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20 December, 19:51

The position of a 2.75*105n training helicopter under test is given by r⃗ = (0.020m/s3) t3i^ + (2.2m/s) tj^ - (0.060m/s2) t2k^. part a find the net force on the helicopter at t=5.0s.

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  1. 20 December, 23:14
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    F (t) = 1.7⋅10^4i^-3.4⋅10^3k

    Here we have the components in the x-direction and the y-direction

    Fx = 17000 N

    Fz = 3400 N

    The resultant force is given by

    F = sqrt (Fx2 + Fz2)

    = sqrt (170002 + 34002)

    = 17336.67 N

    Net force on the helicopter at t = 5.0 s is 17336.67 N
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