Block hanging from cord submerged in water, find density

1. Aug 4, 2016

Taniaz

1. The problem statement, all variables and given/known data
A block hangs by a cord from a spring balance and is submerged in a liquid contained in a beaker. The beaker in turn rests on a kitchen scales. The mass of the beaker is 1 kg, the mass of the liquid is 1.5 kg. The spring balance read 2.5 kg and the kitchen scales reads 7.5 kg. The volume of the block is 0.0030 m^3.
(a)What is the mass per unit volume of the liquid?
(b) What will the spring balance and the kitchen scales read if the block is pulled up out of the liquid?

2. Relevant equations
Fb = weight of displaced liquid = ρgV = mg
mass of displaced liquid=mass of block
Volume of displaced liquid = volume of block

3. The attempt at a solution
For part a I'm a little confused with respect to the masses shown by the scales, how is the kitchen scales showing such a big reading? Where is it coming from? Part a wants the density, I already know the volume of the block = volume of displaced liquid, just need to figure out the mass.
Not sure of part b.

I have another hunch. If we subtract the mass of the beaker and that of the liquid from the kitchen scales mass we get 7.5-(1.5+1) = 5kg and I presume this is the mass of the block and the displaced liquid? Then we subtract the mass of the block which is 2.5 kg as shown by the spring balance I presume from 5 kg to get 5-2.5 = 2.5 kg as the mass of the displaced liquid?

Last edited: Aug 4, 2016
2. Aug 4, 2016

Staff: Mentor

What determines the kitchen scale reading?

Try doing an analysis of all forces acting on the "system" (beaker + liquid + block). What must those forces add to?

3. Aug 4, 2016

Taniaz

The kitchen scales reading is the mass of all 3 combined and the displaced liquid?
The downward force must be the weight of all 3 combined and the upward force is the upthrust or the buoyancy?

4. Aug 4, 2016

Staff: Mentor

What about that cord, which is pulling up?

Do this: Draw an imaginary box around the "system". Identify all forces acting on the system. (One of them will be the force exerted by the kitchen scale.)

5. Aug 4, 2016

Taniaz

Oh yes you're right, so the kitchen scale reading is the sum of 5 things?

So one will be the force exerted by the kitchen scale on the box, the weight and the tension in the string?

6. Aug 4, 2016

Staff: Mentor

Yes. And what must the net force on the system be? Set up an equation that adds up all those forces. (There will only be one unknown--the one you want to solve for.)

7. Aug 4, 2016

Taniaz

The net force should be 0?
So T+N = w where t is the tension, N is the force by the kitchen scales on the system and w is the combined weight
T will be 25 N? N will be 75N? and weight is the weight of the block, liquid and beaker? With the only unknown being the weight of the block?

8. Aug 4, 2016

Staff: Mentor

Good! (Since they are expressing the forces in terms of kilograms, that means that you'd multiply them all by g to get Newtons. But since they are all multiplied by g, the gs cancel.)

9. Aug 4, 2016

Taniaz

So we simply say
2.5 kg+ 7.5kg= (m of block + 1 kg + 1.5 kg)
so m=7.5 kg
since mass of block =mass of displaced liquid
and volume of block=volume of displaced liquid for density=m/V= 7.5/0.003 =2500 kg/m^3.
but how can the mass of the block equal the mass on
the kitchen scale? That doesn't make sense for part b.

Edit: for part b, I think the spring balance will read the same but I'm not sure what the kitchen scale will read

Last edited: Aug 4, 2016
10. Aug 4, 2016

Staff: Mentor

Looks good to me.

Why not? Just happens that the upward force of the tension cancels the weight of the beaker + liquid. So what?

Why do you think the spring balance will read the same?

Do the same kind of force analysis for part b. Note that the block is totally removed from the liquid now, so there's no longer an interaction between them.

11. Aug 4, 2016

Taniaz

No actually once it's removed, for the cord and the block there will be just 2 forces: Tension and weight of the block and for the beaker, liquid and kitchen scales, there will be the weight of the beaker and the liquid and the force by the kitchen scales so N= w which implies that 1.5+1=2.5 kg for the kitchen scales and T=w so 7.5 kg for the spring balance?

12. Aug 4, 2016

Staff: Mentor

Yes, you've got it.

13. Aug 4, 2016

Taniaz

Thank you! :)

14. Aug 4, 2016

TSny

OK

Why would this be true?

15. Aug 4, 2016

Staff: Mentor

Good catch. Luckily that statement wasn't used in the calculation.

The calculation is fine, but that verbiage surrounding it needs correction. (I should have done that earlier.)

The mass of the block you had just calculated. That has nothing to do with the mass of displaced liquid.

And the volume of the block was given. So you correctly calculated the density using m/V. No need for displaced liquid at this point.

16. Aug 4, 2016

TSny

But part (a) asks for the density of the fluid.

17. Aug 4, 2016

Staff: Mentor

Yikes. You're right. I noted that, but then forgot all about it.

That's the only part of the question that requires understanding buoyancy too! D'oh!

18. Aug 4, 2016

TSny

I'm finding myself doing the same sort of thing more and more lately. I'm gonna blame the hot, humid weather we're having.

19. Aug 4, 2016

Staff: Mentor

@Taniaz, I'm going to mark this one unsolved. Sorry that I was asleep at the wheel. (That's why we pay TSny the big bucks! )

OK, you correctly solved for the mass of the block. Now do a force analysis of the submerged block. Set up another equation. This time you'll need to make use of your knowledge of buoyant forces. The only unknown will end up being the density of the liquid.

Sorry for dropping the ball!

20. Aug 5, 2016

Taniaz

Buoyant force=weight
ρgV=mg
so ρV=m but I'll get the same thing with this? V=volume of block and m=mass of block?
Oh there will be tension too?
so T+ρgV=mg?

Last edited: Aug 5, 2016