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Figure \(\mathrm{P} 9.27\) shows the essential parts of a hydraulic brake system. The area of the piston in the master cylinder is \(1.8 \mathrm{~cm}^{2}\) and that of the piston in the brake cylinder is \(6.4 \mathrm{~cm}^{2}\). The coefficient of friction between shoe and wheel drum is \(0.50 .\) If the wheel has a radius of \(34 \mathrm{~cm}\), determine the frictional torque about the axle when a force of \(44 \mathrm{~N}\) is exerted on the brake pedal.

Short Answer

Expert verified
The frictional torque about the axle when a force of \(44 N\) is exerted on the brake pedal is \(2659.14 Ncm\).

Step by step solution

01

Find the Pressure on the Master Cylinder

First, calculate the pressure exerted on the master cylinder by using the formula \(P = F/A\), where \(P\) is the pressure, \(F\) is the force and \(A\) is the area. This gives \(P = F/A = 44 N/ 1.8 cm^2 = 24.44 N/cm^2\).
02

Calculate the Force in the Brake Cylinder

Next, using the same principle of pressure being constant throughout (Pascal's principle), calculate the force in the brake cylinder by rearranging the formula \(F = P \times A\). This gives \(F = 24.44 N/cm^2 \times 6.4 cm^2 = 156.42 N\).
03

Find the Frictional Force at the Wheel Drum

The frictional force between the shoe and the wheel drum is given by the product of the force in the braking cylinder and the coefficient of friction, \(f = \mu F\), where \(f\) is the frictional force, \(\mu\) is the coefficient of friction and \(F\) is the force. This gives \(f = 0.5 \times 156.42 N = 78.21 N\).
04

Calculate the Frictional Torque

Finally, calculate the frictional torque about the axle (which is the amount of turning force being applied at the wheel) by multiplying the frictional force by the radius of the wheel. The formula for torque (\(T\)) is \(T = f \times r\), where \(f\) is the frictional force and \(r\) is the radius of the wheel. This gives \(T = 78.21 N \times 34 cm = 2659.14 Ncm\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pascal's Principle
Understanding how a hydraulic brake system works can be greatly aided by grasping Pascal's Principle. This principle is a basic law in fluid mechanics and states that when pressure is applied to a confined fluid, the pressure change is transmitted undiminished to all portions of the fluid and to the walls of its container. This means that by exerting force on a small piston, you cause an equal change in pressure throughout the entire fluid, which can then exert a larger force through a bigger piston.

Pascal's Principle is critical in the operation of hydraulic systems used in car brakes, hydraulic jacks, and other devices. In the context of our exercise, we apply a force to the smaller piston in the master cylinder, which creates pressure. That pressure is then transferred through the brake fluid to the larger piston in the brake cylinder, ultimately exerting a greater force to apply the brakes. The real magic lies in the fact that the force is magnified proportionally to the ratio of the areas of the two pistons, thereby providing a significant amplification of the braking force from a relatively small input force.
Frictional Torque
The term 'frictional torque' essentially refers to the twisting force caused by friction. This force tends to slow down the rotation of objects like wheels. In the context of our hydraulic brake system, when the brake pads contact the wheel drum, they create a frictional force that opposes the wheel's motion, thus slowing down or stopping the vehicle.

In our exercise, after calculating the frictional force that acts at the point of contact between the brake shoe and the wheel, this force acts over the radius of the wheel, a distance from the point of rotation. Torque has two essential components: the amount of force applied and the distance from the point of rotation (also known as the lever arm). So, when we multiply the frictional force by the wheel's radius, we find the torque exerted by the brakes on the wheel, which enables us to determine how effectively the brakes can stop the vehicle.
Coefficient of Friction
The 'coefficient of friction' is a number that represents the degree of interaction between two surfaces and the force of friction that is created when they slide against each other. It's a unitless value that essentially measures the 'stickiness' or 'grippiness' of surfaces in contact. The higher this coefficient, the stronger the grip and the greater the frictional force that can be generated.

In braking systems, this coefficient plays a crucial role because it impacts how well the brake pads can cling to the wheel drums and produce the necessary friction to stop the vehicle. A high coefficient of friction means more stopping power. In the example we are discussing, the coefficient is given as 0.50, which would indicate a moderate level of friction. This value is used to calculate the actual frictional force that the brake system can exert, which is critical for determining the efficiency of the braking process.

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Most popular questions from this chapter

A large storage tank, open to the atmosphere at the top and filled with water, develops a small hole in its side at a point \(16.0 \mathrm{~m}\) below the water level. If the rate of flow from the leak is \(2.50 \times 10^{-3} \mathrm{~m}^{3} / \mathrm{min}\), determine (a) the speed at which the water leaves the hole and (b) the diameter of the hole.

A hypodermic syringe contains a medicine with the density of water (Fig, \(\mathrm{P} 9.47)\). The barrel of the syringe has a cross-sectional area of \(2.50 \times 10^{-5} \mathrm{~m}^{2}\). In the absence of a force on the plunger, the pressure everywhere is \(1.00 \mathrm{~atm} . \mathrm{A}\) force \(\mathbf{F}\) of magnitude \(2.00 \mathrm{~N}\) is exerted on the plunger, making medicine squirt from the needle. Determine the medicine's flow speed through the necdle. Assume the pressure in the needle remains equal to \(1.00\) atm and that the syringe is horizontal.

Water flowing through a garden hose of diameter \(2.74 \mathrm{~cm}\) fills a \(25.0-\mathrm{L}\) bucket in \(1.50 \mathrm{~min}\). (a) What is the speed of the water leaving the end of the hose? (b) A nozzle is now attached to the end of the hose. If the nozzle diameter is one-third the diameter of the hose, what is the speed of the water leaving the nozzle?

A child slides across a floor in a pair of rubber-soled shoes. The friction force acting on each foot is \(20 \mathrm{~N}\), the footprint area of each foot is \(14 \mathrm{~cm}^{2}\), and the thickness of the soles is \(5.0 \mathrm{~mm}\). Find the horizontal distance traveled by the sheared face of the sole. The shear modulus of the rubber is \(3.0 \times 10^{6} \mathrm{~Pa}\).

An iron block of volume \(0.20 \mathrm{~m}^{3}\) is suspended from a spring scale and immersed in a flask of water. Then the iron block is removed, and an aluminum block of the same volume replaces it. (a) In which case is the buoyant force the greatest, for the iron block or the aluminum block? (b) In which case does the spring scale read the largest value? (c) Use the known densities of these materials to calculate the quantities requested in parts (a) and (b). Are your calculations consistent with your previous answers to parts (a) and (b)?

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