Newton’s laws are empirical laws, deduced from experiments. They were clearly stated for the first time by Sir Isaac Newton, who published them in 1687 in his famous book called “Principia”. Newton’s laws are adequate for speeds that are low compared with the speed of light. For very fast moving objects, such as atomic particles in an accelerator, relativistic mechanics developed by Albert Einstein is applicable.
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2.1 Force
2.1.1 The Concept of Force
If you pull on a spring, the spring stretches. If you pull hard enough on a box, the box moves. When you kick a ball, it deforms briefly and is set in motion. These are all examples of contact forces, so named because they result from physical contact between two objects. Another class of forces doesn’t involve any direct physical contact. It is field forces, for instance, gravity, the electric force and the force exerted by a bar magnet on a piece of iron. Forces exerted on an object can change the object’s state.
The known fundamental forces in nature are all field forces. These are, in the order of decreasing strength, (1) the strong nuclear force between subatomic particles; (2) the electromagnetic forces between electric charges; (3) the weak nuclear force, which arises in certain radioactive decay processes; and (4) the gravitational force between objects. Classical physics deals only with gravitational and electromagnetic forces, which have infinite range.
The SI unit of force is Newton, defined as the force that, when acting on an object that has a mass of 1 kg, produces an acceleration of 1 m/s2. Thus,
2.1.2 The Gravitational Force
The gravitational force is the mutual force of attraction between any two objects in the Universe. It is interesting that, although the gravitational force can be very strong between macroscopic objects, it is the weakest of the fundamental force. For example, the gravitational force between the electron and proton in the hydrogen atom is about , whereas the electromagnetic force between these same two particles is about .
Newton’s law of universal gravitation states that every particle in the Universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them. The magnitude of the gravitational force is
(2.1)
where and are masses of particles, is the distance between them, and is the universal gravitational constant.
2.1.3 Friction and normal Force
When an object rests on a surface, there is always a component of force perpendicular to the surface. We call this component a normal force, denoted by . There may also be a component of force parallel to the surface. We call this a friction force, denoted by .
Consider a box sitting on a table, and assuming that the box isn’t accelerating. The normal force is perpendicular to the table’s surface. The gravitational force (downward) and the normal force (upward) are equal and opposite. The force of friction always acts to oppose the sliding of two surfaces past each other. The magnitude of the friction force, which depends directly on the normal force, is given by
where , the coefficient of friction, depends on the nature of the surfaces and is found experimentally. There are two kinds of sliding friction: static friction and kinetic friction. In general, the force of static friction is greater than the force of kinetic friction. In other words, it is more difficult to begin moving an object at rest than it is to move an object already in motion. When an object is static, we substitute the coefficient of static friction, , in the friction equation. When an object is in motion, we use the coefficient of kinetic friction, .
Sometimes friction forces help us and sometimes they hinder us. Without friction, it would be impossible to make a car start, stop, or turn. However, if we were able to turn friction off once our car was traveling at a constant velocity, we wouldn’t need the engine anymore, because, according to Newton’s first law, we would travel at a constant speed in a straight line.
2.1.4 Free-Body Diagrams
In more formal physics, the object in a force diagram is reduced to a dot representing the object’s centre mass. We draw all the force vectors connected to that point as arrows. We use the length of the arrow to indicate the vector’s magnitude and the direction of the arrow to show which direction it acts in.
Let us consider an example as shown in Figure 2.1. A person pulls on a box with a force of 3N. When we draw the force diagram we represent the box by a dot. The force is represented by arrow, with its tail on the dot.
T = 3N
(a)
T = 3N
(b)
(c)
T
Fg
N
Figure 2.1 A person pulls on a box with a force of 3N.
When we study the motion of an object, we are interested only in those forces that act on the object. For example, in Figure 2.1 (a), the external forces acting on the box are the gravitational force , the normal force N exerted by the floor, and force T. Its force diagram is show in Figure 2.1 (c). Such a force diagram is referred to as a free-body diagram. We call it a “free-body diagram” because the environment is replaced by a series of forces on an otherwise free body. Free-body diagrams are very useful conceptual tools because they help us isolate the object we wish to study from its environment so that we can examine the forces acting on it.
