the density of water is 1000 kg/m cubed. a flat-bottomed wooden box 3m long by 1.5m wide by 1m high is used as a boat to carry five people on a lake. the total mass of the boat and people are 1200 kg (11,760 N).
how do i solve this probelm?
Boat Construction Resources and Suppliers
who knows?! do you have a book with examples in it by chance?
1200kg/1000kg/m^3 = 1. 2m cubed
The mass of water displaced must be equal to or greater than the mass of the boat + cargo for it to float. So a 1200kg boat would need to displace 1200kg of water, so 1. 2 cubic metres of water would need to be displaced.
Archimede’s principle:
“an object is immersed in a fluid is buoyed up by a force equal to the weight of the fluid displaced by the object”
The weight of the displaced fluid can be found mathematically. The fluid displaced has a weight W = mg. The mass can now be expressed in terms of the density and its volume,
ρ – water density
m = ρVwater
Hence
W = ρ Vwater g
the weight of the boat with the load is G = m g =11 7600 N
when the boat loaded is in echilibrium you have the condition:
W = G
ρ Vwater g = m g
Vwater = m / ρ
Vwater = 1. 2 m^3 final result
you also can exprime this results function of the boat volume
Vboat = 3 * 1. 5 * 1 m^3 = 4. 5 m^3
Vwater = 0. 266 Vboat
Are you sure the density of water is 1000 kg/m^3 ???
Sounds too convenient. . . . . . maybe just a book easy
problem. . . . . Fresh or salt??
Now real water weighs 62. 2 lbs / ft^3 fresh.
1200 kg * 2. 21 = 2652 lbs >>>>>
2652 lbs / 62. 2 lbs/ft^3 >>>>> 42. 63665595 etc. . ft^3
now take 3. 1 ft / m cube-ify it get >. . . . . 29. 791 ft^3 / m^3
annnnnnd
42. 63665595 / 29. 791 >>> givith >>> 1. 43119 m^3
now we knowse thay be talkin salt or it B rigged. . . . . bye
btw it is very hard to find in a book but 1 ft^3 is 7. 49 gallons
btw #2 NOBODY checked that the volume displaced, take their 1. 2 m^3, is LESS than the volume of the boat ( 4. 5 m^3) otherwise it sinketh. . . . .