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**Forces and Newton’s Laws of Motion**

Chapter 4 Forces and Newton’s Laws of Motion

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**4.1 The Concepts of Force and Mass**

A force is a push or a pull. Contact forces arise from physical contact . Action-at-a-distance forces do not require contact and include gravity and electrical forces.

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**4.2 Newton’s First Law of Motion**

An object continues in a state of rest or in a state of motion at a constant velocity along a straight line, unless compelled to change that state by a net force. The net force is the vector sum of all the forces acting on an object.

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**4.2 Newton’s First Law of Motion**

Inertia is the natural tendency of an object to remain at rest or in motion at a constant velocity along a straight line.

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**4.3 Newton’s Second Law of Motion**

Mathematically, the net force is written as where the Greek letter sigma denotes the vector sum.

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**4.3 Newton’s Second Law of Motion**

When a net external force acts on an object of mass m, the acceleration that results is directly proportional to the net force and has a magnitude that is inversely proportional to the mass. The direction of the acceleration is the same as the direction of the net force.

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**4.4 The Vector Nature of Newton’s Second Law**

The direction of force and acceleration vectors can be taken into account by using x and y components. is equivalent to

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**4.5 Newton’s Third Law of Motion**

Whenever one body exerts a force on a second body, the second body exerts an oppositely directed force of equal magnitude on the first body.

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**4.6 Types of Forces: An Overview**

In nature there are two general types of forces, fundamental and nonfundamental. Fundamental Forces 1. Gravitational force 2. Strong Nuclear force 3. Electromagnetic force 4. Weak Nuclear force

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**4.6 Types of Forces: An Overview**

Examples of nonfundamental forces: friction tension in a rope normal or support forces

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**4.7 The Gravitational Force**

Newton’s Law of Universal Gravitation Every particle in the universe exerts an attractive force on every other particle. A particle is a piece of matter, small enough in size to be regarded as a mathematical point. The force that each exerts on the other is directed along the line joining the particles.

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**4.7 The Gravitational Force**

For two particles that have masses m1 and m2 and are separated by a distance r, the force has a magnitude given by

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**Definition of the Normal Force**

The normal force is one component of the force that a surface exerts on an object with which it is in contact – namely, the component that is perpendicular to the surface.

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**4.9 Static and Kinetic Frictional Forces**

When an object is in contact with a surface there is a force acting on that object. The component of this force that is parallel to the surface is called the frictional force.

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**4.9 Static and Kinetic Frictional Forces**

When the two surfaces are not sliding across one another the friction is called static friction.

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**4.9 Static and Kinetic Frictional Forces**

The magnitude of the static frictional force can have any value from zero up to a maximum value. is called the coefficient of static friction.

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**4.9 Static and Kinetic Frictional Forces**

Static friction opposes the impending relative motion between two objects. Kinetic friction opposes the relative sliding motion that actually does occur. is called the coefficient of kinetic friction.

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**4.9 Static and Kinetic Frictional Forces**

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**4.11 Equilibrium Application of Newton’s Laws of Motion**

Definition of Equilibrium An object is in equilibrium when it has zero acceleration.

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**4.12 Nonequilibrium Application of Newton’s Laws of Motion**

When an object is accelerating, it is not in equilibrium.

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Example A man can exert a force of 700N on a rope attached to a sled. The rope is at an angle of 30⁰ with the horizontal. If the coefficient of kinetic friction between the sled and ground is 0.4, what is the maximum load on the sled that the man can pull at constant speed?

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**Solution ƩFx = Fcos30⁰ - fk = 0 …………(i)**

ƩFy = FN + Fsin30⁰ - mg = 0 ………..(ii) From (i): μkFN = Fcos30⁰ Therefore FN = [Fcos30⁰]/μk = [700×0.866]/0.4 = N

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**From (ii): m = [FN+ Fsin30⁰]/g**

= [ ×0.5]/9.8 = kg

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Problems to be solved 4.34, 4.40, 4.58, 4.61, 4.91, 4.93, 4.107 B4.1 A subway train has three cars, each weighing 1.2×105N. The frictional force on each car is 103N, and the first car, acting as an engine, exerts a horizontal force of 4.8×104N on the rails. (a) What is the acceleration of the train? (b) What is the tension in the coupling between the first and the second cars?

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**(c) What is the tension between the second and third cars. Ans: (a) 1**

(c) What is the tension between the second and third cars? Ans: (a) 1.3m/s2 (b) 32000N (c) 16000N B4.2 Block 1 of mass m1= 8.0kg is moving on a frictionless 30.0⁰ incline. This block is connected to block 2 of mass m2= 22.0kg by a massless cord that passes over a massless and frictionless pulley. Find the acceleration of each block and the tension in the cord. Ans: 5.88m/s2 86.24N

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B4.3 What minimum force is required to drag a carton of books across the floor if the force is applied at an angle of 45⁰ to the horizontal. Take the mass of carton as 40kg and the coefficient of friction as Ans: N

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