Fluid Mechanics - Hydrostatics Sections 11 5 and 6
A closed system If you take a liquid and place it in a system that is CLOSED like plumbing for example or a car s brake line, the PRESSURE is the same everywhere. Since this is true, if you apply a force at one part of the system the pressure is the same at the other end of the system. The force, on the other hand MAY or MAY NOT equal the initial force applied. It depends on the AREA. You can take advantage of the fact that the pressure is the same in a closed system as it has MANY applications. The idea behind this is called PASCAL S PRINCIPLE.
Pascal s Law Pascal s Law: An external pressure applied to an enclosed fluid is transmitted uniformly throughout the volume of the liquid. Pressure in = Pressure out F in A in F out A out F A in in F A out out
Pascal s Principle
Another Example - Brakes P 1 F A P 2 brake pedal brake pedal F A brake pad/ shoe brake pad/ shoe In the case of a car's brake pads, you have a small initial force applied by you on the brake pedal. This transfers via a brake line, which had a small cylindrical area. The brake fluid then enters a chamber with more AREA allowing a LARGE FORCE to be applied on the brake shoes, which in turn slow the car down.
Example 1. The smaller and larger pistons of a hydraulic press have diameters of 4 cm and 12 cm. What input force is required to lift a 4000 N weight with the output piston? F F F A ; Fin A A A in out out in in out out R in = 2 cm; F in A in R out = 6 cm F out A outt D R ; Area R 2 2 F in (4000 N)( )(2 cm) 2 (6 cm) 2 F = 444 N
Example 2 - The drawing shows a hydraulic chamber with a spring, k = 1600 N/m, attached to the input piston, area 15 cm 2 and a rock of mass 40.0 kg resting on the output plunger, area 65 cm 2. The piston and plunger are nearly at the same height, and each has a negligible mass. By how much is the spring compressed from its unstrained position?
Solution - Since the piston and the plunger are at the same height, Equation 11.5, F2 F 1( A2 / A1 ), applies, and we can find an expression for the force exerted on the spring. Then F = kx, can be used to determine the amount of compression of the spring. F 1 A A 2 1 kx From the drawing in the text, we see that the force on the right piston must be equal in magnitude to the weight of the rock, or F1 mg, Therefore, A mg 2 kx A Solving for x, we obtain mg k A A (40.0 kg)(9.8 m/s 1600 N/m 1 ) 15 cm x 65 cm 2 2 2 x 2 1 0.0565 m 5.65 cm
Buoyancy When an object is immersed in a fluid, such as a liquid, it is buoyed UPWARD by a force called the BUOYANT FORCE. When the object is placed in fluid is DISPLACES a certain amount of fluid. If the object is completely submerged, the VOLUME of the OBJECT is EQUAL to the VOLUME of FLUID it displaces.
Archimedes Principle " An object is buoyed up by a force equal to the weight of the fluid displaced." In the figure, we see that the difference between the weight in AIR and the weight in WATER is 3 lbs. This is the buoyant force that acts upward to cancel out part of the force. If you were to weight the water displaced it also would weigh 3 lbs.
Archimedes Principle F B ( mg) FLUID m V F B ( Vg ) Fluid V object V Fluid
Archimedes Principle An object that is completely or partially submerged in a fluid experiences an upward buoyant force equal to the weight of the fluid displaced. 2 lb The buoyant force is due to the displaced fluid. The block material doesn t matter. 2 lb
Calculating Buoyant Force The buoyant force F B is due to the difference of pressure ΔP between the top and bottom surfaces of the submerged block. Area F B h 1 FB P P2 P1 ; FB A( P2 P1 ) A mg h 2 F A( P P) A( gh gh ) B 2 1 f 2 f 1 Buoyant Force: F ( g) A( h h ); V A( h h ) B f 2 1 f 2 1 F B = f gv f V f is volume of fluid displaced.
Example 3 - A 2-kg brass block is attached to a string and submerged underwater. Find the buoyant force and the tension in the rope. All forces are balanced: F B + T = mg F B = w gv w b mb mb ; Vb V b b 2 kg 8700 kg/m V b = V w = 2.30 x 10-4 m 3 F b = (1000 kg/m 3 )(9.8 m/s 2 )(2.3 x 10-4 m 3) F B = 2.25 N 3 T mg F B = gv Force diagram
Example (Cont.): A 2-kg brass block is attached to a string and submerged underwater. Now find the the tension in the rope. F B = 2.25 N F B + T = mg T = mg - F B T = (2 kg)(9.8 m/s 2 ) - 2.25 N T = 19.6 N - 2.25 N T = 17.3 N This force is sometimes referred to as the apparent weight. T mg F B = gv Force diagram
Floating objects: When an object floats, partially submerged, the buoyant force exactly balances the weight of the object. F B F B = f gv f m x g = x V x g ρ f gv f = ρ x V x g, eliminate gravity mg Floating Objects: ρ f V f = ρ x V x If V f is volume of displaced water V wd, the relative density of an object x is given by: Relative Density: r x w V V wd x
Example 4 - A student floats in a salt lake with one-third of his body above the surface. If the density of his body is 970 kg/m 3, what is the density of the lake water? Assume the student s volume is 3 m 3. V s = 3 m 3 ; V wd = 2 m 3 ; s = 970 kg/m 3 1/3 ρ w V wd = ρ s V s 2/3 V 3 3 3 s wd 2 m 3 ; s w w Vs 3 m 2 3 3(970 kg/m ) w s 2 2 ρ w = 1460 kg/m 3
Example 5 - A frog is a hemispherical pod finds that he just floats without sinking in a blue-green sea (density is 1.35 g/cm 3 ). If the pod has a radius of 6 cm and has negligible mass, what is the mass of the frog? Without sinking, the buoyant force pushing up on the frog and pod is equal to the weight of the frog and the pod. Since the pod has negligible mass, we will ignore it. Therefore, the buoyant force is equal to the weight of the frog. mg Vg, m Vg g V (1.35 g/cm 3 ) 4 ( (6cm) 3 3 )(.5) 610.73 g
Assignment Chapter 11 Pages 352-353, #33, 34, 40, 43, 45
Problem Solving Strategy 1. Draw a figure. Identify givens and what is to be found. Use consistent units for P, V, A, and r. 2. Use absolute pressure P abs unless problem involves a difference of pressure DP. 3. The difference in pressure DP is determined by the density and depth of the fluid: m P2 P1 gh; = ; P = V F A
Problem Strategy (Cont.) 4. Archimedes Principle: A submerged or floating object experiences an buoyant force equal to the weight of the displaced fluid: FB mf g f gv f 5. Remember: m, r and V refer to the displaced fluid. The buoyant force has nothing to do with the mass or density of the object in the fluid. (If the object is completely submerged, then its volume is equal to that of the fluid displaced.)
Problem Strategy (Cont.) 6. For a floating object, F B is equal to the weight of that object; i.e., the weight of the object is equal to the weight of the displaced fluid: mg F B m g m g or V V x f x x f f