Physics 102 Part II Thermal Physics Moza M. Al-Rabban Professor of Physics Fluids (2)

Outline 15-1 Density 15-2Pressure 15-3Static Equilibrium in Fluids: Pressure and Depth 15-4Archimedes' Principle and Buoyancy 15-5Applications of Archimedes' Principle 15-6Fluid Flow and Continuity 15-7Bernoulli's Equation 15-8Applications of Bernoulli's Equation *15-9Viscosity and Surface Tension

Archimedes' Principle and Buoyancy A fluid exerts a net upward force on any object it surround. It is a buoyant force. The surrounding fluid exerts normal forces on all of its faces. Clearly, the horizontal forces pushing to the right and left are equal, hence they cancel each other. The vertical forces, the downward force exerted on the top face is less than the upward force exerted on the lower face, since the pressure at the lower face is greater. This difference in forces gives rise to a net upward force  the buoyant force.

Archimedes' Principle and Buoyancy Let’s calculate the buoyant force acting on the block. We assume that the cubical block is of length L on a side, and that the pressure on the top surface is P 1. The downward force on the block, then, is The pressure on the bottom face is The upward force exerted on the bottom face of the cube is The buoyant force is

Archimedes' Principle and Buoyancy Archimedes’ Principle This is a special case of Archimedes’ Principle. = the weight of fluid that displaced by the cube. More generally, if any volume V of an object is immersed in a fluid, the bouyant force can be expressed as: SI unit : N

Applications of Archimedes’ Principle Complete Submersion Example 5: Measuring the body’s Density A person who weighs N in air is lowered into a tank of water to about chin level. He sits in harness of negligible mass suspended from a scale that reads his apparent weight. He now exhales as much air as possible and dunks his head underwater, submerging his entire body. If his apparent weight while submerged is 34.3 N, find (a) his volume and (b) his density. Picture the Problem:

Strategy: (a) When the person is submerged, the surrounding water exerts an upward buoyant force given by Archimedes’ principle This relation, and Newton’s second law, can be used to determine V p (b) The weight of the person in air is Combining this relation with the volume V p, found in part (a) allows us to determine the density. Solution: Part (a) 1. Apply Newton’s second law. Note that the person remains at rest, and therefore the net force acting on him is zero:

Solution: Part (a) 1. Apply Newton’s second law. Note that the person remains at rest, and therefore the net force acting on him is zero: But, Part (b)

Floatation When an object floats, the buoyant force acting on it equals its weight. An object floats when it displaces an amount of fluid equal to its weight. Let’s use this condition to determine just how much of a floating object is submerged.

Is the solid weight Is the weight of the fluid displaced by V f Definition: Submerged Volume V sub for a solid of volume V s and density  s floating in fluid of density  f

Example: 6 The Tip of the Iceberg What percentage of a floating chunk of ice projects above the level of the water? Assume a density of 917 kg/m 3 for the ice, and 1.00x10 3 kg/m 3 for the water. The fraction of the ice that is above water:

End of Lecture 14