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# 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,

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- NCSLv 73 months agoFavorite 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!

Source(s): https://en.wikipedia.org/wiki/Torque - oubaasLv 73 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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