Chapter 4 Forces and Newton’s Laws of Motion PDF

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CHAPTER 4

FORCES and NEWTON’S LAWS of MOTION

in previous chapters we used displacement, velocity, & acceleration to describe motion of an object but “What causes the motion?” and “What determines the acceleration of an object?” forces cause an object to move and determine the acceleration much of the physics and relationships between the forces acting on an object and causing its motion are based on the observations of Sir Isaac Newton in the 17th century Kinematics deals with concepts that are needed to describe motion. Dynamics deals with effect that forces have on motion. Together, kinematics and dynamics form branch of physics known as Mechanics. For objects larger than atomic dimensions and moving at speeds much less the speed of light, referred to as classical mechanics. 4.1 Concepts of Force and Mass a force represents an interaction of an object with its environment resulting in a push or a pull; a force is a vector quantity contact forces arise from direct physical contact, e.g., hand pushing a block field forces (action-at-a-distance forces) do not require contact and include gravitation, magnetic, and electrical forces 4 fundamental forces of nature: • gravitational force between two bodies • weak nuclear force (radioactive decay processes) • electromagnetic force • strong nuclear force between subatomic particles in the nucleus mass of an object, on the other hand, is a measure of its inertia, i.e., the tendency of object to resist a change in its motion

4.2 Newton’s First Law of Motion A body at rest remains at rest, and a body in motion tends to stay in motion (with a constant velocity and direction),, unless it is acted on by a net external force. net force on an object is vector sum of all forces acting on that object SI unit of force is the Newton (N)

Will the golf ball or the bowling ball have the greater acceleration when hit with the same force?

tendency of an object (such as golf ball and bowling ball) to resist change in motion called the inertia of the object with the bowling ball having a greater inertia mass is a quantitative measure of an object’s inertia SI Unit of Mass: kilogram (kg) FYI -- Newton’s laws are valid in an inertial reference frame, i.e., a non-accelerating reference frame.

4.3 Newton’s Second Law of Motion When a net external force acts on an object of mass m, the acceleration is directly proportional to the net force and inversely proportional to the mass. The direction of the acceleration is the  same as the direction of the net force.     Fnet a Fnet   Fi  m a m SI Unit for Force: kg   m2   kg 2 m  N s s 

If the net force is zero, the acceleration is zero and correspondingly the object is either at rest or moving with a constant velocity in a straight line according to Newton’s 1st Law direction of force and acceleration vectors can be taken into account by using x and y components such that F  ma F  ma



y

y



x

x

Example Hank and Fred are pushing a stalled car with 295 N and 375 N of force, respectively, as shown at the right. A frictional force of 560 N opposes their push. Find the acceleration of the car assuming the car’s mass to be 1650 kg.

Draw a free-body-diagram = a diagram that represents the object and the forces that act on it

4.5 Newton’s Third Law of Motion Whenever one body exerts a force on a second body, the second body exerts a force of equal magnitude on the first body and in the opposite direction. Alternative statement of 3rd Law: For every action, there is an equal and opposite reaction. Example If an astronaut is pushing a spacecraft with a force of 36 N, find the accelerations of the spacecraft and of the astronaut assuming that the mass of the spacecraft is 11,000 kg and the mass of the astronaut is 92 kg.

4.7 The Gravitation Force Newton’s Law of Universal Gravitation For two objects that have masses m1 and m2 and which are separated by a distance r, the two masses are attracted towards each other with a force given by

F G

m1 m2 r2

where G  6.673  1011 N  m 2 kg 2 G is the universal gravitatio nal constant

Example Find the gravitational attractive force between two students separated by a distance of 1.0 m. Assume the students have masses of 50 kg (~ 110 lb) and 90 kg (~200 lb) .

Compare this gravitational attraction between two students to the weight of the 50-kg person on the surface of the Earth. Note that the weight w of person on Earth is the gravitational force between the person and the Earth.

24 ME kg) 11 2 2 (5.98 10 G 2  (6.67  10 N  m /kg ) 9.80 m/s 2 6 2  RE (6.38  10 m)

G

ME  g  accelerati on due to gravity RE2

w  mg Example Find the gravitational force on a 50-kg person when he is an equivalent distance of the two times the radius of the Earth above the Earth’s surface. 2 RE

If the action force is a 54.4 N force of the Earth pulling down on the person, what is the reaction force?

4.10 Tension

Pull

Cables and ropes transmit forces through tension. We will consider that cables and ropes have zero mass and do not stretch, and thus the tension is uniform throughout the cable and rope. A massless rope will transmit tension undiminished from one end to the other.

+y

4.8 The Normal Force normal force FN is the force that a surface exerts on an object with which it is in contact – namely, the component that is perpendicular to the surface

+x

If the concrete block weighs 200 N, determine the magnitude of the normal force.

What is the reaction force to the normal force?

Example Determine the magnitude of the normal force if you pushing down with an 11-N force on a 15-N block.

Note that jumping off of the ground requires normal force to be greater than weight

Example (i) If you are pulling up with 11-N force T on a 15-N block, what is its acceleration?

+y

+x (ii) Determine the magnitude of the normal force.

