Wooden Cube submerged length and the carry limit

[vinamas]

Homework Statement

A wooden cube with the length of 10 cm is and a density of 700 kg/m^3 is floating on water
A)Find the submerged parts length of the cube
B) find the maximum added mass to the block before it becomes totally submerged

Homework Equations

FB=density*volume*gravity
FB=Fg
density=m/v

[/B]

The Attempt at a Solution

A)[/B]
mass of the cube = 700*0.1^3=0.7
so Fg = 0.7*9.81=6.867N
FB=Fg=6.867
1000*V*9.81=6.867
V=0.0007
L^3=0.0007
L=0.008. The answer sheet says that L submerged must be 7 cm
B) m2=Fb-m1
m2=1000*0.1^3*9.81-0.7=98.1-0.7=97.4Kg. The answer sheet says that the added mass is 0.3 kg
am lost guys welp

 

[vinamas]

Sorry guys I messesd up on the second question solved it now but still I need A)

 

[SteamKing]

Homework Statement

A wooden cube with the length of 10 cm is and a density of 700 kg/m^3 is floating on water
A)Find the submerged parts length of the cube
B) find the maximum added mass to the block before it becomes totally submerged

Homework Equations

FB=density*volume*gravity
FB=Fg
density=m/v

[/B]

The Attempt at a Solution

A)[/B]
mass of the cube = 700*0.1^3=0.7
so Fg = 0.7*9.81=6.867N
FB=Fg=6.867
1000*V*9.81=6.867

When calculating the mass of the displaced water, you don't need to multiply by g, since later on you're going to divide by g.

V=0.0007
L^3=0.0007
L=0.008. The answer sheet says that L submerged must be 7 cm

You are assuming the cube shrinks on all sides to produce the volume of displacement. This is not so. At least two of the cubes dimensions are going to be the same. The only dimension which varies is the draft. Draw a picture of this cube if you get confused.

B) m2=Fb-m1
m2=1000*0.1^3*9.81-0.7=98.1-0.7=97.4Kg. The answer sheet says that the added mass is 0.3 kg
am lost guys welp

You need to get in the habit today of showing units in your calculations.

What happens when you multiply mass by g? Do you get kilograms as a result?

Always get in the habit of checking your arithmetic for mistakes.

 

[vinamas]

I don'

When calculating the mass of the displaced water, you don't need to multiply by g, since later on you're going to divide by g.

You are assuming the cube shrinks on all sides to produce the volume of displacement. This is not so. At least two of the cubes dimensions are going to be the same. The only dimension which varies is the draft. Draw a picture of this cube if you get confused.

You need to get in the habit today of showing units in your calculations.

What happens when you multiply mass by g? Do you get kilograms as a result?

Always get in the habit of checking your arithmetic for mistakes.

I don't get what you mean I was trying to get the volume of the submerged part by doing that is the volume correct?

 

[vinamas]

When calculating the mass of the displaced water, you don't need to multiply by g, since later on you're going to divide by g.

You are assuming the cube shrinks on all sides to produce the volume of displacement. This is not so. At least two of the cubes dimensions are going to be the same. The only dimension which varies is the draft. Draw a picture of this cube if you get confused.

You need to get in the habit today of showing units in your calculations.

What happens when you multiply mass by g? Do you get kilograms as a result?

Always get in the habit of checking your arithmetic for mistakes.

oh wait I totally get you now so 0.0007=0.1*0.1*L
and L = 0.07 m thank you!