a tank has a mass of 100 kg and a length of 3 m has an area of 30 cm2 is allowed to stand vertically:?

Question

a.) what pressure does the cylinder exert on the floor?
b.) if it is filled with water, what is the total force at the bottom?
c.) find the total force at the walls of the container.

answers in (b) and (c) should be in Newtons

Answer 1

I didn't answer this one first because seeing Helmut's answer just 2 minutes (then) before I found it correct... but now I find the doubt about total force, and want to clarify, if I can, about it.

In effect, this force is an integration of elementary forces:

dF = p dA = ρ g h dA

where ρ is the density of the liquid contained.

But regardless of the container's shape these forces are in equilibrium at each depth or height (horizontally).

Particularly with a cylinder the elementary area
dA = 2¶ R dh
can be divided in infinite elementary segments:
dA' = ds dh
each of which is simetrically opposed to another element of area supporting the same pressure. The resulting elementary forces on these are equal in module and are opposing in sense. Hence the resultant is zero.

Extending this to all elements of area at the same depth/height the integration renders zero for result.

And extending the integration of 0 vertically along all the height results in zero too.

HOWEVER: when it comes to calculating wall thickness we MUST consider the pressure. The cylinder is considered to be cut by it's middle part, the pressure
p = ρ g h
is considered to be applied on the transversal cylinder section: 2R dh
and supported by the radial sections of the container:
2 t dh (where t is the container's wall thinkness).

Now this isn't being asked... so let's leave it here.

==================================

Well, regarding the problem: reading it carefully it suggests that the mass of 100 kg is the empty container's:

a) p = weight / Area = 100 kg * 9.8 m/s² / 0.003 m²
p ≈ 327 kPa

b) This force is the container's weight plus the water's weight:

F = m g + p(w) A = m g + ρ g h A = g (m + ρ h A)
= 9.8 m/s² (100 kg +1000 kg/m³ * 3m * 0.003 m²)

F ≈ 1068 N

c) all the first part of my answer shows that the force within the walls is zero (horizontal forces, bottom not considered as wall).

Hope this helps.
Regards!

Answer 2

Given:
m = 100 kg, mass of cylindrical tank
L = length of the cylindrical tan
A = area (???) = 30 cm^2 (I suppose you're talking about the
area of one of the circular bases?)

Find:
a) P = pressure the cylinder exert on the floor?

Solution:
P = Fg/A
P = mg/A
P = 100 kg x 9.8 m/s^2/(30cm^2)(1 m^2/100^2cm^2)
P = 327000 N/m^2

b) P = [Weight of Water + Fg]/A
P = [Volume of Tank x density of water x g + Fg]/A
P = [30cm^2(1m^2/100^2cm^2)(3m)(10^3kg/m^3)(9.8m/s^2) +
980N]/0.003m^2
P = 356067 N/m^2 ANS

c) Total force at the walls of the container.
I am not sure about this one. I think it involves integration since the pressure at different depths of the tank are not the same. As we go deeper, the pressure increases. The pressure is greatest at the bottom of the tank.
The TOTAL PRESSURE WITHIN the water at different depth
h can be computed using the following formula:
d = p atm + dgh
where, p atm is the pressure exerted by the atmosphere on the surface of the water; d is the density of water, g is acceleration constant due to gravity; h is the depth of water.
So, if h = 3m (that is at the bottom of the tank), then

d = 1.01 x 10^5 N/m^2 + 1000 kg/m^3(9.8 m/s^2)(3m)
d = 130400 N/m^2

teddy boy

Answer 3

a) P = mg/A
P = (100 kg)(9.80665 m/s^2)/(0.0030 m^2)
P ≈ 326,888 Pa ≈ 327 kPa
b) P = P1 + ρgh
P ≈ 326,888 Pa + (1 g/cm^3)(1 kg/10^3 g)(10^6 cm^3/m^3)(9.80665 m/s^2)(3 m)
P ≈ 356,308 Pa ≈ 356 kPa
c) The total force on the walls of the tank is 0