# 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

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 shoulder-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.

## This Post Has 3 Comments

1. Expert says:

it about 600 mph from what i calculated

2. Expert says:

increase masses, decrease separation

3. 10040813 says:

240.4 Nm

Explanation:

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

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

$T = Fs sin\theta$

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

$T = 872*0.276$

$T = 240.4 Nm$