The effort applied = 200 N
Further explanation
Equilibrium :
[tex]\tt F_1.d_1=F_2.d_2[/tex]
Total length = 1.75
The distance between the fulcrum and the load is 0.5 m ⇒ d₁=0.5 m
The distance between the fulcrum and the force applied :
[tex]\tt 1.75-0.5=1.25~m\rightarrow d_2=1.25~m[/tex]
A load of 500N ⇒ F₁=500 N
The force applied :
[tex]\tt 500\times 0.5=F_2\times 1.25\\\\F_2=\dfrac{500\times 0.5}{1.25}=200~N[/tex]
