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.

Answer 1

The output force will be 196.36 N

Step-by-step 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