If a 60.1 kg person jumps onto a seesaw at an angle of 22 degrees at a distance of 6.9 m from the fulcrum.

Question

If a 60.1 kg person jumps onto a seesaw at an angle of 22 degrees at a distance of 6.9 m from the fulcrum.

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Nem 6 months 2021-08-10T06:12:30+00:00 1 Answers 6 views 0

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    2021-08-10T06:14:02+00:00

    Answer:

    The output force will be 196.36 N

    Explanation:

    For the seasaw to be in equilibrium, net torque should be zero

    R₁Fsin∅ = R₂F(output)

    F(output) = R₁Fsin∅ / R₂

                   = 6.9 (60.1 * 9.8) Sin 22° / 7.8

                   = 0.89 (588.98) Sin22°

                   = 0.89 * 588.98 * 0.3746

    F(output) = 196.36 N

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