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

## Homework Statement

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.

## Homework 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

## 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?.