An athlete at the gym holds a 4.33 kg steel ball in his hand. His arm is 75.0 cm long and has a mass of 5.28 kg. What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor?

The torque has to be the weight of the ball x distance plus the weight of the arm x 1/2 the length of the arm(assuming the cg is at the midpoint).

thanx a lot

To find the magnitude of the torque about the athlete's shoulder, we need to multiply the force exerted by the ball by the perpendicular distance from the axis of rotation (shoulder joint) to the line of action of the force.

The torque (T) can be calculated using the equation:

T = r * F * sin(θ)

where:
T = torque
r = perpendicular distance from the axis of rotation to the line of action of the force
F = force exerted by the ball
θ = angle between the line of action of the force and the line perpendicular to the axis of rotation

In this case, since the athlete holds his arm straight out to his side, parallel to the floor, the angle (θ) between the line of action of the force and the line perpendicular to the axis of rotation is 90 degrees (since sin(90) = 1).

The perpendicular distance (r) from the shoulder to the line of action of the force can be calculated by finding the center of mass of the arm and multiplying it by the length of the arm. Since the arm has a mass of 5.28 kg and a length of 75.0 cm, the center of mass can be assumed to be at a distance of half the length of the arm, i.e., 37.5 cm.

Now, we can calculate the torque:

T = (0.375 m) * (4.33 kg) * sin(90)

T = 1.64475 N.m

Therefore, the magnitude of the torque about the athlete's shoulder is approximately 1.645 N.m.