An athlete at the gym holds a 1.2 kg steel ball in his hand. His arm is 64 cm long and has a mass of 4.0 kg. What is the magnitude of the torque about his shoulder if he holds his arm in each of the following ways?

1. straight but 30 degrees below horizonal?

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98 To calculate the magnitude of the torque about the athlete's shoulder, we need to consider the formula:

Torque = force × distance × sin(θ)

Where:
- Torque is the magnitude of the torque
- Force is the force applied by the steel ball
- Distance is the distance between the shoulder and the point where the force is applied
- θ is the angle between the force vector and the line perpendicular to the distance vector

In this case, let's assume the force applied by the steel ball is equal to its weight. The weight can be calculated using the formula:

Weight = mass × gravitational acceleration

Given:
- Mass of the steel ball = 1.2 kg

The gravitational acceleration is approximately 9.8 m/s^2.

Weight = 1.2 kg × 9.8 m/s^2 = 11.76 N

Now, let's calculate the distance from the shoulder to the point where the force is applied. This can be done using the Pythagorean theorem:

Shoulder-arm length = √(arm length^2 + shoulder length^2)

Given:
- Arm length = 64 cm = 0.64 m
- Shoulder length = 0.00 m (assuming the shoulder joint is the origin)

Shoulder-arm length = √(0.64^2 + 0^2) = 0.64 m

Lastly, we need to determine the angle θ. In this case, it is given as 30 degrees below horizontal. We will use the sine function to convert this angle to radians:

θ = 30 degrees = (30 × π) / 180 radians

Now we have all the values we need to calculate the magnitude of the torque:

Torque = force × distance × sin(θ)
= Weight × Shoulder-arm length × sin(θ)
= 11.76 N × 0.64 m × sin((30 × π) / 180)

Calculating this expression will give you the magnitude of the torque about the athlete's shoulder when their arm is held straight but 30 degrees below the horizontal.