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
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
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Chapter1: Units, Trigonometry. And Vectors
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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
### Diagram Description
The diagram illustrates the forces acting on a human shoulder joint, specifically focusing on the deltoid muscle. Labels include:
- **COR** (Center of Rotation) is marked on the shoulder joint.
- **F_D** represents the force exerted by the deltoid muscle.
- **F_J** and **F_G** are other forces acted upon the joint, with **F_G** equal to 24 N and directed downward.
- Angles and distances are indicated:
- \( \beta = 55^\circ \)
- \( \theta = 30^\circ \)
- \( MA_D = 1.8 \, \text{cm} \) (Moment Arm of Deltoid)
- \( R_G = 30 \, \text{cm} \) (Distance from COR to where \( F_G \) acts)
### Equations
#### Sum of Moments (\( \Sigma M = 0 \) at COR)
\[
(F_D)(MA_D) - (F_G)(R_G)\sin(\theta) = 0
\]
Calculating:
\[
F_D = \frac{(24 \, \text{N})(30 \, \text{cm})\sin(30^\circ)}{1.8 \, \text{cm}} = 200 \, \text{N}
\]
#### Sum of Forces in X Direction (\( \Sigma F_X = 0 \))
\[
F_D \cos(\beta) + F_{JX} = 0
\]
Calculating:
\[
F_{JX} = -200 \cos(55^\circ) \approx -115 \, \text{N}
\]
#### Sum of Forces in Y Direction (\( \Sigma F_Y = 0 \))
\[
F_G + F_D \sin(\beta) + F_{JY} = 0
\]
Calculating:
\[
F_{JY} = -F_G - F_D \sin(\beta) = -24 - 200 \sin(55^\circ) \approx -140 \, \text{N}
\]
#### Resultant Joint Force (\( F_J \))
\[
F_J = \sqrt{F_{JX](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcd4d64a5-7d35-43c8-a56a-0e9b43f79ea4%2F07ef2afe-0a5c-436b-8a10-51efd677c9fd%2Fzan4d3_processed.jpeg&w=3840&q=75)
Transcribed Image Text:# Examining the Forces Box 1.5
## Static Equilibrium Equations Considering Only the Deltoid Muscle
### Diagram Description
The diagram illustrates the forces acting on a human shoulder joint, specifically focusing on the deltoid muscle. Labels include:
- **COR** (Center of Rotation) is marked on the shoulder joint.
- **F_D** represents the force exerted by the deltoid muscle.
- **F_J** and **F_G** are other forces acted upon the joint, with **F_G** equal to 24 N and directed downward.
- Angles and distances are indicated:
- \( \beta = 55^\circ \)
- \( \theta = 30^\circ \)
- \( MA_D = 1.8 \, \text{cm} \) (Moment Arm of Deltoid)
- \( R_G = 30 \, \text{cm} \) (Distance from COR to where \( F_G \) acts)
### Equations
#### Sum of Moments (\( \Sigma M = 0 \) at COR)
\[
(F_D)(MA_D) - (F_G)(R_G)\sin(\theta) = 0
\]
Calculating:
\[
F_D = \frac{(24 \, \text{N})(30 \, \text{cm})\sin(30^\circ)}{1.8 \, \text{cm}} = 200 \, \text{N}
\]
#### Sum of Forces in X Direction (\( \Sigma F_X = 0 \))
\[
F_D \cos(\beta) + F_{JX} = 0
\]
Calculating:
\[
F_{JX} = -200 \cos(55^\circ) \approx -115 \, \text{N}
\]
#### Sum of Forces in Y Direction (\( \Sigma F_Y = 0 \))
\[
F_G + F_D \sin(\beta) + F_{JY} = 0
\]
Calculating:
\[
F_{JY} = -F_G - F_D \sin(\beta) = -24 - 200 \sin(55^\circ) \approx -140 \, \text{N}
\]
#### Resultant Joint Force (\( F_J \))
\[
F_J = \sqrt{F_{JX
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