The gravitational force and the normal force N in Figure 2.1 (c) are equal and opposite; that is, the magnitude of the gravitational force is equal to the magnitude of the upward normal force due to the floor. These two forces are an example of balanced forces. When an object is acted on by balanced forces, the forces cancel each other out and the object behaves as though no these forces is acting on it. We will examine the balanced forces with more detail in Chapter 4. Thus, the force T on the box in Figure 2.1 (a) is the net external force, namely, the external unbalance force.
2.2 Newton’s Laws
2.2.1 Newton’s first law of motion: The law of inertia
Before about 1600, scientists felt that the natural state of matter was the state of rest. However, Galileo devised thought experiments (such as an object moving on a frictionless surface, as just described) and concluded that it’s not the nature of an object to stop, once set in motion, but rather to continue in its original state of motion. This approach was later formalized as Newton’s first law of motion:
An object moves with a velocity that is constant in magnitude and direction, unless acted upon by the net external force.
This is also known as law of inertia. The property of an object tending to maintain the state of rest or state of uniform motion is referred to as the object’s inertia. The more inertia, the stronger is this tendency in the presence of a force. Thus, the mass of the object is a quantitative measure of its inertia.
The frame of reference, in which Newton’s first law of motion holds, is known as inertial frame of reference. A frame of reference stationed on Earth is approximately an inertial frame of reference.
2.2.2 Newton’s second law of motion
Newton’s first law explains what happens to an object that has no unbalanced external force acting on it: The object either remains at rest or continues moving in a straight line with constant speed. Newton’s second law answers the question of what happens to an object that does have unbalanced external force acting on it.
A force applied on an object produces acceleration in its own direction. It has been found that acceleration depends only on the net external force and mass. Combining the two proportionalities just given yields Newton’s second law of motion?s
The acceleration of a system is directly proportional to and in the same direction as the net external force acting on the system, and inversely proportional to its mass.
Mathematically, it is expressed as
(2.1)
where the net force () is the vector sum of all forces acting on the object.
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Example 2.1
As shown in Figure 2.1(a), what is the acceleration of a 1.0 kg box on the ground that experiences a 3.0 N force east? Assume that east is positive.
Given
The box’s mass: m = 1.0 kg;
The force acts on the box: F = 3.0 N.
Find
The acceleration of the box.
Solution
The free-body diagram is shown in Figure 2.1 (c). From Newton’s second law,
(East)
(East)
Therefore, the acceleration of the ball is east. In the vertical direction, .
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2.2.3 Newton’s third law: Action and reaction
Whenever an interaction occurs between two objects, as show in Figure 2.2, each object exerts the same force on the other, but in the opposite direction. In fact, forces in nature always exist in pairs. Newton described such paired forces with his third law:
Whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the one it exerts.
Figure 2.2 Action and reaction
Newton’s third law constantly affects our activities in everyday life. When walking, for example, we exert a frictional force against the ground. The reaction force of the ground against our foot propels us forward. For another example, the force acting on a freely falling object is the force exerted by Earth on the object, and the magnitude of this force is its weight mg. The reaction force is the force exerted by the object on Earth. The reaction force must accelerate the Earth towards the projectile, just as the action force accelerates the projectile towards the Earth. Because the Earth has such a large mass, however, its acceleration due to this reaction force is negligibly small.
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Newton’s Third Law is easy to understand but it can be quite difficult to apply it. An important thing to realize is that the action and reaction forces can never act on the same object and hence cannot contribute to the same resultant.
2.3 Applications of Newton’s Second Laws
2.3.1 Application 1: Objects in Equilibrium
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Example 1
A box weighing 200 N hangs from two cables that are fastened to a support, as in Figure 2.3 (a). The upper cables make angles of 45° with the horizontal. Find the tension in each of the two cables.
(a)
x
y
(b)
Figure 2.3 (a) A box suspended by cables. (b) A free-body diagram for the box.
Strategy
There are two unknowns, so we need to generate two equations relating to them, which can then be solved. Two equations can be obtained by applying the second law to the box: one equation from the x-component and one equation from the y- component.
Solution
Using Figure 2.3 (b), apply the conditions for equilibrium to the box:
(2.4a)
(2.4b)
There are two equations and two remaining unknowns. Solve Equations (2.4):
Example 2
Suppose a block with a mass of 3.0 kg is resting on a ramp. If the coefficient of static friction between the block and ramp is 0.50, what maximum angle can the ramp make with the horizontal before the block starts to slip down?
