Archimedes principle

[tandoorichicken]

A wooden block 20 * 20 *10 cm^3 has a density of .6 g/ cm^3
(a) How much iron (density = 7.86 g/cm^3) can be placed on to p of the block if the top of the block is to be level with the water around it?
(b) If iron were attached to the bottom of the block instead, what mass of iron would it take to bring the top of the wooden block down to the level of the water?

[ShawnD]

Originally posted by tandoorichicken
A wooden block 20 * 20 *10 cm^3 has a density of .6 g/ cm^3
(a) How much iron (density = 7.86 g/cm^3) can be placed on to p of the block if the top of the block is to be level with the water around it?
(b) If iron were attached to the bottom of the block instead, what mass of iron would it take to bring the top of the wooden block down to the level of the water?

A).

find the mass of the wood block. iirc, density is P or something like that.
m = pv
m = (0.6)(20)(20)(10)
m = 2400g

find mass of the water if it were the same size
m = pv
m = (1)(20)(20)(10)
m = 4000g

difference of mass:
4000 - 2400 = 1600g

volume of iron:
v = m/p
v = (1600)/(7.86)
v = 203.56 cm^3




B). This is a neat question.

Just as before, the difference between the wood and the water is 1600g but now we have to take into account the boyant force on the iron. To do that, we have to make a new equation to represent the weight (or mass) of the iron.

Gravity is a constant on every term so I can just leave it out. The mass in the equation represents the mass of iron which has an effective downward force. In both terms, the v is the same but the p is different.
m = pv (iron) - pv (water)
1600 = (7.86)v - 1v
1600 = 6.86v
v = 233.236 cm^3