Question

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 78 cm long and has a mass of 3.8 kg, with the center of mass at 40% of the arm length from the shoulder. What is the magnitude of the torque about his shoulder due to the ball and the weight of his arm if he holds his arm.
a. Straight out to his side, parallel to the floor?
b. Straight, but 45° below horizontal?

Answer 1

Answer:

a. 34.6 Nm

b.24.4 Nm

Explanation:

a.

78 cm = 0.78 m

W = F =mg

m1 = mass of steel ball = 3 kg

m2 = mass of long arm = 3.8 kg

moment due to steel ball = Fd =(m1*g)*(0.78)= (3*9.81)(0.78)=22.95 = 23 Nm

moment due to arm =Fd=(m2*g)*(0.78*0.4)= (3.8*9.81)(0.312)=11.63 = 11.6 Nm

net moment = 23 +11.6 = 34.6 Nm

b. now in this the angle will change the perpendicular moment arm

   moment due to steel ball = (3*9.81)*(0.78cos45) = 16.23 =16.2 Nm

   moment due to arm = (3.8*9.81)(0.4*0.78cos45) = 8.22 = 8.2 Nm

net moment = 16.2 +8.2 = 24.4Nm