Force on object located between two liquids

[songoku]

Homework Statement

A cube has side as shown. If the atmospheric pressure is p, find the force acting at the bottom of the cube!

Homework Equations

P = F / A
P = ρ.g.h

The Attempt at a Solution

F_bottom = P_bottom . A
= (p + Pressure by upper liquid + P by lower liquid) . 4/9 b^2
= (p + Pressure by upper liquid + d.g.1/3 b). 4/9 b^2

I don't understand how to find the pressure by upper liquid. I guess I have to use P = ρ.g.h but I don't know what the 'h' is.

By definition, h is depth so we measure the depth from the surface, but we don't know how deep is the cube from the upper liquid surface.

I think I'm missing something...

 

[cupid.callin]

Homework Statement

A cube has side as shown. If the atmospheric pressure is p, find the force acting at the bottom of the cube!

Homework Equations

P = F / A
P = ρ.g.h

The Attempt at a Solution

F_bottom = P_bottom . A
= (p + Pressure by upper liquid + P by lower liquid) . 4/9 b^2
= (p + Pressure by upper liquid + d.g.1/3 b). 4/9 b^2

I don't understand how to find the pressure by upper liquid. I guess I have to use P = ρ.g.h but I don't know what the 'h' is.

By definition, h is depth so we measure the depth from the surface, but we don't know how deep is the cube from the upper liquid surface.

I think I'm missing something...

I'm not sure but can't we use buoyant force? viz = weight of liquid displaced ...

 

[songoku]

I'm not sure but can't we use buoyant force? viz = weight of liquid displaced ...

I don't know. As far as I understand, bouyant force is the difference between the force acting on the bottom and the force acting on the top of the object; and the question is asking only about the force on the bottom.

btw, what is viz?

 

[RTW69]

Do a Google search for "Object floating in two liquids". I think the key to this problem is to find the density of the object and you can do that if you know the density of the two fluids and the distance submerged in both liquids (which happens to be the same distance).

Then F_boy=ρ_obj*g*vol_displaced and you know the dimensions of the cube

 

[songoku]

Do a Google search for "Object floating in two liquids". I think the key to this problem is to find the density of the object and you can do that if you know the density of the two fluids and the distance submerged in both liquids (which happens to be the same distance).

density of object = 1/2 D + 1/2 d


I don't understand this :

Then F_boy=ρ_obj*g*vol_displaced and you know the dimensions of the cube

I think the density should the the density of liquid, not object.
F_boy = ρ_liquid.g.V_immersed; because therea re 2 liquids, I think there are 2 bouyant forces

F_boy = D.g.4/27 b³ + d.g.4/27 b³

I am not sure I found the bouyant correctly, and also how to find the force acting at the bottom part after finding bouyant force?

Another question, how to take atmospheric pressure into account?

Thanks

 

[RTW69]

My error, it should be ρ_Liq. I think your answer is correct. I think you can neglect the atmospheric pressure component because the density of air is so much less than the density of the fluids.

 

[cupid.callin]

One thing is troubling me ... what will be the direction of buoyant force ?

They will depend on the densities of the liquids and block (maybe) ... Do we know that ?

 

[songoku]

Sorry I didn't give the choices for this question.

a. $$b^3 [(D+d)g + \frac{p}{b}]$$

b. $$\frac{4}{27}b^3 [(3D + d)g + \frac{3p}{b}]$$

c. $$b^3 [(D - d)g + \frac{3p}{b}]$$

d. $$\frac{4}{27}b^3[(3D-d)g+\frac{3p}{b}]$$

e. $$\frac{4}{27}b^3[(D+d)g+\frac{p}{b}]$$Good question cupid, taking the consideration that some of the choices contain negative sign. It means that we should know the direction of the bouyant force; which I don't know...:(

Looking from the choices again, we should take atmospheric pressure into account.

Another question, is bouyant force the same as force acting on the bottom of the cube?

Thanks

 

[songoku]

I think the direction of bouyant force is still upwards. Half of it is immersed in D and half is in d. Both will produce bouyant forces and both are upwards.

But I still don't know how to find the pressure on the bottom...any help please?

 

[gaurav891101]

A section of the pipe with an internal diameter of 10cm tapers to an inner diameter of
6cm as it rises through a height of 1.7m at an angle of 60o with respect to horizontal.
The pipe carries water and its higher end is open to air. (a) If the speed of the water at
the lower end is 15cm/s what are the pressure at the lower end and speed of water as
its exits. (b) If the higher end of the pipe is 0.3m above the ground at what horizontal
distance from the pipe outlet does the water land.

 

[songoku]

A section of the pipe with an internal diameter of 10cm tapers to an inner diameter of
6cm as it rises through a height of 1.7m at an angle of 60o with respect to horizontal.
The pipe carries water and its higher end is open to air. (a) If the speed of the water at
the lower end is 15cm/s what are the pressure at the lower end and speed of water as
its exits. (b) If the higher end of the pipe is 0.3m above the ground at what horizontal
distance from the pipe outlet does the water land.

I don't understand the relation of your post with the question I posted. Are you asking an entirely new question?