Volume and specific weight of hexagonal mass in floating box

[AnnaJa]

Homework Statement

A rectangular prismatic box, which is floating on water has a length, L=1200 mm, breadth, b=800 mm, height, h=500 mm, and a mass of 52.00 kg. When the box is loaded with the suspended hexagonal mass, the depth of its immersion is 300 mm, and when the mass is placed inside the box, the depth of immersion increases to 360 mm (Figure 3). The specific weight of water=9810 N/m3. The volume and weight of the string is negligible.
What is
(a) the volume, and
(b) the specific weight of the hexagonal mass?

Homework Equations

F= ρgv

F$_{b}$=W (weight of floating body)

The Attempt at a Solution

Weight of Box:

W=52x9.81= 510.12 N


W=F$_{b}$=ρgv


v=$\frac{W}{ρg}$


v= $\frac{510.12}{9810}$= 0.052m$^{3}$

v=dx1.2x0.8

d=$\frac{0.052}{1.2x0.8}$=0.0541m = 54.166mm ← submergen due to the self weight of the box.


From First Case (LHS):

300-54.166= 245.83mm ← submerged due to weight pf hexagon


From Second Case (RHS):

F$_{b}$=ρgv

v= 1.2x0.8x0.36=0.3456m$^{3}$


Wtotal= Wbox + Whexagon

Wtotal=1000x9.81x0.3456=3390.3N

Whexagon= Wtotal - Wbox

Whexagon= 3390.3 - 510.12= 2880.2 N

a)

v=$\frac{W}{ρg}$

v=$\frac{2880}{9810}$ =0.29m$^{3}$

b)


ρ=$\frac{Mass}{Volume}$

Mass=$\frac{2880}{9.81}$= 293kg

ρ=$\frac{293}{0.29}$=1012.34

γ= 1012.34x9.81=9931.03Nm $^{2}$

This is what i thought, but i am absolutely not sure if this is right.
Any help would be greatly appreciated,
Thanks guys!

 

[Simon Bridge]

Please show your reasoning - it helps us if we know how you are thinking.

Are you stacking numbers to indicate division?
Have a go using LaTeX: V=\frac{W}{\rho g} gets you: $$V=\frac{W}{\rho g}$$

I was expecting that you'd use maths to describe the two situations then solve the simultaneous equations.

 

[SteamKing]

The units displayed in the calculations show several inconsistencies. Please re-check. For example, v will always be in cu. meters.