#1
hey, need help with this question for class. can't seem to figure it out.

if you could show how you did it that would be awesome.
thanks


A large copper disc is to be lifted. Before hoisting a load weight is required. Calculate the weight of the solid copper disc ( Copper weighs 560 pounds per cubic foot. Assume pi=3.14) Answer to the nearest whole pound

Outer diameter 105.4 inches

Thickness 5.2 inches
#2
isn't it just the formula for a cylinder?

EDIT: i got 25407531.52
then again. i know nothing

EDIT 2: That's the cubic X the weight per cubic unit

EDIT3: 2117294.293333333 i don't even know.
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#3
Quote by misfitsramones
hey, need help with this question for class. can't seem to figure it out.

if you could show how you did it that would be awesome.
thanks


A large copper disc is to be lifted. Before hoisting a load weight is required. Calculate the weight of the solid copper disc ( Copper weighs 560 pounds per cubic foot. Assume pi=3.14) Answer to the nearest whole pound

Outer diameter 105.4 inches

Thickness 5.2 inches

if this wasn't in shitty imperial, it'd be much easier.
convert the inches to feet (i.e. divide by twelve) then half the diameter, square that number then times the resulting number by pi.

then times this by the thickness (in feet) and then multiply the answer by 560.



EDIT: I got 34156.463444...
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Quote by element4433
Yeah. people, like Lemoninfluence, are hypocrites and should have all their opinions invalidated from here on out.
Last edited by Lemoninfluence at Oct 2, 2011,
#4
Quote by Lemoninfluence
if this wasn't in shitty imperial, it'd be much easier.
convert the inches to feet (i.e. divide by twelve) then half the diameter, square that number then times the resulting number by pi.

then times this by the thickness (in feet) and then multiply the answer by 560.



EDIT: I got 34156.463444...



thats exactly what we have been doing but it says its wrong x.x

idk what else to do
#5
use (pi x diameter^2)/4 to find the area of the face of the disc, mulitply that by the thickness making sure you use the same units of distance. that will give you the volume of the disc.

convert squared inches to squared feet

multyply 560 by your answer to get the final weight of the disc
#6
Hooray for the metric system.
Just because I have some strong opinions doesn't mean I agree with everything I say.
#7
your mom weighs 560 pounds per cubic foot
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#8
Outer diameter 105.4 inches = 8.78 ft

Thickness 5.2 inches = 0.43 ft

Volume of a cylinder = (Diameter/2)^2 * pi * thickness

Volume = 104 ft^3

560 lbs/ft^3 * 104 ft^3 = 58,240 lbs

Edit: I made a mistake somewhere in the calculation, but the process should be right.
i don't know why i feel so dry
Last edited by Eastwinn at Oct 2, 2011,
#10
Quote by Eastwinn
Outer diameter 105.4 inches = 8.78 ft

Thickness 5.2 inches = 0.43 ft

Volume of a cylinder = (Diameter/2)^2 * pi * thickness

Volume = 104 ft^3

560 lbs/ft^3 * 104 ft^3 = 58,240 lbs


(8.78/2)^2=4.39^2 =19.272

19.272*3.14=60.514

60.514*0.43 = volume of cylinder = 26.02 cubic feet

where did you get the 104 from?
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Quote by element4433
Yeah. people, like Lemoninfluence, are hypocrites and should have all their opinions invalidated from here on out.
#12
then he'd end up with a value of 240+
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Quote by element4433
Yeah. people, like Lemoninfluence, are hypocrites and should have all their opinions invalidated from here on out.
#14
Curious, what level of difficulty is this meant to be?
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#15
actually it's my second choice for a major. i just ****ed up somewhere when i was plugging in numbers.
i don't know why i feel so dry
#16
I haven't looked at the problem properly (got annoyed by the imperial units ) but maybe everyone's forgetting to multiply by 9.81 for gravity? (that's 9.81 NKg^-1, no idea what the value is in imperial units though)

Remember it wants a load Weight - that's a Force, not a Mass. So we need to take gravity into account.

That get you anywhere?
#17
V=3.14r^2h
V=3.14(105.4^2)(5.2)
V=181390.36448 ft. cubed
V(560)=101578604.1088 pounds
That is a heavy piece of copper
#19
Quote by Gr!t
I haven't looked at the problem properly (got annoyed by the imperial units ) but maybe everyone's forgetting to multiply by 9.81 for gravity? (that's 9.81 NKg^-1, no idea what the value is in imperial units though)

Remember it wants a load Weight - that's a Force, not a Mass. So we need to take gravity into account.

That get you anywhere?

Pounds is a unit of weight

Quote by nicholasv98
V=3.14r^2h
V=3.14(105.4^2)(5.2)
V=181390.36448 ft. cubed
V(560)=101578604.1088 pounds
That is a heavy piece of copper

You're not converting to feet.
#20
Quote by nicholasv98
V=3.14r^2h
V=3.14(105.4^2)(5.2)
V=181390.36448 ft. cubed
V(560)=101578604.1088 pounds
That is a heavy piece of copper

you forgot to halve the diameter to get the radius and you forgot to convert the units.

what you're showing there is the weight of a disc with twice the radius that's made from a material which weighs 560 lbs/cubic inch.
Rhythm in Jump. Dancing Close to You.

Quote by element4433
Yeah. people, like Lemoninfluence, are hypocrites and should have all their opinions invalidated from here on out.
#22
105.4 in diameter means 52.7 in radius
52.7 in * 1ft/12in =4.931666666667 ft

5.2 in * 1ft/12in = 0.4333333 ft

r^2*pi*h = 26.24 ft^3

26.24 ft^3 * 560 lb/ft^3 = 14696 lbs

I fail to see the problem.

Quote by Gr!t
I haven't looked at the problem properly (got annoyed by the imperial units ) but maybe everyone's forgetting to multiply by 9.81 for gravity? (that's 9.81 NKg^-1, no idea what the value is in imperial units though)

Remember it wants a load Weight - that's a Force, not a Mass. So we need to take gravity into account.

That get you anywhere?


It's 32 ft/s^2 btw. I only know this because my Calculus class is dumb and is all "lawl let's do everything the stupid way."
Last edited by NoOne0507 at Oct 2, 2011,
#23
Why such shitty non-SI units? Then assume pi = 3.14? Stupid question.
#25
Quote by Avedas
Why such shitty non-SI units? Then assume pi = 3.14? Stupid question.


It's ~ 0.0507% error from a much closer approximation of pi, so it's not really going to mess up the answer.
#26

Outer diameter 105.4 inches

Thickness 5.2 inches


4 significant figures then 2 significant figures. 3.14 = 3 significant figures.

why.jpg
#28
Quote by Dirge Humani
Its a pure math problem, they almost never care about significant figures in pure math.


Meh, true enough. Inconvenient units, though.
#30
Quote by Dirge Humani
Its a pure math problem, they almost never care about significant figures in pure math.


No it's not. This is totally applied math.
i don't know why i feel so dry
#31
You people across the pond with your imperial stuff...
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