Master and Output Pistons in Car Breaking System: Calculating Force and Movement

[goofnewf]

Homework Statement

the breaking system on a car utilises a master pistion that is 10 cm in dimeter. the four outout pistons (one on each wheel) are 2cm in diameter

a) if 500 N of force is applied to the master piston, what is the output force on each of the wheel pistons?

b) if the master piston moves 5cm, how much does each output piston move?

c)how is this not a free lunch?

Homework Equations

unknown. missed class the day this was done and I cannot foind an example like it.

The Attempt at a Solution

I figured that the force would just be divided by four but looking at C that must not be right or that wouldn't be there.

b) for the output piston i would think it would be 10/2 x 5= 25 cm. but I am not sure about this answer.

 

[gneill]

Think in terms of pressures transmitted by the working fluid from and to the piston surface areas. Consider the pressure in the working fluid to be everywhere the same. Also consider that the total volume of working fluid is constant.

 

[goofnewf]

i still don't get it, what is the formula for it, could you show me a sample calculation?

 

[gneill]

What is the pressure in the working fluid if a 500N force is applied to the 10cm diameter piston? (hint: what is the formula for pressure given force and area?).

 

[rwisz]

I would like to clarify a few things about the breaking system in a car. Firstly, the term "breaking" should be "braking" as the system is used to slow down or stop the car, not break it. Secondly, the master piston is actually called the master cylinder, and it is responsible for generating hydraulic pressure that is transmitted to the output pistons at each wheel.

Now, let's address the questions:

a) To calculate the output force on each of the wheel pistons, we can use the equation F1/A1 = F2/A2, where F1 and A1 are the force and area of the master piston, and F2 and A2 are the force and area of the output piston. Plugging in the values given, we get F2 = (F1 x A2)/A1. So, the output force on each wheel piston would be (500 N x (π x 1 cm^2))/ (π x 5 cm^2) = 20 N.

b) To calculate the movement of each output piston, we can use the equation V1/A1 = V2/A2, where V1 and A1 are the volume and area of the master piston, and V2 and A2 are the volume and area of the output piston. Since the volume of the master piston is not given, we cannot directly calculate the volume of the output piston. However, we can assume that the volume of the master piston remains constant, and use the equation V1 = V2. So, the movement of each output piston would be (5 cm x (π x 10 cm^2))/(π x 2 cm^2) = 25 cm.

c) This is not a free lunch because energy is required to generate the force on the master piston, which is then transmitted to the output pistons. The output pistons are not generating any force on their own, they are simply receiving the force from the master piston. This is similar to a lever system, where the input force is multiplied to produce a greater output force. However, energy is still required to apply the input force. In the case of the braking system, the energy is provided by the car's engine. So, there is no "free" force or movement, it all comes from the energy source.