# Homework Help: Density Of the Object

1. Nov 28, 2008

### Naldo6

An object with density of 831 kg/m^3 and *** of 1003 kg is thrown into te ocean. Find the volume that sticks out of the water. ( use density of water= 1024 kg/m^3)

I use F_b= m'g
m_water*g=m_object*g
p_water*V_water*g=p_object*V_object*g
v_object=(p_water*V_water)/p_object

but what is the volume of the water?....

are my formulas ok?....

2. Nov 29, 2008

### horatio89

Hey Naldo6,

By the Achimedes' Principle, the volume of water displaced = volume of the block submerged. However, since the object is floating, the volume of water displaced is only a fraction of the volume of the whole object. Your equations are correct, however, your problem may have arisen from your definitions of the variables.

V_object=(p_water*V_water)/p_object

Consider this equation - the V_object here is: the volume of the object itself, or the volume of the object submerged? What should be the subject of the equation? (Hint: look at the statement at the beginning of the post)

Regards,
Horatio

Last edited: Nov 29, 2008
3. Nov 29, 2008

### Naldo6

i really dont understand....

4. Nov 29, 2008

### Naldo6

V_object=(1003)(831)=8.33x10^5

but then that is the wrong answer.... how i do this exercise?..help plz

5. Nov 29, 2008

### Staff: Mentor

So far, so good. Now find V_object and V_water and compare.

That's not "the answer", it's a step toward the answer. Good. Now figure out V_water (the volume of water displaced) using your equations. Then calculate out how much of the object's volume sticks out of the water.

6. Nov 29, 2008

### Naldo6

V_water=(V_object*p_object)/p_water
so V_water = (8.33x10^5)(831)/(1024)= 6.76x10^5 m^3

ok but how i can calcuta how much of the objects's volume sticks out of the water

7. Nov 29, 2008

### Staff: Mentor

Hint: The part that sticks out is the part that doesn't displace water.

8. Nov 29, 2008

### Naldo6

ok, but how i know that?....

9. Nov 29, 2008

### Staff: Mentor

You calculated two volumes. One represents the volume under water, the other represents the total volume.

10. Nov 29, 2008

### Naldo6

so that the answer to the problem is the diference between the total volumne and the volume under the water?....

11. Nov 29, 2008

### Staff: Mentor

Yes.

12. Nov 29, 2008

### Naldo6

i put that answer and give me the wrong answer
i did:

8.33x10^5 6.76x10^5= 1.57x10^5 m^3 and that is not the crrect answer...plz can u cheq me in my calculus aboves...

13. Nov 29, 2008

### Staff: Mentor

Oops... this calculation is incorrect.

Density = Mass/Volume, so:

Volume = Mass/Density.

(Redo your values for V_object and V_water.)

14. Nov 29, 2008

### Naldo6

ok ty.... i get my corrects answer... but can u help me with the other exercise i post about density...

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook