1. Friday, November 6, 1998 Chapter 9: shear modulus pressure Pascal’s Principle Archimedes’ Principle

2. Let’s look at a side view of what happens to our copper cylinder under the influence of shear stress. FsFs FsFs xx l Shear stresses produce shear strains “shear force” where A is the area of the top of the cylinder.

3. FsFs FsFs xx l Note: the units of shear stress and shear strain are the same as our tensile and compressive forms of stress and strain.

4. The manner in which an object responds to shear stresses is also a characteristic of the material from which it’s made. Here, we define the ratio of the shear stress to the shear strain as the shear modulus: This quantity has the same units as Young’s modulus: N/m 2.

5. A child slides across a floor in a pair of rubber-soled shoes. The frictional force acting on each foot is 20 N, the cross-sectional area of each foot is 14 cm 2, and the thickness of the soles is 5 mm. Find the horizontal displace- ment of the bottom of the soles relative to the top of the soles. The shear modulus for rubber is 3.0 X 10 6 N/m 2. FfFf xx l

6. A child slides across a floor in a pair of rubber-soled shoes. The frictional force acting on each foot is 20 N, the cross-sectional area of each foot is 14 cm 2, and the thickness of the soles is 5 mm. Find the horizontal displace- ment of the bottom of the soles relative to the top of the soles. The shear modulus for rubber is 3.0 X 10 6 N/m 2. FfFf xx l

7. A child slides across a floor in a pair of rubber-soled shoes. The frictional force acting on each foot is 20 N, the cross-sectional area of each foot is 14 cm 2, and the thickness of the soles is 5 mm. Find the horizontal displace- ment of the bottom of the soles relative to the top of the soles. The shear modulus for rubber is 3.0 X 10 6 N/m 2. FfFf xx l

8. When a force is applied over an area, we say that the object feels pressure. Usually, we talk about pressure of a fluid or a gas (like the atmosphere). “pascal”

9. At the Earth’s surface, we sit at the bottom of an entire column of molecules in the atmosphere. These molecules exert a pressure on us at the surface of about 101.325 kPa. We define this pressure to be 1 atmosphere (atm). As we go up through the atmosphere, what happens to the pressure we feel?

10. WHY? Let’s back track for just a minute. First, we note that fluids are unable to support shear stresses. That means that when we exert a shear stress on the surface of a pond of water, the surface water slips across the water below it. surface water deep water wind

11. Our green fish is completely submerged. The water exerts a pressure on the fish. From which direction does the fish feel the pressure? It’s all around you! In fact, the pressure is exerted in the direction normal to the body of the fish all over the fish.

12. It’s all around you! Think about your own experience walking around outside (when there’s NOT a hard wind blowing). You don’t notice any difference in pressure over the surface of your body. What would happen to you if there was a pressure difference across your hand? P1P1 P2P2

13. P1P1 P2P2 Your hand would feel a net force acting to the RIGHT in this case. Therefore, your hand would start to accelerate to the right, or you would have to exert a force through your arm to counteract this force known as a “pressure gradient” force. F net

14. So, now let’s look at how pressure changes with altitude. We know it’s a lot harder to breathe at the top of a mountain than at sea level--there are fewer oxygen molecules and fewer molecules in general up there.

15. Let’s look at the force balance for a little layer of atmosphere as we head up the mountain. P1P1 P2P2 FgFg The pressure of all the molecules above our layer. The pressure from all the molecules below our layer. The gravitational force acting on the layer itself. A = surface area

16. P1P1 P2P2 FgFg The pressure P 1 on the top surface of area A results in a force downward of F 1 = P 1 A. The pressure P 2 on the bottom surface of area A results in a force upward of F 2 = P 2 A.

17. P1P1 P2P2 FgFg A = surface area If our system is in equilibrium, the net force must be 0. So... But what is the mass of our little layer of atmosphere? hh  is the density of air.

18. If our system is in equilibrium, the net force must be 0. So...

19. NOTE: our derivation here assumes a uniform density of molecules at a given layer in the atmosphere. In the real atmosphere, density decreases with altitude. Nevertheless, our pressure and force balance diagram applies so long as our layer is sufficiently thin so that within it, the density is approximately constant.

20. If the atmosphere is in equilibrium (which would imply a uniform temperature and no winds blowing), the pressure at a given height above the surface would be the same around the Earth. The same arguments can be made for pressure under water. All other things being equal, the pressure at a given depth below the surface is the same.

21. A scuba diver explores a reef 10 m below the surface. Assuming the density of water is 1 g/cc, what is the pressure the diver feels? We know the pressure at the surface of the water is 1 atm = 101.3 kPa. The change in pressure as the diver drops under a 10 m column of water is given by

22. In solving the last problem, we applied a principle that we haven’t even defined yet, but that probably made good sense to us. We said that the surface pressure at the bottom of the atmosphere equaled the pressure in the surface layer of water. If this weren’t true, the water would fly out of the oceans or sink rapidly toward the ocean floor!

23. In fact, Pascal’s Principle guarantees this will be true. It states: The pressure applied to an enclosed liquid is transmitted undiminished to every point in the fluid and to the walls of the container. Which means, that the pressure below the surface of the water is equal to the surface pressure + the pressure due to the column of water above a given level.

24. What happens to a cork when we try to submerge it in water? It shoots right back up to the surface. What is responsible for the motion of the cork? There must be a force acting upward on the cork greater in magnitude than gravity.

25. But what happens to the cork when it gets to the surface? It floats. So what must be the net force on the cork as it’s floating on the surface? ZERO! What’s changed?

26. What have we noticed about our strange underwater force on the cork? It’s greater than gravity when the cork is completely submerged. It’s equal to gravity when the cork floats on the surface, only partially submerged. Our new force relates to the volume of the cork that’s underwater!

27. Archimedes had this whole process figured out some 2000 years ago! He said, A body wholly or partially submerged in a fluid is buoyed up by a force equal to the weight of the displaced fluid. So, the cork naturally float with just the right portion of its volume under the water’s surface so that the buoyant force upward from the water equals the gravitational force.