Figure 2.4 A block on a ramp.
Solution
This is an application of Newton’s second law involving an object in equilibrium. Choose tilted coordinates, as in Figure 2.4. Use the fact that the block is just about to slip when the force of static friction takes its maximum value, .
Write Newton’s laws for a static system in component form. The gravity force has two components:
(2.5a)
(2.5b)
Solve Eqs. (2.5), we can obtain
It’s interesting that the final result depends only on the coefficient of static friction.
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2.3.2 Application 2: Moving a box
Problem
A box (mass 10 kg) on a frictionless flat surface has a 30 N force applied to it in the positive x-direction. In addition, a 15 N force is applied in the negative x-direction. What is the resultant force acting on the box and what is the acceleration of the box?
Solution
According the discription, we are given the block’s mass, two forces F1 = 30 N in the positive x-direction, and force F2 = 15 N in the negative x-direction.
Firstly, we need to determine the net force acting on the block. Since F1 and F2 act along the same straight line, the resultant force is
(x-direction)
According to Newton’s Second Law, the block will be accelerated in the direction of this resultant force, namely,
The final result is then that the block is accelerated at in the positive x- direction.
2.3.3 Application 3: Atwood’s Machine
Problem
Two objects of mass m1 and m2, with , are connected by a light, inextensible cord and hung over a frictionless pulley as in Figure 2.5. Both cord and pulley have negligible mass. Find the magnitude of the acceleration of the system and the tension in the cord.
Figure 2.5 (a) Two hanging objects connected by a light string that passes over a frictionless pulley. (b) Free-body diagrams for the objects.
Strategy
As shown in Figure 2.5, the heavier mass m2 accelerates downwards. Since the cord can’t be stretched, the accelerations of the two masses are equal in magnitude, but opposite in direction. So that a1 is positive and a2 is negative, and |a1| = |a2| = a. Each mass is acted on by a force of tension (T) in the upward direction and a force of gravity in the downward direction.
Solution
Apply the second law to each of the two objects individually:
(2.2a)
(2.2b)
(2.2c)
Assuming upward is positive direction, the Eqs. (2.2) can be written as
(2.3a)
(2.3b)
(2.3c)
Solve for :
(2.4)
Substitute this result into Eq. (2.3a) to find T
(2.5)
Remarks
The acceleration of the second block is the same as that of the first, but with negative direction. When m2 gets very large compared with m1, the acceleration of the system approaches g. As expected, m2 is falling nearly freely under the influence of gravity. Indeed, m2 is only slightly restrained by the much lighter m1.
2.3.4 Application 4: Connected Objects
Problem
A block with mass m1 and a ball with mass m2 are connected by a light string that passes over a frictionless pulley, as shown in Figure 2.6. The coefficient of kinetic friction between the block and the surface is . Find the acceleration of the two objects and the tension in the string.
Strategy
Connected objects are handled by applying Newton’s second law separately to each object. The free-body diagrams for the block and the ball are shown in Figure 4.23 (b), with the x-direction to the right and the y-direction upwards. The magnitude of the acceleration for each object has the same value, . The block with mass m1 moves in the positive x-direction, and the ball with mass m2 moves in the negative y-direction. Using Newton’s second law, we can develop two equations involving the unknowns T and a that can be solved simultaneously.
x
y
(b)
Figure 2.6 (a) Two objects connected by a light string that passes over a frictionless pulley. (b) Free-body diagrams for the objects.
Solution
Write the components of Newton’s second law for the cube of mass m1:
x-direction: (2.5)
y-direction: (2.6)
Substitute this value for n and into Eq. (2.5):
(2.7)
Apply Newton’s second law to the ball
(2.8)
Substitute Eq. (2.8) into Eq. (2.7) and recalling that
2.4 Uniform circular motion
Uniform circular motion is a special kind of two-dimensional problem, which is a motion in a circle with constant speed.
For example, Figure 2.7 shows a ball on a guideline. The speed of the ball is the magnitude of the velocity vector . Since the speed is constant, the vectors in the drawing have the same magnitude at all points on the circle. As the ball moves along the circular path from A to B, the direction of its velocity vector changes, so the ball undergoes a centripetal acceleration.
A
B
r
O
Figure 2.7 The motion of a ball at a constant speed on a horizontal circular path. As the ball moves along the circular path from A to B, the direction of its velocity vector changes, so the ball undergoes a centripetal acceleration.
