# PVg- volume weight

1. Nov 9, 2005

### Izekid

I have a problem that I can't solve,

A bathball with the volume 3,0dm^3 is floating on water and has 6,0% of it's volume down in the water.

a) Count the bathball's mass , Easy : 0,06*3,0= 1,8hg = 0,18 kg

Now comes the real problem

b) how much power do you need to use to force the whole bathball down under water?

Well for this I though i'd use p*v*g /p (where p is water density)
Water density 1,0 * 10^3 = 1000*0,3m^3*9,82 = 2946 / 1000 = 2,946 N

Which totaly wrong it should be 28 N how do I solve this?

2. Nov 9, 2005

### Päällikkö

Draw a diagram. You should have three forces.

"p*v*g /p"
How did you get this?

3. Nov 9, 2005

### Izekid

Eh????

Don't know what you're suggesting please understand I'm no science freak :surprised :rofl: :zzz:

Yeah I have 3 forces gravity is one my power pressing the bathball down is one and the will to go up is the third. But I do not not the force pressing the item down!?
And how do I get to know that ?
Give me a formula or something?

4. Nov 9, 2005

### Päällikkö

There isn't much help I can give without solving the problem. Here goes, though:

Well, the forces should equal zero (for the minimal force).
So:
$$mg + F = \rho _{water} Vg$$