(iii) What is the acceleration when a 20-N tension is applied?

(iv) Determine the magnitude of the tension in the rope in order for the block to be moving at a constant speed of 5 m/s upwards.

What are the forces acting on each block shown below?

The weight of an object according to the reading of a scale (sometimes called apparent weight) is equal to the normal force that scale exerts on the man

F

y

 FN  mg  ma

apparent weight  FN  mg  ma

Now consider the man and scales inside an elevator which can move up or down depending on the tension T exerted by the elevator cable. What are the forces acting on the man and the elevator?

Note how the apparent weight (the normal force) changes when the elevator is moving at a constant velocity, an upward acceleration, a downward acceleration, and under free-fall.

Can you determine the acceleration given the apparent weights shown in the above figure?

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 force that is parallel to the surface is called the frictional force.

When two surfaces are not sliding across one another, the friction is static friction. magnitude of the static frictional force can have any value from zero up to a maximum value sFN

f s  f smax   s FN s = coefficient of static friction 0  s  1 Note that the magnitude of the frictional force does not depend on the contact area of the surfaces. Static friction opposes the impending relative motion between two objects. Kinetic friction opposes the relative sliding motion once the motion between two objects actually occurs.

f k  k FN

k = coefficient of kinetic friction

0  k 1

Example A sled moving at 4.0 m/s eventually comes to rest as a result of kinetic friction. If the combined mass of the child and sled is 40 kg and the coefficient of kinetic friction is 0.05 between the sled and snow, find the kinetic frictional force and the acceleration of the sled?

4.11 Applications of Newton’s Laws of Motion Problem Solving Strategy • Select an object(s) which the equations of motions (Newton’s Laws) are to be applied. • Draw a free-body diagram for each object chosen above. Include only forces acting on the object, not forces the object exerts on its environment. • Choose a set of x, y axes for each object and resolve all forces in the free-body diagram into components that point along these axes. • Apply the equations and solve for the unknown quantities.

Example Assume that child and sled are on a frictionless snow-covered hill. Find the speed of the sled & child at the bottom of the hill.

If the sled is initially at rest, then the velocity at the bottom of the hill is:

Example Now assume that child and sled are on a snow-covered hill with friction. Find the speed of the sled & child at the bottom of the hill if the coefficient of kinetic friction is 0.20.

If the sled is initially at rest, then the velocity at the bottom of the hill is:

4.10 Tension  Cables and ropes have zero mass and do not stretch, and thus the tension is uniform throughout the cable and rope.  If a rope passes around a massless, frictionless pulley, the tension will be transmitted to the other end of the rope undiminished.  A pulley essentially changes the direction of the tension.

Example Consider a 10-kg block resting on a frictionless table connected by a massless rope to a 5-kg block hanging over a frictionless and massless pulley at the edge of a table. Find the acceleration of both blocks and the tension in the rope.

Example Now assume there is friction between the10-kg block and the table with k=0.3 and s=0.6 for the coefficients of friction. Find the acceleration of both blocks assuming the blocks are initially at rest.

Example Find the tensions T1 and T2 and the force F when a suspended mass of 20.0 kg is applied. Assume equilibrium conditions, i.e., F = 0.

Example A block with mass m1 is being pulled by a force P on a frictionless table. A second block with mass m2 rests on the lower block. What is the maximum force P that can be applied and still have the second block not slip if the coefficient of static friction between the two blocks is 0.50?

Forces of friction are important for walking, running, and the motion of cars. 1. What is the direction of the person walking? 2. What must be the direction of the net force? 3. How does one move forward when one walks? - You exert a backward force on the ground from your foot. - The reaction to this backward force is forward force exerted by the floor on your foot. The greater this backward force that your foot exerts on the ground, and the greater the reaction force exerted by the ground on your foot, the faster you walk (and run). - This reaction is realized because of friction between your foot and the ground. At the point of contact, this frictional force is static.

A similar mechanism is responsible for your car to move forward. - Your tires exert a backward force on the road from the tires rotating. -

The reaction to this backward force is a forward force exerted by the road on your tires and thus the car moves forward. The greater the tire rotation, the greater the backward force that the tires exert on the road, and the greater the reaction force exerted by the road on the tires and the faster you go.

-

This reaction is realized because of friction between your tires and the ground. At the point of contact, this frictional force is static. Kinetic friction occurs when your tires slip or when tires lock up in a skid. Recall thatk < s and thus static frictional force is larger than kinetic frictional force.

Air resistance (or drag) is a different type of frictional force: -

the drag force is always opposite to the direction of motion the magnitude of the drag force is strongly dependent on the speed of the object and depends on size and shape of object

Since the drag force increases as the speed increases, a falling object reaches a maximum speed called the terminal velocity when the drag force is equal in magnitude to the weight of the object as they are in opposite directions. That is, the acceleration equals zero and thus the speed no longer increases.

In the presence of air resistance, does a ball thrown vertically upward take longer to reach its maximum height or to fall back down from its maximum height to its original position? Is the magnitude of the velocity upon returning to the original greater, the same, or smaller than it was when it was thrown initially vertically upwards with velocity vo?...