UNIT 4
Density and Pressure
Example 1
Example 2
Example 3
Example 4
Example 5
Example 6
Example 7

 

Density and Pressure in Fluids
Example 4

example1

(a) Two teams of eight horses could not pull apart two evacuated brass hemispheres. Assuming that the thickness of the hemisphere walls is negligible compared to their diameters, derive the expression Fh = πR2Δp relating the force required to pull the hemispheres apart to their radii and the pressure difference between inside and outside the hemispheres.
(b) If R = 60 cm and the pressure inside the is 0.1 of the pressure outside find the force that would be required to pull them apart.
(c) In the original demonstration two teams of horses were used. Explain how one team can be used with a sturdy post to achieve the same outcome.

What
example1

(a) If we slice the hemispheres into a number of infinitely thin layers we have to calculate the area of the slice and the horizontal component of the force produced by the pressure.
The radius of the hemisphere = R
At any angle θ:
The radius of the slice r = R Sin θ
Thickness of the slice = R dθ --- (measuring θ in Radians)
The surface area of the slice exposed to the atmosphere = dA
dA = (2πr) R dθ
dA = (2π R Sin θ) R dθ
dA = 2π R2Sin θ dθ
The force exerted by the horses is exactly opposed to force exerted by the the air pressure = Air pressure difference X Area
= Δp X dA
= Δp (2π R2 Sin θ dθ)
But only half of this force pulls to the right so it should really be [Δp (π R2 Sin θ dθ)], however, the force on the lower right half of the hemisphere is also the same so adding the upper and lower element it goes back to [Δp (2π R2 Sin θ dθ)]
The horizontal component of the force for each element ΔFh = [ (Sin θ dθ)] Cosθ
So the total force pulling to the right Fh is the Integral of all the elements.

Fh
 
=
  2πR2 Δp π/2
  Sin θ Cos θ dθ
0

Using the identity ∫Sin θ Cos θ dθ = ½ Sin θ2

Fh
 
=
  2πR2 Δp [½ Sin θ2] | π/2
   
0

Fh= 2πR2 Δp {(½) [(1)- (0)]}

Fh= πR2 Δp

(b) R = 0.60 m
Δp = 9.09 × 104 Pa)
Fh = π(0.60 m) (9.09 × 104 Pa) = 5.45 × 104 N.

(c) One team of horses could be used if one half of the sphere is attached to a sturdy post. The force of the post on the sphere would oppose the force of the horses by Newton's Third Law. As the team force increases, the opposing reaction from the post will increase, provided the post does not break or uproot.

   
<
>
Concept by Kishore Lal. Programmed by Kishore Lal... Copyright © 2015 Kishore Lal. All rights reserved.