¿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: ∑FAy=0, ∑FBy=0, ∑Cy=0. And then add them to explain how the system works 3. The attempt at a solution ∑FAy = -mAg + FBAysenθ + FCAysenθ = 0 ∑FBy = -mBg - FABysenθ + FBN = 0 ∑FCy = -mCg - FACysenθ + FCN = 0 Where FBAysenθ is the vertical component of the force B exerts on A; -mAg is the weight of A; FBN is the normal force grounds exerts on B. ∑FAy + ∑FBy + ∑FCy = 0 = -mAg -mBg - mCg + FBN + FCN. In conclusion, normal forces over B and C support all the weight. ¿Is it right?.