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.

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  1. 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]

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