How Does Adding a Steamer Trunk Affect the Plane's Center of Gravity?

[mathguy2]

Homework Statement

A partially loaded plane weighs 115,000 pounds. A famous actor's steamer trunk, weighing 300 pounds, is put into the rear cargo hold 250 inches behind the center of gravity. How much does this extra hold shift the center of gravity.

Homework Equations

W₁ = W x (L₂ / L)

W₂ = W x (L₁ / L)

W = 115300
W₁ = 115000
W₂ = 300
L = 250

The Attempt at a Solution

I did this:
W₂ = W x (L₁ / L)
so...
300 = 115300 x (L₁ / 250)
so...
300 = 461.2L₁
divide both sides by 461.2

L₁ = .6504770165

The center of gravity of the plane shifts .6504770165 inches toward the rear. Your input is encouraged.

 

[PhanthomJay]

Looks good, but get rid of some of those figures after the decimal point, maybe just say 0.65 inches

 

[mathguy2]

Thank you for the input. Would you mind taking a look at the more advanced center of gravity problem I posted. It's really frustrating me at the moment.