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Free-Body Diagrams. The first two steps in the solution of Newton's second-law problems are to select an object for analysis and then to draw frec-body diagrams for that object. Draw free-body diagrams for the following situations: (a) a mass \(M\) sliding down a frictionless inclined plane of angle \(\alpha,\) and (b) a mass \(M\) sliding up a frictionless inclined plane of angle \(\alpha ;\) (c) a mass \(M\) sliding up an inclined plane of angle \(\alpha\) with kinetic friction present.

Short Answer

Expert verified
Draw free-body diagrams showing forces for each situation.

Step by step solution

01

Understanding the Problem

We have three scenarios involving a mass sliding on an inclined plane. In each situation, we need to identify and illustrate all forces acting on the mass as it either slides down or up the plane. For Scenario (a) and (b), the plane is frictionless, meaning no frictional force is present. In Scenario (c), there is kinetic friction acting against the motion of the mass. Let’s draw the free-body diagrams step by step for each case.
02

Free-Body Diagram for (a) Mass Sliding Down a Frictionless Inclined Plane

Draw a rectangle to represent the mass. Identify the forces:- The gravitational force acting downwards, labeled as \(Mg\), where \(g\) is the acceleration due to gravity. - The normal force \(N\), acting perpendicular to the surface of the inclined plane.The gravitational force can be decomposed into two components:- \(Mg \cos(\alpha)\) perpendicular to the incline.- \(Mg \sin(\alpha)\) parallel to the incline, acting downwards.
03

Free-Body Diagram for (b) Mass Sliding Up a Frictionless Inclined Plane

Again, draw a rectangle for the mass. The forces it experiences are:- The gravitational force \(Mg\) acting downwards.- The normal force \(N\), again perpendicular to the incline.Since it's sliding up, the gravitational component \(Mg \sin(\alpha)\) acts downwards along the plane direction.
04

Free-Body Diagram for (c) Mass Sliding Up an Inclined Plane with Kinetic Friction

Draw the rectangle for the mass. Identify these forces:- Gravitational force \(Mg\), acting directly downward.- Normal force \(N\), perpendicular to the inclined plane.- Frictional force \(f_k = \mu_k N\), where \(\mu_k\) is the coefficient of kinetic friction, acting down the incline, opposing the motion.Decompose the gravitational force into:- \(Mg \cos(\alpha)\) perpendicular to the incline.- \(Mg \sin(\alpha)\) acting downwards along the incline plane.

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Key Concepts

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

Newton's second law
Newton's second law of motion forms the foundation of many physics problems, especially those involving forces and motion. It states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. In equation form, this is represented as \( F = ma \). This means that when you apply a force to an object, it will accelerate depending on its mass. Understanding this law helps us predict how an object will move when different forces act on it.
To apply Newton's second law, you must first identify all the forces acting on the object. This includes gravitational force, normal force, and in some cases, frictional force.
  • Gravitational force: Downward force proportional to mass and gravity \( Mg \).
  • Normal force: Perpendicular force from a surface supporting the object.
  • Frictional force: Force opposing the motion, dependent on surface characteristics.
By considering these forces, you can draw an accurate free-body diagram, a visual representation that aids in the calculation of the resulting acceleration.
frictionless inclined plane
A frictionless inclined plane is an idealized surface where no friction acts on the sliding mass. This simplification helps focus on other forces, such as gravitational and normal forces. Without friction, the only forces on the inclined plane are gravity pulling the object downward and the normal force exerted by the plane surface.
On a frictionless inclined plane, the gravitational force decomposes into two components:
  • Parallel to the slope: This component, \( Mg \sin(\alpha) \), causes the mass to slide down the incline.
  • Perpendicular to slope: The other component, \( Mg \cos(\alpha) \), holds the mass against the plane.
Since there's no friction, the normal force is equal to \( Mg \cos(\alpha) \). This ensures that the mass does not penetrate the plane, maintaining equilibrium in the perpendicular direction. Knowing these components is key to solving problems involving frictionless inclined planes, as it allows for predictions about acceleration and motion.
kinetic friction
Kinetic friction comes into play when an object slides over a surface with resistance. It acts against the direction of motion, slowing down the object. The force of kinetic friction is calculated by the equation \( f_k = \mu_k N \), where \( \mu_k \) is the coefficient of kinetic friction and \( N \) is the normal force.
When an object slides up or down an inclined plane, kinetic friction opposes the motion, whether it’s moving upwards or downwards. This force needs to be considered alongside gravitational components:
  • Movement up the plane: Both gravitational and kinetic friction forces act downward.
  • Movement down the plane: Kinetic friction acts upwards, opposing the gravitational pull.
The presence of kinetic friction adds an element of complexity, as it reduces net acceleration and adjusts the force calculations. This makes kinetic friction an essential skill to master, affecting how quickly or slowly an object moves.
gravitational force decomposition
Gravitational force decomposition is a crucial technique used in physics to analyze forces when dealing with inclined planes. This involves breaking down the gravitational force \( Mg \) into two perpendicular components.
In the context of an inclined plane, these components are:
  • Parallel to the plane: \( Mg \sin(\alpha) \), which drives the motion along the incline.
  • Perpendicular to the plane: \( Mg \cos(\alpha) \), which interacts with the normal force.
This decomposition is useful because it allows us to simplify calculations by separating forces affecting motion from those canceling out due to normal force. It's the key to solving many physics problems involving inclined planes.
Understanding how to decompose gravitational force enhances one's ability to predict the trajectory and behavior of objects on sloped surfaces, whether it's on a frictionless incline or one with friction present.

