# ¿Why the forces in the arch are carried to the ground?

**1. The problem statement, all variables and given/known data**

Hello, I want to know if my argue is right. The question is to justificate why the forces in the arch are carried to the ground.

I suppose a three voussoir’s rounded arch. Just the arch, no structure supported. I call A to the keystone, B the springer down on the left, and C to the springer down on the right.

I place the origin of coordinates down in the middle.

This time I only care about vertical forces.

**2. Relevant equations**

First law of static equilibrium. I must prove that the vertical forces on every part of the system is zero: ∑F_{Ay}=0, ∑F_{By}=0, ∑_{Cy}=0. And then add them to explain how the system works

**3. The attempt at a solution**

∑F_{Ay} = -m_{A}g + F_{BAy}senθ + F_{CAy}senθ = 0

∑F_{By} = -m_{B}g – F_{ABy}senθ + F_{BN} = 0

∑F_{Cy} = -m_{C}g – F_{ACy}senθ + F_{CN} = 0

Where F_{BAy}senθ is the vertical component of the force B exerts on A; -m_{A}g is the weight of A; F_{BN} is the normal force grounds exerts on B.

∑F_{Ay} + ∑F_{By} + ∑F_{Cy} = 0 = -m_{A}g -m_{B}g – m_{C}g + F_{BN} + F_{CN}.

In conclusion, normal forces over B and C support all the weight.

¿Is it right?.

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