Anatomy and Torque Question (please help! I am confused!)

Question

An athlete at the gym holds a 3.5 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg.

1. What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor?
2. What is the magnitude of the torque about his shoulder if he holds his arm straight, but 30' below horizontal?

Answer 1

Torque (τ) is the perpendicular component of the force (F cosθ) acting on the arm times the distance (L) from the pivot point (shoulder).

There are two sources of torque on the arm:

The weight of the arm: W₁ = m₁g, producing the torque τ₁ = (m₁g cosθ)L₁

where m₁ = 4 kg and L₁ = the distance to the center of mass of the arm = r/2 = 0.35 m

The weight of the ball: W₂ = m₂g, producing the torque τ₂ = (m₂g cosθ)L₂,

where m₂ = 3.5 kg and L₂ = r (arm length) = 0.7 m

1. When the arm is held straight out to the side, θ = 0 and cosθ = 1,

So τ = (4 kg)(9.8 m/s²)(0.35 m) + (3.5 kg)(9.8 m/s²)(0.7 m) = (13.72 + 24.01) N⋅m

τ = 37.73 N⋅m ≈ 38 N⋅m (rounded to 2 significant figures)



2. When the arm is held 30° below horizontal, θ = 30° and cosθ = √3/2,

So τ = (4 kg)(9.8 m/s²)(√3/2)(0.35 m) + (3.5 kg)(9.8 m/s²)(√3/2)(0.7 m)

(√3/2)[(4 kg)(9.8 m/s²)(0.35 m) + (3.5 kg)(9.8 m/s²)(0.7 m)] = (√3/2)(37.73) N⋅m

τ ≈ 32.68 N⋅m ≈ 33 N⋅m (rounded to 2 significant figures)

Answer 2

Make a diagram.

Torque = Fd but the force must be perpendicular to the distance. In the first problem it is perpendicular so:

T = (W_arm)(d₁) + (W_ball)(d₂)

W_arm = mg = (4)(9.8) = 39.2 N

W_ball = mg = (3.5)(9.8) = 34.3 N

d₁ = 35 cm = 0.35 m

d₂ = 70 cm = 0.7 m

T = (39.2)(0.35) + (34.3)(0.7) = 38 Nm

The second problem is a bit harder because you must break each force into components to find the force perpendicular to the distance. Note that weight always points straight down so only a portion of the weight will contribute to the torque.

W_arm-y = (W_arm)(cos(30°)) = 33.95 N

W_ball-y = (W_ball)(cos(30°)) = 29.70 N

T = (33.95)(0.35) + (29.70)(0.7) = 33 Nm