# Difficult Bouyancy Question

1. Jun 15, 2017

1. The problem statement, all variables and given/known data

2. Relevant equations
F = ma
F = pVg
p = m/V

3. The attempt at a solution

First i tried finding the volume of displaced water which i did by...

(50/1000) * 10 = 1000 * 10 * V ---> V = 0.00005 m^3
Next as the metal sits on the bottom its weight force minus the bouyant force contributes to the 100 g mass increase (1 N increase) in weight force on the scales. Using this we can find Mass
Buoyant force provided by 50g water = 0.5 N
Therefore weight force of metal = 1 N - 0.5 N = 0.5 N

F = ma
0.5 = m *10
m = 0.05 kg

p = m/V = 0.05 / 0.00005 m^3
p = 1000...
Answer says its B.) but i dont know how.
I'm sorry to ask but i urgently need the answer to this one.
Is it ok to ask for working out, or a really big hint?

2. Jun 15, 2017

### Orodruin

Staff Emeritus
1 N is the weight minus the buoyant force ... so what do you need to do with the buoyant force and 1 N to get the weight?

Also, computing the buoyant force is to overcomplicate things ...

3. Jun 15, 2017

hello sorry for the late reply...
oh i see. so you need to add the bouyant force?
you then get 1 + 0.5 = 1.5N
1.5 N ---> 0.15 kg
0.15/0.00005 = 3000
thanks a lot!!!
but i dont get why you added, ill have to think about it

4. Jun 15, 2017

ok i understand now, the mass of the ball at the bottom already has buoyant force accounted for and so if it was out of the water it would have that weight force added back to it.
thanks a lot for helping me with my stupidity

5. Jun 15, 2017

### Orodruin

Staff Emeritus
You will always need to do some computation. However, the easiest way is to note that the weight of the ball is 3 times that of the water removed and the volume is the same. Therefore, the density is three times that of water.