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
