Download presentation

Presentation is loading. Please wait.

Published byTaryn Tozer Modified about 1 year ago

1
1 Clicker Question Room Frequency BA By Pascal’s Principle, the applied pressure is the same every in the fluid (at the same depth). A force of 200 lbs is applied as F IN. Assume that A IN is 1 cm 2 and A OUT is 20 cm 2. What’s P OUT ? A)200 lb/cm 2 B) 400 lb/cm 2 C) 2000 lb/cm 2 D) 4000 lb/cm 2

2
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 Announcements 2

3
3 Pressure as a function of depth Pressure only depends on the depth; at a given depth the pressure is the same.

4
4 Forces on a Small Box of Water h Δy 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. F net = F top + F bottom = 0 Using our depth formula we can find that The pressure difference between top and bottom is just the weight of the water in the box

5
5 Forces on a Small Box of Gold h Δy 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. F net = F top + F bottom Using our depth formula we can find that The pressure difference between top and bottom is just the weight of the water in the box

6
6 Buoyant Force h Δy So we conclude that the net force is The fluid exerts an upward force of m water g on the gold cube. This is known as the buoyant force. You can show that this is true for any shape volume V! For some general fluid, we can write

7
7 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!

8
8 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….. Clicker Question Room Frequency BA The weight of displaced fluid does not depend on depth

9
9 A solid piece of plastic of volume V and density ρ plastic 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 ρ water.) What is the buoyant force on the plastic? A)Zero B)ρ plastic V C) ρ water V D) ρ water V g E) ρ plastic V g Clicker Question Room Frequency BA FBFB

10
10 Motion of Submerged objects When will an object sink or float? Object will rise: F buoy > F g (m fluid > m object ) Object will sink: F buoy < F g (m fluid < m object ) Object will be in equilibrium: F buoy = F g (m fluid = m object ) F buoyant FgFg F buoy = m fluid g F g = m object g Δy

11
11 Motion of Submerged objects F buoyant FgFg Δy Since F buoy and F g both depend on the same volume V we can also write Object will rise: ρ fluid > ρ object Object will sink: ρ fluid < ρ object Object will be in equilibrium: ρ fluid = ρ object Diet vs Regular Pepsi!

12
12 Example. A block of copper (Cu) with mass 400 g and density ρ CU = 8.9 g/cm 3 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.

13
13 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 F B compare to the weight mg of the sphere? A)F B > mg A)F B = mg A)F B < mg mg FBFB Clicker Question Room Frequency BA Net force = 0

14
14 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 F B on the balloon? A) ρ air V g B) ρ helium V g C) Mg D) (ρ air – ρ helium ) V g Clicker Question Room Frequency BA Weight of displaced fluid

15
15 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? A) ρ air V g -Mg B) (ρ air + ρ helium ) V g - Mg C) ρ helium V g - Mg D) (ρ air – ρ helium ) V g - Mg Clicker Question Room Frequency BA Helium has weight!

16
16 Floating Object: Displaces a mass of water equal to its mass. Floating objects V displ

17
17 What fraction of a piece of Aluminum will be submerged if it is placed in a tank of Mercury? (ρ Al = 2.7 x 10 3 kg/m 3, ρ Hg = 13.6 x 10 3 kg/m 3 ) A) 20% B) 40% C) 60% D) 80% The volume of the Mercury displaced is the amount of the Aluminum that will be submerged. Clicker Question Room Frequency BA

18
18 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.

Similar presentations

© 2016 SlidePlayer.com Inc.

All rights reserved.

Ads by Google