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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?

Respuesta :

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

onyebuchinnaji

(a) The magnitude of the torque about his shoulder due to the ball and the weight of his arm parallel to the floor is 34.55 Nm.

(b) The magnitude of the torque about his shoulder due to the ball and the weight of his arm straight, but 45° below horizontal is 24.43 Nm.

The given parameters;

  • mass of the ball, m = 3.0 kg
  • length of the arm, L = 78 cm = 0.78 m
  • mass of his arm, = 3.8 kg

A sketch of the position of the ball and the arm;

 

---------------------------------------------------------------------------78cm

↓           40%               ↓        

3 kg                            3.8 kg        

Take moment about the arm;

The moment due to arm (clockwise), is calculated as follows;

M₁  = Fd = (3.8 x 9.8) x (0.4 x 0.78) = 11.62 Nm

The moment due to ball (clockwise), is calculated as follows;

M₂ = Fd = (3 x 9.8) x (1 x 0.78) = 22.93 Nm

The magnitude of the torque about his shoulder due to the ball and the weight of his arm parallel to the floor is calculated as;

τ = M₁ + M₂

τ =  11.62 + 22.93 = 34.55 Nm  

(b) The moment at angle 45⁰ below the horizontal is calculated as follows;

The moment due to arm (clockwise), is calculated as follows;

M₁  = Fd = (3.8 x 9.8) x (0.4 x 0.78) x (cos45) = 8.22 Nm  

                   

The moment due to ball (clockwise), is calculated as follows;

M₂ = Fd = (3 x 9.8) x (1 x 0.78) x cos(45) = 16.21 Nm

The magnitude of the torque about his shoulder due to the ball and the weight of his arm straight, but 45° below horizontal is calculated as;

τ = M₁ + M₂

τ = 8.22 + 16.21

τ = 24.43 Nm.

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