Stability - calculations taken for granted in books

[MatsNorway]

Homework Statement

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(Solved. through the work on this post i found the mistakes)
Most basic stability problems is introduced simply as a box with a load placed on its side. And you do the typical momentum calculation around the opposite edge. Most books/pdfs stops there.

I have however some troubles understanding questions such as this: They state the density of the box(20kN/m³) and then the Stability wanted, Sw=(2). Find the sides (x).
F in X direction on top of box is 15.78kN

In this case the two sides are identical, in other words. The box has a squared footprint. Height (7) times width and length -> 7*x*x equals volume. So Surely the mass of the box is just volume times density.

Add safety margin or stability.. and you get a formula as such ->

Sw(2) = Ms/Mv

Ms=Volume times density times half the length
Mv= Force times arm

Homework Equations

How is x = 1.467? What is it i am missing?

The Attempt at a Solution

Stability(2) = (stabilizing Momentum(7*x²)*(x/2)*20))/(vaulting Momentum(15.78*7))

Ms((7*x²)*(x/2)*20)))=S(2)*Mv(15.78*7)

(7*x³)*(1/2)*20))=15.78*7*2

x³=(15.78*7*2)/(7*10)

Simplified= x³=(15.78*2)/(10)

X= 1.467

 

[jbsteele]

The correct answer is x = 1.256.My mistake was in calculating the stabilizing momentum, Ms. The correct formula for Ms is as follows: Ms = (7*x2)*(x)*20)Using this formula, I got the correct answer of x = 1.256.