A pitcher is in 85° of abduction, holding a 1.4 N baseball at point C, 65 cm from the joint axis at point O • The center of gravity of his arm is 25 cm from the joint axis of shoulder abduction at point O • The weight of the arm W is 0.06 of the pitcher’s weight of 100 N • Deltoids muscles are at an angle θ of 15° with respect to the humerus and insert 15 cm from the joint axis at point A • Determine the force applied by the Deltoid muscles and the joint reaction force at the shoulder joint and its orientation β

Respuesta :

I attached a Diagram for this problem.

We star considering the system is in equlibrium, so

Fm makes [tex]90-(\theta+5)[/tex] with vertical

Fm makes 70 with vertical

Applying summatory in X we have,

[tex]\sum F_x = 0[/tex]

[tex]W+1.4-Fm cos(70)[/tex]

We know that W is equal to

[tex]W= 0.06*100N = 6N[/tex]

Substituting,

[tex]Fm cos (70) = W+1.4N[/tex]

[tex]Fm cos (70) = 6N + 1.4N[/tex]

[tex]Fm = \frac{7.4}{cos(70)}[/tex]

[tex]Fm = 21.636N[/tex]

For the second part we know that the reaction force Fj on deltoid Muscle is equal to Fm, We can assume also that [tex]\beta  = \theta[/tex]

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