Hydrostatic equilibrium of a cylindrical area

In summary, the conversation discusses the formula for volume of a cylindrical region at a distance from the center of the sun, as well as the net force on the cylinder due to gravity and pressure. The net pressure equation shows that the force is the difference between the pressure at two points multiplied by the area. It is important to note the sign of the force, as it determines the direction of the force towards or away from the center of the star. The negative sign in the gravity equation and the positive sign in the pressure equation ensure that the forces are in opposite directions. The value of G is always positive and the negative sign in the force equation is necessary for the direction of the force towards the center.
  • #1
shounakbhatta
288
1
Hello,

Consider a cylindrical region (length dr, end area dA) at a distance r from the center of the sun

Density =ρ(r)
Volume = Length x area = dr.dA (How is the formula for volume of a cylinder works here?)
Mass= Density x Volume = ρ(r).dr.dA

Now, computing F(grav) = -G.m(r).dm/r^2...we get F(grav)=-g(r)ρ(r)drdA. I understand up to that.

But the net pressure equation shows:

F(press)=(P(r+dr))- P(r))dA)

Can you please explain me how the above equation arrives?

Thanks
 
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  • #2
F(press)=(P(r+dr))- P(r))dA)

Remember that force is pressure multiplied by area. This appears to be saying that the net force on the cylinder is the force on the top of the cylinder pushing downwards minus the force on the bottom of the cylinder pushing up.
 
  • #3
Hello colin456,

Thanks a lot for the help. I could now understand the equation.
 
  • #4
Just one thing to be careful about. Do not forget about the sign of the force. Your gravity equation has a minus in it, which means the force is negative. Since gravity points towards the middle of the star, that means that positive force is by definition away from the centre of the star. However, the force from the pressure equation as you wrote it is going to be negative since P(r) is greater than P(r+dr) which means that it also will be directed towards the centre of the star, which would be wrong. I am sure you would notice this when you put the two equations together, but it is worth being clear about.
 
  • #5
Hello colin456,

Sorry for the late reply. As you have written the 'minus' sign '-' means the force is negative. Correct me if I am wrong: the '-' sign is also as gravity is attractive not repulsive. Right? Yes the force from the pressure eqn.is -ve.
 
  • #6
There is no rule that says that negative force is attractive. The minus sign does not mean that gravity is attractive. It means that gravity is pointing in the negative r direction. This is what you want since you are using spherical coordinates with r=0 is defined as the centre of the star, and positive r values are pointing away from the centre of the star. The force in the pressure equation needs to be positive since it must be in the opposite direction to the gravity force.
 
  • #7
Can you tell me one thing: F=-G(m1.m2)/R^2. What does -G denotes? Pointing downwards?
 
  • #8
G is just a number. It stands for 6.67384 × 10-11 m3 kg-1 s-2 and is therefore always positive. The masses, m1 and m2, are also always positive, and R^2 must be positive because it is the square of a real number. Therefore, in order for the force to be negative, which it needs to be to point towards the centre of the coordinate system, the negative sign is necessary.
 
  • #9
Ok, clear.

Thanks a lot.
 

1. What is hydrostatic equilibrium?

Hydrostatic equilibrium refers to the balance of forces acting on a fluid within a confined space, such as a cylindrical area. It occurs when the weight of the fluid is evenly distributed and balanced by the pressure exerted by the surrounding walls or other forces.

2. How does a cylindrical area maintain hydrostatic equilibrium?

A cylindrical area maintains hydrostatic equilibrium by ensuring that the weight of the fluid is evenly distributed along the vertical axis. This is achieved through the balance of gravitational force and pressure exerted by the fluid's weight, as well as any external forces acting on the fluid.

3. What factors affect hydrostatic equilibrium in a cylindrical area?

Some of the factors that can affect hydrostatic equilibrium in a cylindrical area include the density and viscosity of the fluid, the shape and size of the cylindrical area, and any external forces or objects present in the space.

4. What happens when hydrostatic equilibrium is disturbed in a cylindrical area?

If hydrostatic equilibrium is disturbed in a cylindrical area, the fluid will respond by moving and adjusting its shape to regain balance. This may result in changes in pressure and flow within the fluid, as well as changes in the position of the fluid's surface.

5. How is hydrostatic equilibrium important in various applications?

Hydrostatic equilibrium is important in many areas of science and engineering, including fluid mechanics, geology, and oceanography. It helps us understand and predict the behavior of fluids in different environments, such as in pipelines, reservoirs, and natural bodies of water.

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