# Finding the weight of water required to float object

Find the weight of water required to float the pontoon in the casting dock below when the gap all around is 0.1m. The total weight the pontoon is 500 tons (1ton=8896N). Is this consistent with Archimedes principle (upthrust=weight of displaced fluid)?

The pressure on the base to support the weight is $$F_b = \rho \times g \times z \times$$, Using vertical equilibrium, $$4.448 MN = F_b$$, therefore $$z = 2.83m$$. My confusion lies the next part. Here is a solution from my lecturer.

I am really confused at what I am looking at , especially finding the volume of water. Is the volume of water we are after the volume of water before the block has been placed?

• My interpretation of the question is that its the same volume of water before and after Mar 21 at 20:41
• Where did "Pressure in small gaps gives large buoyancy forces" come from? That looks like a bogus statement to me. Mar 21 at 21:52

If not, try this (see Fig. below):

$$W$$ = Weight of the pontoon

$$A$$ = Horizontal project area of the pontoon

$$p_V = W/A$$ = Applied pressure

$$p_U = \rho gz$$ = Uplift pressure due to buoyancy

Setting $$p_V = p_U$$, and get

$$z = \dfrac{W}{\rho gA}$$ = Depth of submergence

Think of it this way. If the tank was filled to the height of 2.934 m and a drain tube was installed that would empty all water above that level, you would find that the volume of water that passed through the drain tube is equal to the weight of the pontoon or 453.44 m^3.

The volume of water remaining in the tank would be equal to 39.47 m^3 or 43.52 tons of water. So if you only have 39.47 m^3 in the dock then placed the pontoon in, it would then float to the prescribed clearances of 0.1 m from bottom and sides without spilling any water.

Now I would say your lectures volume is slightly off, but I could be wrong.

(4+2*2.934)*0.1*40


This is calculating the volume of water along the 40m long sides. Basically it is the area of the rectangles of water times the 40 m length

A1= 4*0.1 (area under pontoon)
A2= 2.934*0.1 (sidewall area to left)
A3= 2.934*0.1 (sidewall area to right)
L1= 40

(A1+A2+A3)*L1
(4*0.1+2.934*0.1+2.934*0.1)*40

pull out the 0.1 and simplify
(4+2.934+2.934)*0.1*40
(4+2*2.934)*0.1*40


What I believe is missing is the 0.1 clearance in the front and the back which has not been accounted for.

(4+0.1+0.1)*2.934*0.1*2=2.46 m^3


So the volume remaining in the tank should really be

39.47+2.46 = 41.93 m^3


Based on the calculations provided.

An important note here is that the height of the pontoon is not given. But the question states that it will float, so you have to assume the height is greater than 2.834m. Otherwise the pontoon is too dense and a denser medium than water would be required to make it float.

My version of the calculations. Density of water may be slightly different.