An athlete at the gym holds a 3.5 kg steel ball in his hand. His arm is 76 cm lo
ID: 1466718 • Letter: A
Question
An athlete at the gym holds a 3.5 kg steel ball in his hand. His arm is 76 cm long and has a mass of 4.1 kg . Assume the center of mass of the arm is at the geometrical center of the arm.
“What is the magnitude of the torque about his shoulder due to the weight of the ball and his arm if he holds his arm straight out to his side, parallel to the floor?
“What is the magnitude of the torque about his shoulder due to the weight of the ball and his arm if he holds his arm straight, but 45 below horizontal?
Explanation / Answer
Known Variables:
Mass of Arm: 4.1 kg
Mass of Ball: 3.5 kg
arm length: 76 cm
First you have to find the weight of both the ball and the arm. Multiply them by the force of gravity.
weight of arm: 4.1kg x 9.8 = 40.18 =40.2
weight of ball: 3.5kg x 9.8 = 34.3
We are looking for the Torque
Torque, = r1 F1 sin +r2 F2 sin
We need to find r1 and r2
r1= 76/2= 38 cm or 0.38m (arm: to find the center of gravity, divide by 2)
r2= 76cm or 0.76m (ball)
Since the arms are pointed straight out, the angle is 180°
=m1g(L/2) + m2gL.
L is the length of the system.
=4.1 * 9.8 *(0.76/2) + 3.5 * 9.8 * 0.76
= 41.34Nm
Now the arms are pointed out at an 45° angle.
you would be using cosine since its the horizontal distance.
= 0.38 x 40.2cos(45) + 0.76 x 34.3cos(45)
= 29.2 Nm