Question

An athlete at the gym holds a 2.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg .

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor?

What is the magnitude of the torque about his shoulder if he holds his arm straight, but 35 degrees below horizontal?

Answer 1

given

m_a = 4 kg

m_b = 2 kg

L = 70 cm

= 0.7 m

a )

W_a = 4 x 9.8

= 39.2 N

r_a = L = 0.7 m = r_b

W_b = 2 x 9.8

= 19.6 N

_b,1 = - 90^o

_a,1 = 0.7 x 39.2 x sin(-90)

_a,1 = - 27.44 N-m

_b,1 = 0.7 x 19.6 x sin(-90)

_b,1 = - 13.72 N-m

₁ = _a,1 + _b,1

= - 27.44 - 13.72

₁ = - 41.16 N-m clockwise direction

the magnitude of the torque about his shoulder if he holds his arm straight out to his side,

parallel to the floor is ₁ = - 41.16 N-m clockwise direction

b )

_b,2 = - 35^o

_a,2 = 0.7 x 39.2 x sin(-35)

_a,2 = - 15.73 N-m

_b,2 = 0.7 x 19.6 x sin(-35)

_b,2 = - 7.86 N-m

₂ = _a,2 + _b,2

= - 15.73 - 7.86

₂ = - 23.59 N-m clockwise direction

the magnitude of the torque about his shoulder if he holds his arm straight,

but 35 degrees below horizontal ₂ = - 23.59 N-m clockwise direction