Calculating Force of Wheel Cylinders: Pascal Puts on the Brakes · Physics Playground

Example 11.6 from College Physics, 11.5 Pascal’s Principle

Calculating Force of Wheel Cylinders: Pascal Puts on the Brakes

Consider the automobile hydraulic system shown in Figure 11.13.

When the driver applies force on the brake pedal the pedal cylinder transmits the same pressure to the wheel cylinders but results in a larger force due to the larger area of the wheel cylinders.

Figure 11.13 Hydraulic brakes use Pascal’s principle. The driver exerts a force of 100 N on the brake pedal. This force is increased by the simple lever and again by the hydraulic system. Each of the identical wheel cylinders receives the same pressure and, therefore, creates the same force output

F_{2}

. The circular cross-sectional areas of the pedal and wheel cylinders are represented by

A_{1}

and

A_{2}

, respectively

A force of 100 N is applied to the brake pedal, which acts on the pedal cylinder—called the master—through a lever. A force of 500 N is exerted on the pedal cylinder. (The reader can verify that the force is 500 N using techniques of statics from Applications of Statics, Including Problem-Solving Strategies.) Pressure created in the pedal cylinder is transmitted to four wheel cylinders. The pedal cylinder has a diameter of 0.500 cm, and each wheel cylinder has a diameter of 2.50 cm. Calculate the force $F_{2}$ created at each of the wheel cylinders.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Strategy
We are given the force $F_{1}$ that is applied to the pedal cylinder. The cross-sectional areas $A_{1}$ and $A_{2}$ can be calculated from their given diameters. Then $\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}}$ can be used to find the force $F_{2}$ . Manipulate this algebraically to get $F_{2}$ on one side and substitute known values:
Solution
Pascal’s principle applied to hydraulic systems is given by $\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}}$ :

$F_{2} = \frac{A_{2}}{A_{1}} F_{1} = \frac{{πr}_{2}^{2}}{{πr}_{1}^{2}} F_{1} = \frac{{(1.25 cm)}^{2}}{{(0.250 cm)}^{2}} \times 500 N = 1 . 25 \times 10^{4} N .$
11.27
Discussion
This value is the force exerted by each of the four wheel cylinders. Note that we can add as many wheel cylinders as we wish. If each has a 2.50-cm diameter, each will exert $1 . 25 \times 10^{4} N .$

0 of 3 steps open

How did it go?

Strategy

Solution

Discussion