of L2=0.15m from the scapula. The deltoid muscle makes an angle of theta = 17 degrees with the horizontal, as shown. (Intro 1 figure) (Intro 2 figure) Use g=9.8m/s^2 throughout the problem.
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).
