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Chapter 4: Problem 1

4.1 A car of mass M travels in a straight line at constant speed along a level road with a coefficient of friction between the tires and the road of \(\mu\) and a drag force of \(D\). The magnitude of the net force on the car is a) \(\mu M g\). c) \(\sqrt{(\mu M g)^{2}+D^{2}}\) b) \(\mu M g+D\)

Short Answer

Expert verified
Short Answer: None of the given formula options correctly represent the net force on the car, which should be zero when the car is traveling at a constant speed. The forces acting on the car are the friction force (\(\mu M g\)) and the drag force (D), and they must be equal but opposite in direction. Therefore, none of the options (a, b, or c) correctly depict the net force as zero.

Step by step solution

01

Determine the Friction Force

The friction force between the tires and the road can be determined using the formula: \(F_{friction} = \mu M g\), where \(M\) is the mass of the car, \(g\) is the gravitational acceleration, and \(\mu\) is the coefficient of friction.
02

Analyze the Net Force on the Car

Since the car is moving at a constant speed, the net force acting on it is zero. Thus, the friction force and drag force must be equal but opposite in direction. Mathematically, this can be represented as: \(F_{friction} = D\), which implies \(\mu M g = D\).
03

Test the Given Formula Options to Determine the Net Force

We will test each of the given formula options to check if it represents the net force as zero: a) \(\mu M g\): This option represents the friction force and not the net force. So, this is incorrect. b) \(\mu M g + D\): Since the friction force and drag force are equal and opposite, their sum should be zero. But this option represents their sum which cannot be equal to zero since both forces are nonzero. So, this is incorrect. c) \(\sqrt{(\mu M g)^{2}+D^{2}}\): The net force on the car is given by the vector sum of the friction force and the drag force. Since they are equal and opposite in direction, their sum is equal to zero. Using the Pythagorean theorem, we can find the net force as \(\sqrt{(\mu M g)^2 + (-D)^2} = \sqrt{D^2 +D^2} = \sqrt{2D^2}\). This option represents the magnitude of the net force, which should be equal to zero. So, this option is also incorrect. Based on these tests, none of the given formula options represent the net force on the car correctly.

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Key Concepts

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

Friction
Friction is a force that resists the motion of two surfaces sliding past one another. In the context of our problem, it is the force that acts between the car's tires and the road surface. Friction is crucial for vehicle movement because it prevents tires from sliding and allows the car to "grip" the road.
  • Friction depends on two main factors: the nature of the surfaces in contact and the normal force pressing the surfaces together.
  • The formula for calculating friction force is given by: \( F_{friction} = \mu M g \), where \( \mu \) is the coefficient of friction, \( M \) is the mass of the car, and \( g \) is the gravitational acceleration (approximately \( 9.81 \text{ m/s}^2 \) on Earth).
In this problem, we assume that friction is static since the car is not sliding but is moving at a constant speed. The friction force balances other forces like the drag force, ensuring the car maintains its speed on a straight road.
Net Force
The net force is the overall force acting on an object when all individual forces are combined. In simpler terms, it determines how an object will accelerate.
  • For an object to move at a constant speed, like the car in our exercise, the net force must be zero. This is due to Newton's First Law of Motion, which states that an object in motion will stay in motion at a constant velocity if no net force acts upon it.
  • Mathematically, if an object is in dynamic equilibrium, the sum of all forces must satisfy \( \sum F = 0 \).
In the case of the car, the friction force and the drag force are equal and opposite to each other, thereby cancelling out. Consequently, their sum is zero, resulting in no net force acting on the car. This explains why the car continues moving at a constant speed.
Equilibrium
Equilibrium occurs when all the forces acting on an object are balanced, resulting in a net force of zero. An object in equilibrium does not accelerate, meaning its velocity is constant. There are two types of equilibrium:
  • Static Equilibrium: The object is at rest and remains at rest.
  • Dynamic Equilibrium: The object is moving at a constant velocity without accelerating.
The car, traveling at a constant speed in our problem, is in dynamic equilibrium. The forces acting on it - the friction force and the drag force - are equal in magnitude but opposite in direction. Consequently, they offset each other completely, ensuring that the car remains in stable motion without changing speed.
Drag Force
Drag force, often referred to as air resistance, acts opposite to the direction of an object's motion through a fluid, such as air. This force is generally experienced by objects moving at considerable speeds.
  • It depends on several factors including the speed of the object, the density of the fluid through which the object is moving, the cross-sectional area, and the drag coefficient, a number that represents the object's aerodynamic characteristics.
  • In practical terms, for cars, drag force can significantly affect fuel efficiency, as it requires additional power to maintain speed.
When the car in our scenario is said to be experiencing a drag force \( D \), it means there is a resistance that must be countered for the car to maintain its constant speed. The frictional force from the tires matches this drag force, allowing the car to stay in equilibrium and move without acceleration.

