How Do You Calculate Buoyancy and Density for Different Blocks?

[JJBrian]

Homework Statement

1. A block is 7 cm wide, 8 cm long, and 2 cm tall and has a mass of 67 g.
a) What is its volume?
b) What is its density?
c) Floating in water (with the 7x8 face down), how deep will the bottom be (what is x in
figure 2)?
d) How much extra weight can it support without sinking?
2. A 2 cm by 2 cm by 2 cm metal block has a mass of 80 g.
a) What volume of water does it displace when it is submerged?
b) What is the weight of the water displaced?
c) What is the apparent weight of the block when measured under water?

Homework Equations

A=Lw.
V=Ah=Lwh.
density = m/v
Fbuoyancy = (Mass displaced water)*g
mblock*g = Fbuoyancy =mdisplaced water*g
---->density of block* volume of block = density water* Volume displaced water
----> DENSITY BLOCK*H = DENSITY WATER X
Fbouyancy = density water*Volume of block

The Attempt at a Solution

1a)V=Ah=Lwh.
(.08m)(.07m)(.02m)
V=0.000112m^3

1b)density = m/v
.067kg/0.000112m^3 = 598.214kg/m^3

c) Density block* h = densitywater*x
(598.214kg/m^3)(.02m) =(1000kg/m^3) x
x=0.019964m

d)m displaced water = Vblock* density water
Mdw = (.000112m^3)(1000kg/m^3)
Mdw = 0.112kg Mblock = .067kg
mass block - massdw = .045kg

2a)
V=Lwh
V=(.02)(.02)(.02) = .000008m^3
Density = 10,000kg/m^3
Pbmetal*Vmetal = density water*Vdisplaced water
(10,000kg/m^3)(.000008m^3) = (1000kg/m^3)( Vdw)
Vdw = 0.00008m^3

b)F bouyancy = density water * vblock*g
=(1000kg/m^3)(.000008m^3)(9.8m/s^2)
=0.0784N
dw =(1000kg/^3)(.000008m^3) =.008kg
Fbouyancy = Mdw*g= .0784N
c) I don't know...
F bouyancy = density water * vblock*g
=(1000kg/m^3)(.000008m^3)(9.8m/s^2)
=0.0784N



Can someone check my work?
Question 2a-c, I am unsure of. I need some explanation or some ideas to solve 2b and 2c.

 

[Doc Al]

1a)V=Ah=Lwh.
(.08m)(.07m)(.02m)
V=0.000112m^3

1b)density = m/v
.067kg/0.000112m^3 = 598.214kg/m^3

c) Density block* h = densitywater*x
(598.214kg/m^3)(.02m) =(1000kg/m^3) x
x=0.019964m

d)m displaced water = Vblock* density water
Mdw = (.000112m^3)(1000kg/m^3)
Mdw = 0.112kg Mblock = .067kg
mass block - massdw = .045kg

All good.

2a)
V=Lwh
V=(.02)(.02)(.02) = .000008m^3
Density = 10,000kg/m^3
Pbmetal*Vmetal = density water*Vdisplaced water
(10,000kg/m^3)(.000008m^3) = (1000kg/m^3)( Vdw)
Vdw = 0.00008m^3

Careful here. Since the metal cube is totally submerged, the volume of water displaced is just equal to the volume of the cube. Do not try to compute the volume of displaced water by setting its weight equal to the weight of the cube--it's not floating!

b)F bouyancy = density water * vblock*g
=(1000kg/m^3)(.000008m^3)(9.8m/s^2)
=0.0784N
dw =(1000kg/^3)(.000008m^3) =.008kg
Fbouyancy = Mdw*g= .0784N

OK.

c) I don't know...
F bouyancy = density water * vblock*g
=(1000kg/m^3)(.000008m^3)(9.8m/s^2)
=0.0784N

Imagine you suspended this metal cube by a string. The tension in the string is its "apparent" weight. In air, that tension just equals the weight of the cube. But when the cube is submerged in water, the apparent weight is reduced by the upward buoyant force. (Something floating would have an apparent weight of zero.)

 

[Borek]

In air, that tension just equals the weight of the cube.

In vacuum to be precise. It may seem as a nitpicking, but bouyancy in the air means you will be off by about 0.1% (order of magnitude, exact value depends on the density of the object). If you need high accuracy that's not a thing to forget

See http://www.titrations.info/volumetric-glass-calibration for example calculation in the real lab situation (scroll down to the text on the grey background).

--
methods

 

[Doc Al]

In vacuum to be precise.

Good point. Shame on me!

 

[Borek]

LOL it is the first time that I pointed out at yor mistake, so far it was always the other way around