The arm in the figure below weighs 42.0 N. The force of gravity acting on the ar

Question

The arm in the figure below weighs 42.0 N. The force of gravity acting on the arm acts through point A. Determine the magnitudes of the tension force in the deltoid muscle and the force exerted by the shoulder on the humerus (upper-arm bone) to hold the arm in the position shown.

Ft=

Fs=

The arm in the figure below weighs 42.0 N. The force of gravity acting on the arm acts through point A. Determine the magnitudes of the tension force in the deltoid muscle and the force exerted by the shoulder on the humerus (upper-arm bone) to hold the arm in the position shown. Ft= Fs=

Explanation / Answer

Summing moments about O we have

Ft*0.080*sin(12) - 42N*0.290m = 0

So Ft = 42*0.290/(0.080*sin(12)) = 732.28N

Now summing vertical forces

-42 + 732.28*sin(12) - Fs*sin(?) = 0
So Fs*sin(?) = -42 + 732.28 *sin(12) = 110.25N------------------- (eqn 1)

Now sum horizontal forces

Fs*cos(theta) - Ft*cos(12) = 0

Fs*cos(theta) = 732.28*cos(12) = 716.27 N ---------------------(eqn 2)

Now square both eqns and add them

Fs^2*cos^2(?) + Fs^2*sin^2(?) = 716.27^2 + 110.25^2 =>


So Fs = sqrt(513398) = 724.7 N

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