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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.

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MechEngineer

Answer:

240.4 Nm

Explanation:

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

[tex]T = \vec{F} \times \vec{s}[/tex]

[tex]T = Fs sin\theta [/tex]

[tex]T = 16*(25.5 + 7 + 22)* sin16^o [/tex]

[tex]T = 872*0.276[/tex]

[tex]T = 240.4 Nm[/tex]

The magnitude of the torque about the elbow produced by the weight is 240.4 Nm.

Calculation of the magnitude of the torque:

Since there is 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

So, here the magnitude should be

= 16 * (25.5 + 7 + 22) * sin 16 degrees

= 872* 0.276

= 240.4Nm

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