Answer:
(a) τ = 26.58 Nm
(b) τ = 18.79 Nm
Explanation:
(a)
First we find the torque due to the ball in hand:
τ₁ = F₁d₁
where,
τ₁ = Torque due to ball in hand = ?
F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N
d₁ = perpendicular distance between ball and shoulder = 60 cm = 0.6 m
τ₁ = (29.4 N)(0.6 m)
τ₁ = 17.64 Nm
Now, we calculate the torque due to the his arm:
τ₁ = F₁d₁
where,
τ₂ = Torque due to arm = ?
F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N
d₂ = perpendicular distance between center of mass and shoulder = 40% of 60 cm = (0.4)(60 cm) = 24 cm = 0.24 m
τ₂ = (37.24 N)(0.24 m)
τ₂ = 8.94 Nm
Since, both torques have same direction. Therefore, total torque will be:
τ = τ₁ + τ₂
τ = 17.64 Nm + 8.94 Nm
<u>τ = 26.58 Nm</u>
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(b)
Now, the arm is at 45° below horizontal line.
First we find the torque due to the ball in hand:
τ₁ = F₁d₁
where,
τ₁ = Torque due to ball in hand = ?
F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N
42.42 cm = 0.4242 m
τ₁ = (29.4 N)(0.4242 m)
τ₁ = 12.47 Nm
Now, we calculate the torque due to the his arm:
τ₁ = F₁d₁
where,
τ₂ = Torque due to arm = ?
F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N
d₂ = perpendicular distance between center of mass and shoulder = 40% of (60 cm)(Cos 45°) = (0.4)(42.42 cm) = 16.96 cm = 0.1696 m
τ₂ = (37.24 N)(0.1696 m)
τ₂ = 6.32 Nm
Since, both torques have same direction. Therefore, total torque will be:
τ = τ₁ + τ₂
τ = 12.47 Nm + 6.32 Nm
<u>τ = 18.79 Nm</u>
