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PHYS2001_Ch. 11

# PHYS2001_Ch. 11 - Ch 11 Fluids 11.1 Mass Density We will...

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Ch. 11 Fluids 11.1 Mass Density We will deal mostly with liquids in this chapter. But gases can also be considered as fluids. So can solids. Consider lava (molten rock) or sand, for example. Let’s define the volume Mass Density : V M = ρ Units? [kg/m 3 ] The density of pure water at 4 o C: 3 3 kg/m 10 000 . 1 × = water 11.2 Pressure Push a piece of chalk and a nail into your hand with the same force. F F 3 3 kg/m 10 06 . 1 × = blood The nail hurts much worse, but why??? The reason is understood by looking at the ends of the two objects:

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The areas are different! Area of the nail tip is small. Area of the end of the chalk is much bigger. F F There must be a way to quantify this relationship between Force and Area. We call this Pressure : A F P = Notice, the much smaller area on the tip of the nail creates a much greater pressure for the same force. The smaller the area --- The greater the pressure Units? [N/m 2 ] = [Pa] (Pascal) Pressure is not a vector. F is the magnitude of the force. The force in a static fluid acts like a normal force – it is always perpendicular to the surface of contact. You, I, and everything in this room is immersed in a fluid right now – the Air ! We are all existing in a fluid that is at 1 atmosphere of pressure (1 atm). 1 Pa is actually a very small pressure. 1 atm = 1.013 × 10 5 Pa
Force can also be measured in pounds (lbs.) and area in square inches (in. 2 ). Thus, pressure can have units of lbs./in. 2 (pounds per square inch), or psi . 1atm = 14.7 psi

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Compared to the mass of a dozen eggs (0.75 kg), the mass of air inside a standard size refrigerator is 1. Negligible 2. About 1/10 as much 3. About the same 4. More Clicker Question air air air V m ρ = 3 3 m 6 . 0 ft. 20 ~ = air V 3 kg/m 29 . 1 = air kg 77 . 0 ) 6 . 0 )( 29 . 1 ( = = air m So why don’t we notice the weight of the air!!?? D
Example : High-heeled shoes can cause tremendous pressure to be applied to the floor. Suppose the radius of a circular heel is 6.00 × 10 -3 m. At times during normal walking motion, nearly the entire body weight acts perpendicular to the surface of such a heel. Find the pressure that is applied to the floor under the heel due to the weight of a 50.0- kg woman. r = 0.006 m A F P = A mg A W = = A here, is the area of a circle, so: 2 r mg P π = Pa 10 3 . 4 ) 006 . 0 ( ) 8 . 9 )( 50 ( 6 2 × = = psi 629 atm 43 = = Ouch!

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11.3 Pressure and Depth When you dive deeper under the water in a swimming pool, for example, you feel more pressure against you. Why??? P = F/A , and your area doesn’t change, so the force on you must change. The deeper you go, the more pressure you feel. Because you and the fluid are in a gravitational field, the fluid at greater depths must support the weight of the fluid above it.
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PHYS2001_Ch. 11 - Ch 11 Fluids 11.1 Mass Density We will...

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