Dee asked in Science & MathematicsPhysics · 3 months ago

Torque question ?

An athlete at the gym hold a 3kg steel ball in his hands. His arm is 70 cm long and has a mass of 4.0kg. What is the mangintude of torque about his shoulder if he holds his arm...

a) Straight out to his side, parallel to floor

b) Straight, but 45 degrees below horizontal? 

Ty,

2 Answers

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  • NCS
    Lv 7
    3 months ago
    Favorite Answer

    torque = r x F

    or more simply, when available,

    τ = F * d * sinΘ

    where Θ is measured between the directions of F and d.

    a) assuming the mass in his arm is uniformly distributed,

    τ = (4.0kg * ½*0.70m + 3kg * 0.70m) * 9.8m/s² * sin90º

    τ = 34 N·m

    b) now Θ = 45º, and so

    τ' = 34N·m * sin45º = 24 N·m

    (the same as if held at 45º ABOVE horizontal)

    Hope this helps!

    • NCS
      Lv 7
      2 months agoReport

      Thanks for awarding!

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  • oubaas
    Lv 7
    3 months ago

    T1 = g(4*0.7/2+3*0.7) = g*0.7*(2+3) = 3.5g N*m ( ≅ 34.3)

    T2 = (T1*√2)/2 = (1.75√2)*g N*m ( ≅ 24.3)

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