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Most popular questions from this chapter

A box of books is initially at rest a distance \(D=0.540 \mathrm{~m}\) from the end of a wooden board. The coefficient of static friction between the box and the board is \(\mu_{s}=0.320\), and the coefficient of kinetic friction is \(\mu_{k}=0.250 .\) The angle of the board is increased slowly, until the box just begins to slide; then the board is held at this angle. Find the speed of the box as it reaches the end of the board. -4.55 A block of mass \(M_{1}=0.640 \mathrm{~kg}\) is initially at rest on a cart of mass \(M_{2}=0.320 \mathrm{~kg}\) with the cart initially at rest on a level air track. The coefficient of static friction between the block and the cart is \(\mu_{s}=0.620\), but there is essentially no friction between the air track and the cart. The cart is accelerated by a force of magnitude \(F\) parallel to the air track. Find the maximum value of \(F\) that allows the block to accelerate with the cart, without sliding on top of the cart.

A bowling ball of mass \(M_{1}=6.0 \mathrm{~kg}\) is initially at rest on the sloped side of a wedge of mass \(M_{2}=9.0 \mathrm{~kg}\) that is on a frictionless horizontal floor. The side of the wedge is sloped at an angle of \(\theta=36.9^{\circ}\) above the horizontal. a) With what magnitude of horizontal force should the wedge be pushed to keep the bowling ball at a constant height on the slope? b) What is the magnitude of the acceleration of the wedge, if no external force is applied?

-4.44 A mass \(m_{1}=20.0 \mathrm{~kg}\) on a frictionless ramp is attached to a light string. The string passes over a frictionless pulley and is attached to a hanging mass \(m_{2}\). The ramp is at an angle of \(\theta=30.0^{\circ}\) above the horizontal. \(m_{1}\) moves up the ramp uniformly (at constant speed). Find the value of \(m_{2}\)

A bosun's chair is a device used by a boatswain to lift himself to the top of the mainsail of a ship. A simplified device consists of a chair, a rope of negligible mass, and a frictionless pulley attached to the top of the mainsail. The rope goes over the pulley, with one end attached to the chair, and the boatswain pulls on the other end, lifting himself upward. The chair and boatswain have a total mass \(M=90.0 \mathrm{~kg}\). a) If the boatswain is pulling himself up at a constant speed, with what magnitude of force must he pull on the rope? b) If, instead, the boatswain moves in a jerky fashion, accelerating upward with a maximum acceleration of magnitude \(a=2.0 \mathrm{~m} / \mathrm{s}^{2},\) with what maximum magnitude of force must he null on the rone?

4.32 A hanging mass, \(M_{1}=0.50 \mathrm{~kg}\), is attached by a light string that runs over a frictionless pulley to a mass \(M_{2}=1.50 \mathrm{~kg}\) that is initially at rest on a frictionless table. Find the magnitude of the acceleration, \(a,\) of \(M_{2}\)
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