Ch0134

# Ch0134 - Chapter 34 Pascals Principle, the Continuity...

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Chapter 34 Pascal’s Principle, the Continuity Equation, and Bernoulli’s Principle 247 34 Pascal’s Principle, the Continuity Equation, and Bernoulli’s Principle There are a couple of mistakes that tend to crop up with some regularity in the application of the Bernoulli equation constant 2 2 1 = + + h P g r v . Firstly, folks tend to forget to create a diagram in order to identify point 1 and point 2 in the diagram so that they can write the Bernoulli equation in its useful form: 2 2 2 2 1 2 1 2 1 2 1 1 h P h P + + = + + . Secondly, when both the velocities in Bernoulli’s equation are unknown, they forget that there is another equation that relates the velocities, namely, the continuity equation in the form 2 2 1 1 A A = which states that the flow rate at position 1 is equal to the flow rate at position 2. Pascal’s Principle Experimentally, we find that if you increase the pressure by some given amount at one location in a fluid, the pressure increases by that same amount everywhere in the fluid. This experimental result is known as Pascal’s Principle. We take advantage of Pascal’s principle every time we step on the brakes of our cars and trucks. The brake system is a hydraulic system. The fluid is oil that is called hydraulic fluid. When you depress the brake pedal you increase the pressure everywhere in the fluid in the hydraulic line. At the wheels, the increased pressure acting on pistons attached to the brake pads pushes them against disks or drums connected to the wheels. Example 34-1 A simple hydraulic lift consists of two pistons, one larger than the other, in cylinders connected by a pipe. The cylinders and pipe are filled with water. In use, a person pushes down upon the smaller piston and the water pushes upward on the larger piston. The diameter of the smaller piston is 2 . 20 centimeters. The diameter of the larger piston is 21 . 0 centimeters. On top of the larger piston is a metal support and on top of that is a car. The combined mass of the support-plus-car is 998 kg. Find the force that the person must exert on the smaller piston to raise the car at a constant velocity. Neglect the masses of the pistons.

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Chapter 34 Pascal’s Principle, the Continuity Equation, and Bernoulli’s Principle 248 Solution We start our solution with a sketch. Now, let’s find the force R N exerted on the larger piston by the car support. By Newton’s 3 rd Law, it is the same as the normal force F N exerted by the larger piston on the car support. We’ll draw and analyze the free body diagram of the car-plus-support to get that. = 0 F 0 N = g m F m F = N = kg newtons 80 9 kg 998 N . F newtons 9780 N = F D S = 2 . 20 cm D L = 21 . 0 cm F g = m F N Table of Forces Symbol Name Agent Victim F g =m Gravitational Force on Support-Plus-Car The Earth’s Gravitational Field The Support- Plus-Car F N Normal Force The Large Piston The Support- Plus-Car
Chapter 34 Pascal’s Principle, the Continuity Equation, and Bernoulli’s Principle

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## This note was uploaded on 02/29/2012 for the course PHYS 227 taught by Professor Rabe during the Fall '08 term at Rutgers.

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Ch0134 - Chapter 34 Pascals Principle, the Continuity...

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