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Math question (water displacement)

Edd

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I know there's some smarties on the board. I'm studying for a test and even Google isn't helping me figure out this formula.

A displacer has a diameter of 4 inches and a length of 30 inches. It it is submerged to a depth of 20 inches in a liquid with a specific gravity of 0.8, how many pounds of upward buoyancy force will be exerted on the displacer?

The answer to this is 7.238. I have no idea how to arrive at it. Formulas I'm finding seem to factor the weight of the displacer in this and that is not given in this problem.

Thanks for any help.
 

o3jeff

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This is what google says.

Formula for buoyancy force,

F = V * (L[SUB]w[/SUB]/L) * (B) * (SG)
Where F = buoyant force (in lb)
V = Total Displacer Volume (in cubic inches)
L[SUB]w [/SUB]= Working Length of displacer (inches)
L = Total Displacer Length (inches)
B = constant (weight of unit volume of water) (0.036 lb[SUP]f[/SUP]/in[SUP]3[/SUP])
For a cylindrical displacer, V = (π/4) * d[SUP]2[/SUP] * L = (3.142/4) * 16 * 30 = 377.04 cubic inch
So, F = 377.04 * (20/30) * 0.036 * 0.8 = 7.238 lb
 

Puck it

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This is what google says.

Formula for buoyancy force,
F = V * (L[SUB]w[/SUB]/L) * (B) * (SG)
Where F = buoyant force (in lb)
V = Total Displacer Volume (in cubic inches)
L[SUB]w [/SUB]= Working Length of displacer (inches)
L = Total Displacer Length (inches)
B = constant (weight of unit volume of water) (0.036 lb[SUP]f[/SUP]/in[SUP]3[/SUP])
For a cylindrical displacer, V = (π/4) * d[SUP]2[/SUP] * L = (3.142/4) * 16 * 30 = 377.04 cubic inch
So, F = 377.04 * (20/30) * 0.036 * 0.8 = 7.238 lb

You beat me.
 

Edd

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I'm confused about the Lw figure. You're calling it the working length of the displacer but for that you're using 20, which is the depth.

How is it that the depth is called the working length of the displacer?

Thanks very much for helping me.
 

Cannonball

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I'm confused about the Lw figure. You're calling it the working length of the displacer but for that you're using 20, which is the depth.

How is it that the depth is called the working length of the displacer?

Thanks very much for helping me.

Picture your displacer as a 30" long cylinder held vertically and only submerged down to 20". Working length = the length that is in the liquid.
 

Edd

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Picture your displacer as a 30" long cylinder held vertically and only submerged down to 20". Working length = the length that is in the liquid.

Ah, got it. For some reason I was picturing it fully submerged like sideways or something.
 

ctenidae

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Ah, got it. For some reason I was picturing it fully submerged like sideways or something.

Yeah, the question isn't too specific there- if you put it lengthwise 20 inches down, you'd get a different answer. Of course, in that configuration, the depth doesn't matter.
 

Edd

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This is what google says.

Formula for buoyancy force,

F = V * (L[SUB]w[/SUB]/L) * (B) * (SG)
Where F = buoyant force (in lb)
V = Total Displacer Volume (in cubic inches)
L[SUB]w [/SUB]= Working Length of displacer (inches)
L = Total Displacer Length (inches)
B = constant (weight of unit volume of water) (0.036 lb[SUP]f[/SUP]/in[SUP]3[/SUP])
For a cylindrical displacer, V = (π/4) * d[SUP]2[/SUP] * L = (3.142/4) * 16 * 30 = 377.04 cubic inch
So, F = 377.04 * (20/30) * 0.036 * 0.8 = 7.238 lb

To be crystal clear, is the d2 in this formula "displacer length squared"? Just spent 10 minutes Googling this and am not finding anything quite the same as this formula. I've gone stupid.
 

hammer

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To be crystal clear, is the d2 in this formula "displacer length squared"? Just spent 10 minutes Googling this and am not finding anything quite the same as this formula. I've gone stupid.
The d in this case is the diameter of the base circle.

Little confused by the formula since I usually calculate the volume of a cylinder as pi * r^2 * L but with d=2*r it's equivalent to (pi/4) * d^2 * L​. I also carry out pi to more digits and then round the result as needed...
 

Edd

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The d in this case is the diameter of the base circle.

Little confused by the formula since I usually calculate the volume of a cylinder as pi * r^2 * L but with d=2*r it's equivalent to (pi/4) * d^2 * L​. I also carry out pi to more digits and then round the result as needed...

Good lord, I was fried from studying when I posted this earlier today. Thanks Hammer.
 
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