Calculating Relative Density and Apparent Weight Using Archimedes Principle

[lionely]

Homework Statement

A piece of sealing-wax weighs 0.27 N in air and 0.12 N when immersed in water. Calculate :
a) It's relative density
b) It's apparent weight in a liquid of density 800 kg/m^3




The attempt at a solution
I did part a) I got 1.8 I have no idea how to attempt part b) this part of physics is hard my teacher at school didn't teach it well, all he did was give notes. :(

 

[mr.me]

Next time try a little harder with posting relevant equations. Homework's often confusing and I make post for help that are sparse too but it helps show your earnest

Nonetheless, I believe that you must first find the volume of the wax and then the mass of the liquid that is being moved, Remember that mass=density*Volume
or m=d*V

Then find the apparent mass of the block

and then finally find the apparent weight of the block

weight is a "type of force" which can be given with the equation F=m*g where g is gravity's constant of 9.8ms^2

There may be a shortcut to all this but I can't think of it of the top of my head

 

[lionely]

But how would you find the volume here?

 

[mr.me]

Well, if mass=Volume*density and Volume=mass/density and you said you already have d=1.8

--and f=m*g

Then can you say how you would find the Volume?

 

[lionely]

Hmm Can I calculate the mass by using f=ma and m= f/a? and use that mass in m/d?

 

[mr.me]

Yes

You already have the force F Newtons stated in the original problemsolve f=m*a for the mass because you know that a in this case is equal to 9.8ms^2 and the force F Newtons is given

then you have the mass to use in the m/d

 

[lionely]

Alright here's what I did, f=ma, m= f/a, .27/10= .027kg.

volume = md = .027kg x 1800kgm-3 = 48.6 m^3

Apparent mass = 800 x 48.6 = 38,880kg?

:S This seems wrong.

 

[mr.me]

volume doesn't = m*d

volume=m/d

you might need to review solving for a variable in linear equations

Ex.) V = m/d = 32 / 2700 = 0.0119 m^3

 

[lionely]

Oh sorry! I'm very tired it's 2am here.

 

[lionely]

I ended up getting 0.12 N but in the back of my book it says 0.15N.

 

[mr.me]

That's ok,

Remember to use a calculator to avoid silly mistakes.

Maybe you just need some sleep It might make sense in the morning

 

[mr.me]

Did you use the correct value of the Newtons?

 

[lionely]

yes the weight in air is 0.27 N , so I divided that by 10 to get the mass.

 

[Phrak]

What is "apparent weight"?

This seems to be some vague reference to what the weight 'appears' to be, without clarification as to how the peering is to be done.

 

[lionely]

The weight a body appears to have in a fluid?

 

[Phrak]

The weight a body appears to have in a fluid?

In this context, this seems reasonable. From a) you should have the mass and weight of the object in air, so you have enough information to calculate the weight in the 800 grams/liter fluid to solve part b).

 

[lionely]

The .12 N I got should I subtract it from the actual weight? Sorry for my ignorance, I don't know this part of physics well at all!