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25 May, 22:51

A 76 kg pilot at an airshow performs a loop de loop with his plane. at the bottom of the 52-m radius loop, the plane is moving 48 m/s. determine the normal force acting upon the pilot.

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  1. 25 May, 23:44
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    The solution for this problem is:

    At bottom, both centrifugal force and weight are performing in the same direction.

    So we can say that, force would be computed by f = mg + mv^2 / r

    = (76 x 9.81) + (76 x 48 x 48 / 52)

    = 745.6 + 3367.4

    Normal force = 4113 N is the answer
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