Force exerted on the horizontal bottom of the container

[Hidd]

Homework Statement

In an empty container, having the shape of a Rectangular Parallelepiped, with width l=0.5m and MN=H=0.4m, and NP=L=0.75M. we poureda mass of mercury =2448kg.
T0,T1,T2 have the same section which is 0.2 m^2.
we give g=10N/kg, and density of mercury u=13.6 g/(cm^3).

determine the pressing force exerted on the horizontal bottom of the container?

Homework Equations

Δ[/B]p/Δz = -ρ*g

dF=p*dS
3. The attempt at a solution
Since that we already have the mass, i tried to solve it by using Newton's second law.
F=m*g
=2448*10=24480N
I thought it was an easy problem and than i took a look at the solution here is what i found:

the volume of mercury is 2448/(13.6*1000)=0.18 (m^3)
the volume in the Parallelepiped is H*L*l=0.15 (m^3)
than the volume in the 3 tubes is: V=(0.18-0.15)/3=0.01 (m^3)
so h= V/s=0.5m

the pressure is p=u*g*(H+h)

so F=p*s=p*l*L
=u*g*(H+h)*l*L
=45900N

I understand the solution but i don't understand why i coudn't apply Newton's second law? why did i have a contradiction?

Thanks in advance and All the best,

 

[berkeman]

Sorry, I have trouble following your post. What is the total mass? What is the total weight force? What is the area of the base? Pressure is Force/Area. Can you please fill in these numbers and compare to the solution you are given? Thanks, and sorry if that is all in your post.

 

[Hidd]

The total mass of Mercury 2448kg, the total weight force is 24480N, the area of the base is 0,375 m²,
the pressure (of the weight) is (24480/0,375)=65280Pa

what i did't understand is why the wight force is not equal to the force exerted at the bottom of the container.
if we applied Newton's laws on the fluid (in rest), we should get the same magnitude for the weight force and for the force exerted by the the bottom of the container, right?

 

[Chestermiller]

The total mass of Mercury 2448kg, the total weight force is 24480N, the area of the base is 0,375 m²,
the pressure (of the weight) is (24480/0,375)=65280Pa

what i did't understand is why the wight force is not equal to the force exerted at the bottom of the container.
if we applied Newton's laws on the fluid (in rest), we should get the same magnitude for the weight force and for the force exerted by the the bottom of the container, right?

Well, no. What are the forces that the fluid exerts on the container in the vertical direction? (Hint: don't only look at the bottom)

 

[Hidd]

Well, no. What are the forces that the fluid exerts on the container in the vertical direction? (Hint: don't only look at the bottom)

Well, beside the weight force, I couldn't find other forces!

 

[Chestermiller]

Well, beside the weight force, I couldn't find other forces!

Is the fluid exerting an upward force on the lid, given that the pressure on the bottom of the lid is $\rho gh$?

 

[Hidd]

Is the fluid exerting an upward force on the lid, given that the pressure on the bottom of the lid is $\rho gh$?

I think that the container is exerting a normal force on the fluid, and the pressure at the bottom must be : ρg(h+H).
I guess the other force must be the "air pressure"?!

 

[Hidd]

What i don't understand is that: if I apply Newton's law on the "Fluid" in order to find the normal force which is equal to the "pressing force exerted by the fluid on the horizontal bottom of the container", I don't get the same result as by using the "hydrostatic force equation"

 

[Chestermiller]

What i don't understand is that: if I apply Newton's law on the "Fluid" in order to find the normal force which is equal to the "pressing force exerted by the fluid on the horizontal bottom of the container", I don't get the same result as by using the "hydrostatic force equation"

First let's consider the forces that the fluid exerts on the container. This is minus the force that the container exerts on the fluid. The downward force that the fluid exerts on the base of the container is $\rho g (H+h)A$, where A is the area of the base. The upward force that the fluid exerts on the lid of the container (MQ in the figure) is $\rho g h(A-A_P)$, where $A_P$ is the total area of the three vertical pipes extending above the lid. So the net downward force that the fluid exerts on the container is $$F=\rho g (H+h)A-\rho g h(A-A_P)=\rho g H A+\rho g h A_p$$ But this is just the weight of the fluid (as expected). And this is equal to the net upward force that the container exerts on the fluid.

Now, let's next consider the forces acting on the base of the container. If we do a free body diagram on the base, we find that there are 3 vertical forces acting: (1) The upward normal force of the ground $F_g$ (2) The downward pressure force of the fluid $F_f$ and (3) the upward tension exerted by the vertical sides of the container on the base T. From the previous analysis, we showed that $$F_g=Weight=\rho g H A+\rho g h A_p$$ and $$F_f=\rho g (H+h)A$$So, from a force balance on the base, we find that the tension in the vertical walls of the container is given by: $$T=\rho g (H+h)A-(\rho g H A+\rho g h A_p)=\rho g h(A-A_P)$$
So the tension in the walls is not zero in a case where the fluid in the three tubes extends above the lid.