# Clicker Question Room Frequency BA

## Presentation on theme: "Clicker Question Room Frequency BA"— Presentation transcript:

Clicker Question Room Frequency BA
A force of 200 lbs is applied as FIN. Assume that AIN is 1 cm2 and AOUT is 20 cm2. What’s POUT? 200 lb/cm2 B) 400 lb/cm2 C) 2000 lb/cm2 D) 4000 lb/cm2 By Pascal’s Principle, the applied pressure is the same every in the fluid (at the same depth).

Announcements CAPA assignment #13 is due on Friday at 10 pm.
This week in Section: Assignment #6 Start reading Chapter 11 on Vibrations and Waves Prof. Nagle will have his regular office hours this afternoon at 2-3pm (in his office). Prof. Uzdensky’s office hours Tuesday 11am are canceled this week

Pressure as a function of depth
Pressure only depends on the depth; at a given depth the pressure is the same.

Forces on a Small Box of Water
Consider again a small imaginary cubic box at depth h We argued last time that in static fluids, the net force on the box must be zero. Again we conclude that the side forces all cancel and only the top and bottom forces are left. h Fnet = Ftop + Fbottom = 0 Using our depth formula we can find that Δy The pressure difference between top and bottom is just the weight of the water in the box

Forces on a Small Box of Gold
Now suppose the cubic box is a block of gold at depth h The fluid outside the block is the undisturbed, so the pressures are not changed. The net force however is no longer necessarily zero!!! Again we conclude that the side forces all cancel and only the top and bottom forces are left. h Fnet = Ftop + Fbottom Using our depth formula we can find that Δy The pressure difference between top and bottom is just the weight of the water in the box

Buoyant Force h Δy So we conclude that the net force is
The fluid exerts an upward force of mwaterg on the gold cube. This is known as the buoyant force. h For some general fluid, we can write Δy You can show that this is true for any shape volume V!

Archimedes Principle A solid body, either partially or totally submerged in a fluid, will experience an upward buoyant force equal to the weight of the displaced fluid. Let’s check with a demonstration!

Clicker Question Room Frequency BA
Two identical bricks are held under water in a bucket. One of the bricks is lower in the bucket than the other. The upward buoyant force on the lower brick is….. The weight of displaced fluid does not depend on depth

Clicker Question Room Frequency BA
A solid piece of plastic of volume V and density ρplastic would ordinarily float in water, but it is held under water by a string tied to the bottom of bucket as shown. (The density of water is ρwater.) What is the buoyant force on the plastic? Zero ρplastic V ρwater V D) ρwater V g E) ρplastic V g FB

Motion of Submerged objects
Fbuoyant When will an object sink or float? Fbuoy = mfluidg Fg = mobjectg Δy Fg Object will rise: Fbuoy > Fg (mfluid > mobject) Object will sink: Fbuoy < Fg (mfluid < mobject) Object will be in equilibrium: Fbuoy = Fg (mfluid = mobject)

Motion of Submerged objects
Fbuoyant Since Fbuoy and Fg both depend on the same volume V we can also write Object will rise: ρfluid > ρobject Object will sink: ρfluid < ρobject Object will be in equilibrium: ρfluid = ρobject Δy Fg Diet vs Regular Pepsi!

Example. A block of copper (Cu) with mass 400 g and density ρCU = 8
Example. A block of copper (Cu) with mass 400 g and density ρCU = 8.9 g/cm3 is suspended by a string while under water. How does the tension in the string compare with the weight of the copper block? The buoyant force helps support the weight of the block.

Clicker Question Room Frequency BA
A carefully-made sphere, when placed under water, remains at rest, in equilibrium as shown above. How does the magnitude of the upward buoyant force FB compare to the weight mg of the sphere? FB > mg FB = mg FB < mg FB mg Net force = 0

Clicker Question Room Frequency BA
A helium-filled balloon of volume V can carry a total mass M (M includes the mass of the rubber balloon but not the mass of the Helium inside). What is the correct expression for the buoyant force FB on the balloon? ρair V g ρhelium V g Mg (ρair– ρhelium) V g Weight of displaced fluid

Clicker Question Room Frequency BA
A helium-filled balloon of volume V can carry a total mass M (M includes the mass of the rubber balloon but not the mass of the Helium inside). What is the correct expression for the net force on the balloon? ρair V g -Mg (ρair+ ρhelium) V g - Mg ρhelium V g - Mg (ρair– ρhelium) V g - Mg Helium has weight!

Floating objects Floating Object: Displaces a mass of water equal to its mass. Vdispl

Clicker Question Room Frequency BA
What fraction of a piece of Aluminum will be submerged if it is placed in a tank of Mercury? (ρAl = 2.7 x 103 kg/m3, ρHg = 13.6 x 103 kg/m3) A) 20% B) 40% C) 60% D) 80% The volume of the Mercury displaced is the amount of the Aluminum that will be submerged.

Waterfall Death Trap Every year several people drown near/underneath waterfalls! Why? Foamy, frothy water is much less dense than regular water!!! Hence the force of buoyancy is much less! Humans rapidly sink deep under the surface. Also “explains” mysterious disappearance of ships in the Bermuda Triangle: Seabed releases of large quantities of gas (“avalanches” at the edge of the continental shelf) just under the ship. The ship immediately sinks below the surface.