Truck weight from ferry sinkage of 0.0367 m

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[SOLVED] Floating Weight

When a truck drives onto a river ferry the ferry sinks 0.0367M. The length and width of the ferry are 15.24M and 6.10M respectively. Determine the weight of the truck.
Just not quite sure I'm setting it up correctly, or getting the right answer.
I originally thought of using the pressure equation (p=pgh) to find the pressure at the new "depth" and using that in the force equation (F=P/A) to find the Force. That doesn't really make sense though, right?
 
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Just plain old Archimedes' Principle will do in this case. Extra buoyant force to take the weight of the lorry = extra water displaced.
 
So we are looking at something like:

Buoyant Force only changes once the truck drives onto the ferry. So the displaced water equals the weight of the truck? That would mean...

Buoyant force (wt of truck) = P2A - P1A
P1 - ATM pressure
P2 = [tex]\rho[/tex]gh
A = length X width

so...

= ([tex]\rho[/tex]gh)A - (ATM Pressure)A
= (1000)(9.8)(.0367m)(93m^2) - (1.013 X10^5 Pa)(93m^2)
= 33448.37 - 9420900
But that gives a negative number?
 
"pressure" has nothing to do with this. If the ferry is 125.4 m by 6.10 m and sinks 0.0367 m, what volume of water was displaced? How much did that water weigh?

(If you are using air pressure to calculate the force of the air on the ferry, remember that was their before the truck moved onto the ferry and is already accounted for.)
 
This is making me feel really stupid.
I understand the concept that the water displaced will equal the weight of the truck, but I can't seem to find an equation to find that mass of the water.
I can't determine the volume or weight of the ferry.
 
Could I not find the pressure at that depth by [tex]\rho[/tex]gh
(1000)(9.8)(.0367) = 359.7pa
F = PA
= (359.7)(15.24m X 6.10m)
= 33439.2 / 9.8
= 3412.16 kg ?
 
7starmantis said:
I understand the concept that the water displaced will equal the weight of the truck, but I can't seem to find an equation to find that mass of the water.
I can't determine the volume or weight of the ferry.

The cross-sectional area of the ferry is given. It is sinking by a certain height. So, vol of water displaced = area*height. Vol*density gives you mass, and mass*g gives you weight.

Now you can apply Archimedes' Principle.
 
Ok, so Vol of water displaced is:

vol = a*h
= (15.24X6.10)*(.0367)
= 3.41m^3

Then:
m = v*[tex]\rho[/tex]
= (3.41)*(1000)
= 3410kg

Wt = 3410kg * 9.8 = 33418N

So that is the weight of the water displaced which is also the weight of the truck.
That makes sense to me. Why couldn't I figure that out before I have no idea.
Does that seem right?
 
Last edited:
Thats ok, I just wanted to understand the concept and make sure I was setting it up right.(although it does sound like a big weight)

Thank you very much for your help, you explained it in a way I could understand. I really appreciate it! a lot!