An athlete at the gym holds a 2.2 kg steel ball in his hand. His arm is 70 cm lo
ID: 1457916 • Letter: A
Question
An athlete at the gym holds a 2.2 kg steel ball in his hand. His arm is 70 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? (a) Straight out to his side, parallel to the floor Incorrect: Your answer is incorrect. . N m (b) Straight, but 40° below horizontal Incorrect: Your answer is incorrect. . N m
here is what i tried Fb = 3.4 * 9.81 = 33.35 N
Weight of athlete's arm:
Fa= 4 * 9.81 = 39.24 N
I'll assume that weight of the arm acts in the center of the arm.
1) When hand is parallel to ground, torque is:
T = Fb * L + Fa * L/2
or
T = (Fb + Fa/2) * L
where L is length of the arm, L=0.7m
T = (33.35 + 39.24/2)*0.7 = 25.96 Nm
2) When hand is held at angle A below horizontal:
T = Fb * L * cosA + Fa * (L/2) * cos A
or
T = (Fb + Fa/2) * L * cos A
if A=35°
T = (33.35 + 39.24/2)*0.7 * cos 35° = 21.26 Nm But i still get it wrong
Explanation / Answer
Your answers are wrong, because you have put incorrect values for the mass of the ball and the angle. Please see the correct calculations below:
Weight of the ball:
Fb = 2.2 * 9.8 = 21.56 N
Weight of athlete's arm:
Fa= 4 * 9.8 = 39.2 N
We'll assume that weight of the arm acts in the center of the arm.
1) When hand is parallel to ground, torque is:
T = Fb * L + Fa * L/2
=> T = (Fb + Fa/2) * L
where L is length of the arm, L=0.7m
T = (21.56 + 39.2/2)*0.7 = 28.812 Nm
2) When hand is held at angle A below horizontal:
T = Fb * L * cosA + Fa * (L/2) * cos A
=> T = (Fb + Fa/2) * L * cos A
=> T = (21.56 + 39.2/2)*0.7 * cos 40° = 22.07 Nm