Question

What muscular force F pulling at an angle of 25 degrees would be required to keep the abducted arm in a position of 20 degrees with the horizontal? The muscle inserts 10 cm from the shoulder joint. The arm weighs 50N and its center of gravity is located 30 cm from the shoulder. A 45-N weight is held in the hand 60 cm from the shoulder joint.

Answer 1

To solve this problem, we can break it down into components. The force required to keep the abducted arm in position has both horizontal and vertical components.

First, let's calculate the horizontal and vertical components of the weight of the arm and the weight held in the hand. The horizontal component of the weight is given by:

Horizontal Component = Weight * (Distance from Shoulder / Total Arm Length)

For the arm, it's:
Horizontal Component (Arm) = 50 N * (30 cm / 40 cm) = 50 N * 0.75 = 37.5 N

For the weight held in the hand:
Horizontal Component (Weight) = 45 N * (60 cm / 90 cm) = 45 N * 0.6667 = 30 N

Now, we can calculate the net horizontal force required to keep the arm in position:

Net Horizontal Force = Horizontal Component (Arm) + Horizontal Component (Weight) = 37.5 N + 30 N = 67.5 N

Now, we can calculate the vertical component of the force required to keep the arm in position:

Vertical Component = Net Horizontal Force / tan(25 degrees)

Vertical Component = 67.5 N / tan(25 degrees) ≈ 112.67 N

So, the muscular force required to keep the abducted arm in position is approximately 112.67 N at an angle of 25 degrees with the horizontal.