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Most popular questions from this chapter

An airplane flies in a loop (a circular path in a vertical plane) of radius 150 \(\mathrm{m}\) . The pilot's head always points toward the center of the loop. The speed of the airplane is not constant; the airplane goes slowest at the top of the loop and fastest at the bottom. (a) At the top of the loop, the pilot feels weightless. What is the speed of the airplane at this point? (b) At the bottom of the loop, the speed of the airplane is 280 \(\mathrm{km} / \mathrm{h}\) . What is the apparent weight of the pilot at this point? His true weight is 700 \(\mathrm{N}\) .

Two objects with masses 5.00 \(\mathrm{kg}\) and 2.00 \(\mathrm{kg}\) hang 0.600 \(\mathrm{m}\) above the floor from the ends of a cord 6.00 \(\mathrm{m}\) long passing over a frictionless pulley. Both objects start from rest. Find the maximum height reached by the \(2.00-\mathrm{kg}\) object.

Genesis Crash. On September \(8,2004\) , the Genesis space-craft crashed in the Utah desert because its parachute did not open. The \(210-k g\) capsule hit the ground at 311 \(\mathrm{km} / \mathrm{h}\) and penetrated the soil to a depth of 81.0 \(\mathrm{cm} .\) (a) Assuming it to be constant, what was its acceleration \(\left(\text { in } \mathrm{m} / \mathrm{s}^{2} \text { and in } g^{\prime} \mathrm{s}\right)\) during the crash? (b) What force did the ground exert on the capsule during the crash? Express the force in newtons and as a multiple of the capsule's weight. (c) For how long did this force last?

Rolling Friction. Two bicycle tires are set rolling with the same initial speed of 3.50 \(\mathrm{m} / \mathrm{s}\) on a long, straight road, and the distance each travels before its speed is reduced by half is measured. One tire is inflated to a pressure of 40 psi and goes 18.1 \(\mathrm{m}\) ; the other is at 105 psi and goes 92.9 \(\mathrm{m}\) . What is the coefficient of rolling friction \(\mu_{r}\) for each? Assume that the net horizontal force is due to rolling friction only.

Stopping Distance. (a) If the coefficient of kinctic friction between tires and dry pavement is 0.80 , what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 28.7 \(\mathrm{m} / \mathrm{s}\) (about 65 \(\mathrm{mi} / \mathrm{h} ) ?\) (b) On wet pavement the coefficient of kinetic friction may be only \(0.25 .\) How fast should you drive on wet pavement in order to be able to stop in the same distance as in part (a)? (Note: Locking the brakes is not the safest way to stop.)

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