The effort applied : 200 N
Further explanation
Equilibrium rotation
[tex]\large {\boxed {\bold {\sum \tau = 0}}[/tex]
F₁.d₁-F₂.d₂=0
F₁.d₁=F₂.d₂
F₁=500 N
d₁=0.5 m
d₂=1.75-0.5=1.25 m
[tex]\tt 500\times 0.5=F_2\times 1.25\\\\F_2=\dfrac{500\times 0.5}{1.25}=200~N[/tex]
