Chapter 8 Buoyancy

What are the main differences between the three states of matter?

Why do objects float? Objects float due to buoyant forces or a difference between mass densities of the objects interacting. The water in the picture is exerting an upward force (buoyant force) while gravity is simultaneously pulling down on the object.

By experience most people know that objects “feel” lighter in a pool than on land. The “feel” lighter is known as apparent weight.

Archimedes's Principle An object partially or completely submerged in water will displace a volume of water equal to the volume of the submerged object.

The buoyant force of the object is equal to the mass of the water (fluid) displaced by the submerged object. FB= mfg FB = Fg –apparent weight

Float or Sink??? If the buoyant force is greater than the force of gravity the object will float. In other words, if Fnet is negative we have a floater…if not we have a sinker. If the object is submerged at water level, then Fnet is 0. FB is up (-) Fnet = Fg - FB Fg is down (+) f refers to the displaced fluid Fnet = mfg - mog Fnet = (ΔfVf - ΔoVo)g

Fnet = (ΔfVf - ΔoVo)g The above equations shows us that buoyancy (apparent weight) depends on density. Fg/ FB = Δo/ Δf

Example: Your grandma gives you a “gold” jewelry box for your birthday. You are suspicious of the quality of the gift, so you bring it to school to run some test. You find its mass to be 0.799kg and when completely submerged in water the scale reads 0.699kg. Calculate the density of the gift to find out if the jewelry box is real gold. Hint: Δ of water is 1.00ee3 kg/m3 and the accepted Δ of gold is 19.3 kg/m3

Homework page 279 (practice A) 1-3

Quick Review: FB is the force exerted by the fluid (water) on an object partially or totally submerged. FB = FG – Apparent Weight FB = FG when an object is “floating” Of course F = ma = mg= VΔg

Fluid Pressure Deep-sea divers wear suits in order to resist forces exerted by water deep in the ocean. You may have experienced that same sensation at the bottom of a swimming pool OR flying in an airplane.

Any substance with mass exerts pressure!! P = F/A P = mg/ A A is area P is measured in pascal 1 Pa = 1N/m2 or 1atm=105 Pa

Pascal’s Principal… Pressure applied to a fluid in a closed container is transmitted equally to every point of the fluid and the walls of the container. Hydraulic lifts allow a small (human sized) applied force to lift large masses.

Pascal’s Principal cont… Pincrease = F1/A1 = F2/A2 This equation shows that output force (F2) is larger than the input force by a ratio of the areas of the pistons.

Example: A small piston of a hydraulic lift has an area of 0.20 sq. meters. A car weighing 1.20 ee4N sits on a rack supported by a large piston. The large piston has an area of 0.90sq.meters. How large a force must be applied to the small piston to support the car?

Homework: Page 282 1 - 3

Pressure varies with the depth of the fluid column. Water pressure increases with depth because it must support the weight of the water column above it. The column of water has a volume equal to Ah. A is area and h is the height.

Fluid Pressure as a function of depth… P= (P0 + Δgd) - Δgb P0 is the atmospheric pressure pushing on the surface of the fluid Δ is density of fluid d is depth below surface in meters b is column of water below object in meters When submerging an object, you must consider the water column above and below the object.

Of a submerged object… Pnet = Pbot – Ptop The water column above the object is applying a force in the opposite direction!! Look at page 283 in you book…

Homework…page 283 1-4 due at EOH!!