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AnnaJa

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## 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[itex]_{b}[/itex]=W (weight of floating body)

## The Attempt at a Solution

Weight of Box:

W=52x9.81= 510.12 N

W=F[itex]_{b}[/itex]=ρgv

v=[itex]\frac{W}{ρg}[/itex]

v= [itex]\frac{510.12}{9810}[/itex]= 0.052m[itex]^{3}[/itex]

v=dx1.2x0.8

d=[itex]\frac{0.052}{1.2x0.8}[/itex]=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[itex]_{b}[/itex]=ρgv

v= 1.2x0.8x0.36=0.3456m[itex]^{3}[/itex]

**W**

*total*=

**W**

*box*+

**W**

*hexagon*

**W**

*total*=1000x9.81x0.3456=3390.3N

**W**

*hexagon*=

**W**

*total*-

**W**

*box*

**W**

*hexagon*= 3390.3 - 510.12= 2880.2 N

**a)**

v=[itex]\frac{W}{ρg}[/itex]

v=[itex]\frac{2880}{9810}[/itex] =0.29m[itex]^{3}[/itex]

**b)**

ρ=[itex]\frac{Mass}{Volume}[/itex]

Mass=[itex]\frac{2880}{9.81}[/itex]= 293kg

ρ=[itex]\frac{293}{0.29}[/itex]=1012.34

γ= 1012.34x9.81=9931.03Nm [itex]^{2}[/itex]

This is what i thought, but i am absolutely not sure if this is right.

Any help would be greatly appreciated,

Thanks guys!

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