¿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?.


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