Buoyancy force and floating

1. Feb 23, 2017

vishnu 73

1. The problem statement, all variables and given/known data
A 0.75-m rod has a uniform linear mass density of λ. A small mass m with negligible volume is attached to one end of the rod. The rod with the attached mass is placed in a container of unknown ﬂuid and after oscillating brieﬂy, comes to rest at its equilibrium position. At equilibrium, the rod ﬂoats vertically with 2/3 of its length submerged and mass m in the ﬂuid. If the rod were fully submerged it would displace 7.5×10-4 kg of ﬂuid.

(a) What is the maximum value that the mass m can have?
(b) What is the minimum value that the mass m can have?
(c) Sketch a graph that shows the values of λas a function of m

2. Relevant equations
fbuoyancy = g ρ v

3. The attempt at a solution
fboyant = fweight

g 5 x 10-4 = g (m + 0.75 λ)

how is m supposed to have maximum and minimum values for this equation

2. Feb 23, 2017

BvU

Think of the rest position of the rod. Is there a degree of freedom you overlooked ? Make a drawing.

3. Feb 23, 2017

vishnu 73

isn't there only one degree of freedom as the question specifies it is vertically aligned and and the mass is at the bottom

4. Feb 23, 2017

BvU

My mistake. I thought there was an angle left over, but you are right. In that case I don't see why m isn't fully determined and we're in the same boat

Maybe we should start calculating m and see where we end up ...

5. Feb 23, 2017

vishnu 73

but we need to know λ for that

6. Feb 23, 2017

BvU

Gather the equations - use the symbol $\lambda$

7. Feb 23, 2017

BvU

Oops, I see it's there already. 5 x 10-4 = m + 0.75 λ

Well then all I can think of is a liberal interpretation of 'floats vertically': that it means 'in a vertical plane, but at an angle with the liquid surface'

8. Feb 23, 2017

vishnu 73

so how is that supposed to help even if it floats at an angle how are we supposed to calculate the fluid it displaces

9. Feb 23, 2017

BvU

I think that stays at 2/3 of the volume of the rod. I was thinking about the of the given 'vertical orientation': if it wants to be stable, there is a condition. Perhaps that sets bounds on m ?

10. Feb 23, 2017

haruspex

Quite so, it is a question about stability.
@vishnu 73 , suppose the rod is not quite vertical, but at some small angle to the vertical. What relationship between the forces will lead it to rotate towards vertical?

Edit: to clarify, the stability consideration is for finding the minimum value of m. The maximum value is simpler.

11. Feb 25, 2017

vishnu 73

i did what you said but i got the opposite results meaning i got maximum value for when the rod is tilted and minimum value for m when the rod is perfectly vertical my calculation are as follows

when the rod is perfectly vertical:

fb = fweight
5 x 10-4 g = mg + λ 0.75g
5 x 10-4 = m + λ 0.75
5 x 10-4 - λ 0.75 = m this is the minimum value of m

when the rod is tilted i used torque about the point where the rod intersects the water surface = 0
hence,

fb 1/3 L= fweight 1/6 L

2x(5 x 10-4 g) = (mg + λ 0.75g)
10-3 = m + 0.75λ
10-3 - 0.75λ = m this is the maximum value for m i got

12. Feb 25, 2017

haruspex

That is the condition for it to float 2/3 submerged, vertical or not.
The mass centre of the rod is 1/6 below the surface, but not the mass m.

13. Feb 25, 2017

vishnu 73

so working it out again i got
for tiled rod:

fb = fweight
5 x 10-4(1/3) = (2/3)m + 0.75(1/6)λ
10-3 -0.75λ = 4m
m = (10-3 -0.75λ)/4

is this correct is this lesser than the m for the just floating one
and why is it that the tiled rod can have a lower value for m than the just floating rod.

14. Feb 25, 2017

haruspex

Your equations are right now, but you are not interpreting them correctly.
No, it's not the minimum value of m. It is a relationship between m and λ that must be true, given that it floats 2/3 submerged.
This is the relationship between m and λ when it is borderline whether it will be vertical or not (and if not, it will float horizontally; tilted is unstable).
But you do not care about λ, so combine the two equations to eliminate it.

Similarly, for the maximum value of m, the question wants the absolute maximum over all possible values of λ. This is very easy.

15. Feb 26, 2017

vishnu 73

huh i see what you are saying but when i combine the equations
m = (10-3 -0.75λ)/4
5 x 10-4 - λ 0.75 = m
and solving simultaneously i get one solution and not a range for me to find the minimum
i get m = 10-3/(4+5x10-4) this is just 2/8001

and for the maximum value of value the only thing i can think of is λ = 0;
but how that effectively is no rod in which case the m = 5x10-4

is my above working correct ?
i think i have a problem understanding this question the question already specifies the rod is vertical then i dont understand the need for calculating m when it is horizontal please help me explain what the question is asking thanks!

16. Feb 26, 2017

haruspex

The equations give you the minimum.
You did not calculate m for being horizontal. Your second equation is for the rod being on the borderline between remaining vertical and falling over. If it does fall over it will go horizontal. There is no stable solution between the two.

17. Feb 26, 2017

vishnu 73

i now get m = (10-3 - 5x10-4)/3 = 1/6000

why cant the rod be tilted why must it be vertical or horizontal
i am sorry i still dont get it but why does the above equation get me the minimum value of m
and so is my maximum value in the previous post correct
and if its not too much too ask please walk me through without equation what the is logic behind this entire question that would be very helpful

18. Feb 26, 2017

haruspex

Yes.

Your second equation says the torque is in perfect balance, that it will be stable at any angle. If m is a fraction more, the net torque will push it towards the vertical, and it will keep going until it gets there. If m is a fraction less, the net torque will push it away from the vertical, and it will keep going until it is horizontal. Note that your equation did not depend on the angle.
Because if m were any less the vertical arrangement would be unstable.
No. It needs to be expressed independently of λ. At what value of m is it impossible for the rod to float 2/3 immersed no matter how small λ is?

19. Feb 26, 2017

vishnu 73

ok now i get the minimum value part but idont understand the maximum value part i dont see how it is supposed to be simpler the best i can do is

m = 5x10-4 - 0.75λ

i dont see you can get rid of the λ

20. Feb 26, 2017

haruspex

What is the smallest possible λ?