1)
The pressure at a depth h in a fluid is given by the formula P = ρgh, where ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth.
a) At the top face of the cylinder (0.20m below the surface):
P_top = ρgh + P_atm
P_top = 1000 kg/m³ * 9.81 m/s² * 0.20m
P_top = 1962 Pa
b) At the base of the cylinder:
P_base = ρgh + P_atm
P_base = 1000 kg/m³ * 9.81 m/s² * 0.50m
P_base = 4905 Pa
2)
The force due to pressure on a face of the cylinder is given by the formula F = PA, where P is the pressure and A is the area of the face.
a) Force on the top face:
F_top = P_top * A
F_top = 1962 Pa * 0.50m²
F_top = 981 N
b) Force on the base face:
F_base = P_base * A
F_base = 4905 Pa * 0.50m²
F_base = 2452.5 N
The net upward force due to water pressure is:
F_net = F_base - F_top
F_net = 2452.5 N - 981 N
F_net = 1471.5 N
3)
The upthrust on the cylinder calculated using Archimedes Principle is equal to the net upward force calculated in (2) as both represent the buoyant force acting on the cylinder.
4)
If the cables are removed, the cylinder will float as the buoyant force acting on the cylinder is greater than its weight, so it will rise to the surface of the water.
A solid cylinder, mass 100kg, is suspended in water between two tight cables so that the circular faces are horizontal and the upper face is 0.20m below the surface.The area of a circular cross-section of the cylinder is 0.50m² and it's height is 0.30m. The density of water is 1000kg/m³.
1) Calculate the pressure in excess of that due to the atmosphere at the top and at the base of the cylinder
2) Calculate the corresponding forces acting on each side of these faces and hence the new upward force due to the water pressure
3) Show that the upthrust on the cylinder calculated using Archimedes Principle is equal to that calculated at (2)
4) Will the cylinder sink, float or rise if the cables are removed?
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