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The deltoids exert a force of 100 N at an angle of 45 degrees to the humerus.The tendon inserts 7.5 cm from the shoulder joint axis of rotation.The arm is abducted 45 degrees from the vertical or anatomical position.The combined weight of the upper arm, forearm, and hand is 75 N acts at the center of gravity of the upper extremity which is 15 cm from the shoulder joint axis.What is the torque about the shoulder developed by the deltoids in Nm?

Question

The deltoids exert a force of 100 N at an angle of 45 degrees to the humerus.The tendon inserts 7.5 cm from the shoulder joint axis of rotation.The arm is abducted 45 degrees from the vertical or anatomical position.The combined weight of the upper arm, forearm, and hand is 75 N acts at the center of gravity of the upper extremity which is 15 cm from the shoulder joint axis.What is the torque about the shoulder developed by the deltoids in Nm?

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Solution

To calculate the torque developed by the deltoids, we need to use the formula for torque which is:

Torque = Force x Distance x sin(θ)

where:

  • Force is the force exerted by the deltoids (100 N)
  • Distance is the distance from the shoulder joint axis where the tendon inserts (7.5 cm or 0.075 m)
  • θ is the angle between the force and the lever arm (45 degrees)

So, the torque developed by the deltoids is:

Torque = 100 N x 0.075 m x sin(45 degrees)

First, we need to convert the angle from degrees to radians because the sin function in most calculators uses radians. 45 degrees is equal to 0.7854 radians.

So, the torque is:

Torque = 100 N x 0.075 m x sin(0.7854 rad) = 5.3 Nm

Therefore, the torque about the shoulder developed by the deltoids is 5.3 Nm.

This problem has been solved

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