What are the bodies of the couple's total volume in liters

[chawki]

Homework Statement

Water bath whose walls are vertical and the bottom of which has dimensions as shown in line with a dip in the husband and wife. The pool water level rises then 6.4 cm.

Homework Equations

What are the bodies of the couple's total volume in liters.

The Attempt at a Solution

I don't know from where to start
we can calculate the vloume of the base of that bath by V=(1.5+1.5+1.2+1.2)+Pi*0.6=7.28m

 

[AC130Nav]

Since the walls are vertical, the volume of the tub (and any section thereof, to include the amount generated by the couple entering) is equal to the area of the base times the height.

Your formula for V is incorrect. Rethink the formula for the area of the base.

 

[chawki]

Let x be the height of the bath.
V=(1.5*1.2*x)+((Pi*0.6²*x)/2) ?

 

[AC130Nav]

Let x be the height of the bath.
V=(1.5*1.2*x)+((Pi*0.6²*x)/2) ?

I would have phrased it V = [(1.5*1.2)+.6Pi/2]h; but you have the formula basically. Thing is you don't have the height of the bath nor do you need it. You are only interested in the height of the displacement.

 

[chawki]

how so ? and i still don't get your formula..it should be (0.6²*Pi)/2

 

[AC130Nav]

how so ? and i still don't get your formula..it should be (0.6²*Pi)/2

Sorry, you're right, the .6 should be squared. Typo. I did say what you had was essentially correct. You just didn't seem to realize you already had the value for your x, 6.4 cm. The volume of the bathers is equal to the volume they displace.

 

[chawki]

Ahhhhhhh ok, WOW:!)
so the volume of the bathers is V=(1.5*1.2*0.064)+((Pi*0.62*0.064)/2)
V=0.177m³
in litres, V=177L

 

[AC130Nav]

Ahhhhhhh ok, WOW:!)
so the volume of the bathers is V=(1.5*1.2*0.064)+((Pi*0.62*0.064)/2)
V=0.177m³
in litres, V=177L

It looks like you actually did put 0.62 into your calculator instead of (0.6)^2. I get .151.

 

[chawki]

Oh my lord...i did..
now i get 0.151 cubic meters and that's 151L
Thank you :)

 

[chawki]

What if they asked the weight of the bathers!
is the weight of the bathers=the weight of water they displaced?
is there a buoyancy problem here?

 

[gneill]

What if they asked the weight of the bathers!
is the weight of the bathers=the weight of water they displaced?
is there a buoyancy problem here?

Only if the bathers are floating. If they sink, then there's no telling how much heavier they are than the water they displaced (which is equal to their volume).

 

[chawki]

so they are taking their bath, probably not floating..and thus their mass would be same as their volume? 151Kg?

 

[gneill]

No. Read again what I wrote in post #11.

 

[AC130Nav]

so they are taking their bath, probably not floating..and thus their mass would be same as their volume? 151Kg?

From the statement of the question, we have to make the assumption they completely submerge--the question makes no sense any other way. We can also assume they have no diving weights attached in the bath, so they're not pinned to the bottom, which probably does mean their weight is equal to the water displaced though the average human does float and is partially above water. Probably the reason that question wasn't asked.

If you want more accuracy, you have to say the volume of water is equal to the volume of the bathers under water and the weight of the water displace is equal to their total weight (both above and below the water line).

 

[chawki]

in that case how can we find the weight of the water displaced? the mass at first!

 

[Piyu]

you got the rise of the water level. Use that to calculate the volume displaced and then multiply by water density. Usually density would be given but is often used as 1000kg/m^3.