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Irina-Kira [14]
3 years ago
6

An outstretched arm of a person is held at an angle θ= 16.0° below the horizon and a 16.0 N weight is placed in the hand. The sh

oulder-to-elbow length is 25.5 cm and the an elbow-to-wrist length is 22.0 cm. The center of the weight is 7.0 cm from the person's wrist. Determine the magnitude of the torque about the elbow produced by the weight.
Physics
1 answer:
ElenaW [278]3 years ago
8 0

Answer:

240.4 Nm

Explanation:

The torque is the cross product of the force vector and the moment arm length vector

T = \vec{F} \times \vec{s}

T = Fs sin\theta

T = 16*(25.5 + 7 + 22)* sin16^o

T = 872*0.276

T = 240.4 Nm

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