Engineering How much force will be experienced by the sidewall?

[qnach]

Homework Statement

I have a V-shape depression (2-dimensional) with inelastic but deformable material filled.
If the material is pressed atop, how much force will be experienced by the sides?

Relevant Equations

none.

I have a V-shape depression (2-dimensional) with inelastic but deformable material filled.

If the material is pressed atop, how much force will be experienced by the sides?

 

[mitochan]

Hi.
Say F is force atop, it is equal to $2N\ sin\theta$ where $\theta$ is angle of both the sides from vertical line and N is force from a side.
N=\frac{F}{2 \sin\theta}

 

[Chestermiller]

Homework Statement:: I have a V-shape depression (2-dimensional) with inelastic but deformable material filled.
If the material is pressed atop, how much force will be experienced by the sides?
Relevant Equations:: none.

I have a V-shape depression (2-dimensional) with inelastic but deformable material filled.

If the material is pressed atop, how much force will be experienced by the sides?

View attachment 274375

What do you define as "inelastic but deformable?"

 

[qnach]

What do you define as "inelastic but deformable?"

Yes, your criticism is correct. The force is not large, therefore will not result in large deformation hence can be considered as inelastic. Deformation, such as sands are deformable but inelastic.

 

[qnach]

Hi.
Say F is force atop, it is equal to $2N\ sin\theta$ where $\theta$ is angle of both the sides from vertical line and N is force from a side.
N=\frac{F}{2 \sin\theta}

How did you got the result? I am not sure about the \sin \theta, but the force F should be P presumably.

 

[mitochan]

Please find attached sketch to show balance of vertical forces.

 

[Tom.G]

Hi.
Say F is force atop, it is equal to $2N\ sin\theta$ where $\theta$ is angle of both the sides from vertical line and N is force from a side.
N=\frac{F}{2 \sin\theta}

Then where does the rest of the force go, for all angles and depths?

 

[mitochan]

Then where does the rest of the force go, for all angles and depths?

N comes from area integration of pressure on a side
\mathbf{N}=\int p \ d\mathbf{S}