1. Pascal’s Principle
Lesson 3

2. Pascal’s Principle
A car hoist in an automotive repair is a hydraulic system.
A hydraulic system takes advantage of the property of liquids; they cannot be (easily) compressed. The relationship between pressure and enclosed fluids was formulated by the French scientist Blaise Pascal.

3. Pascal’s Principle
Pressure applied to an enclosed liquid is transmitted to every part of the liquid and to the walls of the container.

4. Pascal used this principle in the design of a device called the hydraulic press.
A device in which a small force on a small piston is transmitted through an enclosed liquid and applies a large force on a large piston.

5. A small downward force applied to the small movable piston can produce a large upward force applied to the large moveable piston.
According to Pascal’s principle, the pressure (pS) on the small piston equals the pressure (PL) on the large piston, or pS = pL.

6. Thus, since
than , where “S” or 1 means small and “L” or 2 means large.

7. The formula can be rearranged to find any unknown.

8. and Ideal Mechanical Advantage (IMA) where .
Just like with the use of pulleys and levers, there is Actual Mechanical Advantage (AMA) with the ratio ,
and Ideal Mechanical Advantage (IMA) where .

9. The use of the hydraulic press is widespread, from medical syringes to a car’s breaking system.
In each of these cases, a small moveable piston must move a greater distance than the large piston in order to move the same volume of liquid in each cylinder.

10. Example 1
A car of mass 1.4 x 103 kg is hoisted on the large cylinder of a hydraulic press. The area of the large piston is 0.22 m2, and the area of the small piston is m2.
Calculate the magnitude of the force of the small piston needed to raise the car (at a constant, slow speed) on the larger piston.
Calculate the pressure, in Pascals and kilpascals, in this hydraulic system.

11. FL = mg
FL = (1.4 x 103 kg) (9.8 m/s2)
FL = 1.37 x 104 N
AL = 0.22 m2
AS = m2
FS = ?

13. FS = 8.1 x 102 N
Therefore, the force on the small piston is 8.1 x 102 N

14. Calculate the pressure, in Pascals and kilopascals, in this hydraulic system.
FS = 8.1 x 102 N
AS = m2
PS = ?

15. PS = 6.2 x 104 pa or 62 kPa
Therefore, the pressure in the hydraulic press is 62 kPa.

16. Car Brakes

17. When you depress a brake pedal, the car transmits the force from your foot to its brakes through a fluid.
Since the actual brakes require a much greater force than you could apply with your leg, your car must also multiply the force of your foot. It does this in two ways:
Mechanical advantage(leverage)
Pascal’s Principle

18. The brakes transmit the force to the tires using friction, and the tires transmit that force to the road using friction.

19. The diagram right, shows the distance from the pedal to the pivot is four times the distance from the cylinder to the pivot, so the force at the pedal will be increased by a factor of four before it is transmitted to the cylinder. (2nd class lever)

20. The diameter of the brake cylinder is three times the diameter of the pedal cylinder. This further multiplies the force by nine. All together, this system increases the force of your foot by a factor of 36
(4 x 9 = 36).

21. If you put 10 kg of force on the pedal, 3600 kg will be generated at the wheel squeezing the brake pads

22. Questions Explain Pascal’s principle C (1)
Explain why Pascal’s principle does not apply to gas. C (1)
Explain how Pascal’s principle is useful for car brake designers. C (1)
Copy the table shown below and calculate all unknowns. T (4)

23. FS (N) AS (m2) FL (N) AL (m2)
Copy the table shown below and calculate all unknowns. T (4)
FS (N)
AS (m2)
FL (N)
AL (m2) ?
0.25
1.8 x 103
1.5 65
5.2 x 102
0.64
5.4 x 102
0.15 ?`
1.2
1.4 x 102
2.0 x 10-2
7.7 x 103

24. In a mining operation, a hydraulic jack can exert a maximum force of magnitude 2.2 x 103 N on the small piston. The surface area of the small piston is 0.10 m2, and the surface area of the large piston is 2.0 m2. Calculate.
The magnitude of the maximum force that can be used to lift a load. T (1) C (1)
The mass of that load. T (1) C (1)