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7 June, 21:15

What is the force required to produce a 1.4 Nm torque when applied to a door at a 60.0º angle and 0.40 m from the hinge?

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  1. 8 June, 00:12
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    Refer to the diagram shown below.

    The component of the applied force perpendicular to the door is

    F * sin (60°) = 0.866F N

    Because the moment arm is 0.40 m, the torque is

    (0.866F N) * (0.4 m) = 0.3464F N-m

    This torque is equal to 1.4 N-m, therefore

    0.3464F = 1.4

    F = 4.04 N

    Answer: 4.04 N
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