With the shoulder flexed at 30 degrees, the moment arm of the deltoid muscle is 1.8 cm. Solve for the force exerted by the deltoid muscle and joint reaction forces at the glenohumeral joint give the following assumptions: ​The deltoid is the only active muscle at the glenohumeral joint ​The weight of the humerus is 24 N.   ​The center of gravity of the humerus is located 30 cm from the shoulder center of rotation

Inquiry into Physics
8th Edition
ISBN:9781337515863
Author:Ostdiek
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Chapter2: Newton's Laws
Section: Chapter Questions
Problem 28P: . A mass of 0.75 kg is attached to a relaxed spring with k = 2.5 N/m. The mass rests on a...
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With the shoulder flexed at 30 degrees, the moment arm of the deltoid muscle is 1.8 cm. Solve for the force exerted by the deltoid muscle and joint reaction forces at the glenohumeral joint give the following assumptions:

​The deltoid is the only active muscle at the glenohumeral joint

​The weight of the humerus is 24 N.  

​The center of gravity of the humerus is located 30 cm from the shoulder center of rotation

EXAMINING THE FORCES BOx 1.5
STATIC EQUILIBRIUM EQUATIONS
CONSIDERING ONLY THE DELTOID MUSCLE
FJ
Fo
MAO = 18Cra
B= 55°.
30°
Ro= 30 cm
Fa = 24 N
EM = 0 (at OcOR)
(Fo(MA) - (Fa(RG)sin(e) = o
Fo = (24 N}(30 cm)sin (30°) = 200 N
1.8 cm
EFx = 0
Fo cos(B) +Fjx = 0
Fjx = -200 cos (55°)= -115 N
EFy = 0
Fo sin(B) + FG + Fgy = 0
FJy = - Fa- Fp sin(B) = - (-24) -200 sin (55°) = -14ON
Fj =F
Py=(-115? + (-14ON?
181N
Transcribed Image Text:EXAMINING THE FORCES BOx 1.5 STATIC EQUILIBRIUM EQUATIONS CONSIDERING ONLY THE DELTOID MUSCLE FJ Fo MAO = 18Cra B= 55°. 30° Ro= 30 cm Fa = 24 N EM = 0 (at OcOR) (Fo(MA) - (Fa(RG)sin(e) = o Fo = (24 N}(30 cm)sin (30°) = 200 N 1.8 cm EFx = 0 Fo cos(B) +Fjx = 0 Fjx = -200 cos (55°)= -115 N EFy = 0 Fo sin(B) + FG + Fgy = 0 FJy = - Fa- Fp sin(B) = - (-24) -200 sin (55°) = -14ON Fj =F Py=(-115? + (-14ON? 181N
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