The magnitude of centripetal acceleration is constant, but its direction changes every instant so that it is always directed toward the centre of the circle. The magnitude of centripetal acceleration is
(2.9)
An object undergoing uniform circular motion experiences centripetal acceleration. From Newton’s second law, the centripetal acceleration must be caused by an net external force. Whenever an object travels in a circle at a constant speed, it must have a force acting on it that is perpendicular to its velocity. The centripetal force is the net external force; that is, the vector sum of all forces acting on the object. If the net force becomes zero, inertia will cause the object to move off at a constant speed in a straight line.
From Newton’s second law, the centripetal force is
.
Its magnitude is
.
Sometimes it is more convenient to describe uniform circular motion by specifying the period of the motion, rather than the speed. The period T is the time required to travel once around the circle. There is a relationship between period and speed, namely,
, (2.10)
where is the radius of the circle path.
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Example 2.2
The wheel of a car has a radius of r = 0.29 m and is being rotated at 600 revolutions per minute on a tire-balancing machine. Determine the speed (in m/s) at which the outer edge of the wheel is moving.
Strategy
The speed v can be obtained directly from Eq. (2.10), as long as the period T is known. The period is the time for one circle, and it must be expressed in seconds, because the problem asks for the speed in meters per second.
Given
The radius of the wheel: r = 0.29 m;
The wheel is rotated at 600 revolutions per minute.
Find
The speed at the outer edge of the wheel.
Solution
Since the tire makes 600 revolutions in one minute, the time required for a single revolution is
According to Eq. (2.9), the speed of tire is
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Example 2.3
The car tracks along the path contained turns with radii of 30 m and 20 m, as Figure 2.8 illustrates. Find the centripetal acceleration at each turn for a speed of 35 m/s.
Given
The radii of the path: r1 = 30 m and r2 = 20 m;
The speed of the car: v = 35 m/s.
Find
The centripetal acceleration at each turn.
Solution
From Eq. (2.10), the centripetal accelerations for each circle paths are
r1 = 30 m:
r2 = 20 m:
Remark
The centripetal acceleration is indeed smaller when the radius is larger.
v = 35m/s
r1 = 30m
r2 = 20m
Figure 2.8 The car travels at the same speed around two curves with different radii.
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Questions
1. A small car collides with a truck. The greater impact force (in magnitude) acts on (a) the car, (b) the truck, (c) neither, the force is the same on both. Which one undergoes the greater magnitude of acceleration? (d) the car, (e) the truck, (f) the accelerations are the same.
2. A baseball of mass m is thrown upward with some initial speed. If air resistance is neglected, the force acting on the ball when it reaches its peak is (a) mg and upward, (b) mg and downward, (c) zero, (d) none of these.
3. You press a book flat against a vertical wall with you hand. What is the direction of the friction force exerted by the wall on the book? (a) downward, (b) upward, (c) out from the wall, (d) into the wall.
4. A 3.0 kg box undergoes an acceleration of . (a) what is the magnitude of the resultant force acting on it? (b) If this same force is applied to a 30.0 kg box, what acceleration is it produced?
5. A car with mass 1000 kg moves through the road with two forces acting on it. One is 1500N forward push propeller, and the other is a 500N friction force. (a) What is the acceleration of the car? (b) If it starts from rest, how far will it move in 5s? (c) What will its velocity be at the end of this time.
6. Find the tension of each cable supporting the 500N box in Figure 2.9.
400
500N
Figure 2.9 (Question 6)
7. An object with mass rests on a frictionless horizontal table and is connected to a cable that passes over a pulley and is then fastened to a hanging object with mass , as shown in Figure 2.10. Find the acceleration of each object and the tension in the cable.
Figure 2.10 (Question 7)
8. Two objects with masses of 8.00 kg and 6.00 kg are connected by a light string that passes over a frictionless pulley, as in Figure 2.11. Determine (a) the tension in the string, (b) the acceleration of each object, and (c) the distance each object will move in the first second of motion if both objects start from rest.
8kg
6kg
Figure 2.11 (Question 8)
9. A raceway is shown in Figure 2.12. The radii are 100 m and 50 m, respectively. A driver travels at a constant speed of 40 m/s for one complete lap. Find the centripetal acceleration at point A and point B.
50m
100m
A
B
Figure 2.12 The raceway.
