Calculating Volume & Density of a Metal Object

[jefgreen]

Homework Statement

A metal object is suspended from a spring scale. The scale reads 920 N when the object is suspended in air, and 750 N when the object is completely submerged in water.

a. Find the volume of the object.
b. Find the density of the metal.

Homework Equations

d=m/v or v=md or m=dv
D of Water=1.00x10³kg/m³
1N=1kg
v=_____³

The Attempt at a Solution

Part A: v=md

(750N)(1.00x10³kg/m³)
V=7.5x10⁵

Honestly, I'm fairly sure that ^ is completely wrong... :uhh:

Part B: d=m/v

I am completely lost..

 

[jefgreen]

BUMP. Help please.

 

[Redsummers]

Do you know about the buoyant force?

 

[jefgreen]

Do you know about the buoyant force?

I have an equation sheet which mentions it.

Buoyant Force: Fbuoyant=DVg

I know:
Dh2o=1.00x10^3 kg/m^3
g=9.80 m/s^2

 

[Redsummers]

Okay, and you know that the buoyant force on a submerged body is directed in the opposite direction to gravity. It's always helpful to make a draw of what's really going on here.

So now you should think about what does the Archimedes principle states:

Real weight (920N) = apparent weight (measured as 750N) + weight of displaced fluid (buoyant force).

If you work out the maths here, you will get the volume. Don't hesitate to ask if you're not sure though.

 

[jefgreen]

Would I set it up like this?

920=750+(1.00x10^3)v(9.80)

and solve for v?

 

[Redsummers]

...

No, that's not the answer sorry.

I just gave you the equation that you have to use to work out the volume. It's not that tedious to plug numbers in.

950 = 720 + DVg

Letting D be the density of the fluid...

 

[Redsummers]

Would I set it up like this?

920=750+(1.00x10^3)v(9.80)

and solve for v?

Oh haha, okay you got it yeah.
And once you have the volume you can use the density equation (D = m/v) to get the density of the body.

Hope it's clear ;)

 

[jefgreen]

Would I set it up like this?

920=750+(1.00x10^3)v(9.80)

and solve for v?

I got .0872 for v.

 

[jefgreen]

What would I plug in for the mass?

d=m/v

D=M/.0872

What would I plug in?

 

[Redsummers]

Oh, when calculating the density, you should think that it is not important in which surroundings are you doing it, because rigid bodies will always have the same density. So for the density you should basically search what's the real mass of the body i.e.:

real weight (920) = M(Kg) · g(9.81m/s^2)

 

[jefgreen]

So...

920/.0872 will give me Density?

 

[Redsummers]

Nope, 920N is the weight. Weight and mass are two different things, the mass can be calculated as:

M = weight/gravity

Hence, M = 93.78Kg. Now you can plug this M in the density formula.

 

[jefgreen]

So the answer Part B is:

D=93.78kg
--------
.0872

which equals 1075.458716 or 1.1x10^3?

 

[jefgreen]

Is v= .0872 correct, by the way?

 

[Redsummers]

So the answer Part B is:

D=93.78kg
--------
.0872

which equals 1075.458716 or 1.1x10^3?

As long as your volume was .0872, that should be right. But check that out concerning the volume:

920-750 = 9.81 x 10^3 x V, are you sure that V is .0872?

 

[jefgreen]

No, v definitely is not .0872.

To find v do I set it up like this?

920-750=(9.80)(1.00x10^3)v

 

[jefgreen]

V=.0173 is what I have now...

 

[Redsummers]

Is v= .0872 correct, by the way?

Yeah, that's what I was wondering, I think you did some kind of algebra error or calculator typo, because I keep getting .0173 for the Volume.

Check now for the density as:

D = 93.78Kg / .0173m^3

 

[Redsummers]

V=.0173 is what I have now...

Bingo!

 

[jefgreen]

Answers:

Part A: V=.0171 m^3
Part B: D= 5.4x10^3

Is this correct? Also, what unit would density be in?

 

[Redsummers]

Yeah, that is numerically correct. Now for the units of the density, if we have:

D = m/V, we should know that mass is in (Kg) and V in (m^3], so density is just (Kg/m^3).

 

[jefgreen]

Yeah, that is numerically correct. Now for the units of the density, if we have:

D = m/V, we should know that mass is in (Kg) and V in (m^3], so density is just (Kg/m^3).

Thank you so much.