Help My Physics Question

Question

What are the reactive force and moment at shoulder?

BW = 78Kg

Upper arm = 0.028BW

distance from shoulder to elbow = 33cm

center of mass = 15 cm from the shoulder

Southeast is elbow.

Answer 1

I am not sure I am understanding your question correctly, but

as nobody has offered their understanding in three days, here is my take on it.

One thing that makes me unsure about my interpretation is that I do not see

the significance of the given distance from shoulder to elbow.

Let

m = 0.028×78 kg = 2.184 kg be the mass of the upper arm,

d = 15 cm = 0.15 m be the distance from shoulder to center of gravity,

θ = 30° be the angle of the arm,

g = 9.81 m/s² be gravitational acceleration.

The weight of the upper arm is mg.

That force of weight is directed straight down.

It can be decomposed into

- a force F1 along the arm, and

- a force F2 perpendicular to the arm.

The force F1 is pulling the arm out of the shoulder, and

the force F2 is moving the arm so as to be parallel to the body.

If you draw the right angle triangles showing the three forces, you can see that

F1 = mgcos(θ)

While the arm is pulling down on the shoulder with force F1,

the shoulder is pulling the arm up with a reactive force.

That reactive force has the opposite direction, and same magnitude as

F1 = mgcos(θ) = 2.184×9.81×cos(30°) = 18.6 N

Secondly you want the moment, which I interpret of moment of force, or torque.

For that we take the force of weight, mg, as acting at the center of gravity,

so the radius of turn is the distance d.

Moment of force is the cross product of the two vectors d×mg.

It is directed perpendicular to the plane of your picture, and its magnitude is

dmgsin(θ) = 0.15×2.184×9.81×sin(30°) = 1.6 Nm