newton’s laws of motion chapter 4. force force newton stated that the change in velocity of an...

Newton’s Laws of Motion

Chapter 4

Force

FORCE Newton stated that the change

in velocity of an object is caused by FORCES.

When the velocity of an object is constant, or if the object is at rest, it is said to be in equilibrium.

FORCE Contact forces: forces that result

from physical contact between two objects. Examples: friction, a push or a pull

Field forces: forces that can act at a distance. Examples: magnetism, gravity

FORCE DIAGRAMS Force is a vector.

Force diagrams – show forces vectors as arrows

Free body diagrams – shows only the forces acting on a single object

Newton’s First Law

Newton’s First Law (Law of Inertia)

A body at rest will remain at rest, a body in motion will remain in motion, traveling with a constant velocity in a straight line, unless an unbalanced force acts on it.

INERTIA = a measure of a body’s ability to resist changes in velocity. INERTIA ≈ MASS (the greater the mass of a body, the

less it will accelerate under the action of an applied force)

It’s easier to push a Volkswagen than a Mack Truck

But, once it’s moving, it’s harder to stop the truck

That’s because the Mack Truck has more

mass, and thereby more inertia.

Warm Up 2/25• If you are standing on a scale

in an elevator and the elevator is accelerating upwards, would the scale show your true weight?

• What if it was travelling at a constant velocity upwards?

• What if it was accelerating downwards?

• Explain your reasoning.

Newton’s Second and Third Laws

Newton’s Second LawF = ma

The acceleration of an object is directly proportional to the resultant force acting on it and inversely proportional to its mass.

The direction of the acceleration is in the direction of the resultant force.

The SI unit of force is the NEWTON. 1N = 1 kg•m/s2

Weight: w = mg

Any acceleration requires a force.

Q

Acceleration can be…

Newton’s Third Law

If two bodies interact, the force exerted on a body 1 by body 2 is equal in magnitude, but opposite in direction to the force exerted on body 2 by 1.

OR…. For every action there is an equal and opposite reaction.

Action Reaction Pairs

Ball being thrown

Car hitting wall

Elevator pulled up by cable

http://ed.ted.com/lessons/joshua-manley-newton-s-3-laws-with-a-bicycle

Implicit in these laws are the following ideas.

• If no unbalanced force acts on a body, its acceleration must be zero.

• An unbalanced force is one whose vector sum does not equal zero.

Implicit in these laws are the following ideas.

If an unbalanced force acts on a body, it must accelerate. It will continue to accelerate for as long as the force(s) are unbalanced.

Implicit in these laws are the following ideas.

If a body has no acceleration, the vector sum of all the forces acting on it must be zero.

What is the vector sum of all the

forces acting on this point?

Zero

A 1 kg rock is thrown at 10 m/s straight upward. Neglecting air resistance, what is the net force that acts

on it when it is half way to the top of its path?

1. -9.8 N 2. -9.8/2 N 3. 0 N 4. -9.8/4 N

Everyday Forces

Weight vs Mass

The weight of an object is the force of gravity acting on the mass of that object. Measured in pounds in the USA Measured in Newtons elsewhere (and

in this class) Weight = mg or (mass) x (9.8m/s2)

On Earth

Normal Force

Normal force: the upward force acting on an object on a surface; the normal force is perpendicular to the surface on which the object is sitting. W = mg

N

In this case, N = mg;But not always

Tension

Tension (T) is a force acting on a rope. When a rope is taught there is tension in the rope.

Friction

Friction (Ff) is a resistive force that opposes motion.

How to draw a FBD

All the forces on one object are drawn as vectors (arrows)

The tail of the arrow starts on the object and point off in the direction of the force.

If the force is at an angle, include the angle.

Example: A car hits a wall. Draw the FBD of the car

Normal Force (N)

Weight (mg)

Force from wall (Fwall)

Multi-Object Force Diagrams

Draw all objects as boxes. Create a table of interactions

within the system. Using the table of interactions,

draw the force pairs acting on all objects.

Example 1: Car hits wall

Wall Car

Earth

Car Wall Earth

Car 0

Wall 0

Earth

0

+

+ +

+

++

Example 2: Elephant vs. Man

Elephant Rope

Earth

Man

Eleph

Rope

Man Earth

Eleph

0

Rope

0

Man 0

Earth

0

++ -++

-+++

+ + +

Example 3: Hanging Box

Rope

Box

Earth

Ceiling Ceiling

Rope Box Earth

Ceiling

0

Rope 0

Box 0

Earth 0

++ -

++-

+++

+ ++

How to solve problems with Newton’s Laws

Draw the picture or Free Body Diagram

Label ALL forces (in x and y direction)

Pick +X (or +y) to be in the direction of acceleration

Sum the forces in each direction Solve

Example 1: A car engine will push a 2500 kg car forward with 800. N of force on flat ground. If the air resistance is 300. N, what is the acceleration of the car? X Y

Fcar

Fair

w

N

800.N300.N

N

mg

Fx = max

800. - 300. = (2500kg)ax

500. = (2500kg)ax

ax = 0.20m/s2

800. 0

-300. 0

0 -mg

0 N

m = W/g = 8800/9.8

= 898 kg

Example: What is the tension in the cable of an elevator with a weight of 8800 N that ascends with an acceleration of 1.30 m/s2?

T

W=mgW=8800N

0 T

0 -8800

X Y

Fy = may

T - 8800 = (m)(1.30)

T - 8800 = (898kg)(1.30)

T = 9,967 NT = 1.0 x 104 N

A coach pulls a 35 kg bag of soccer balls across the field with a force of 84 N directed at an angle of 35o above the horizontal. If the bag travels at a constant velocity, (a) how strong are the resistive forces from the field on the bag and (b) how strong is the normal force on the bag?

N

mg

Fr = ?

84 N

35o

Fx = max

68.81 – Fr = 0

Fr = 68.81 N

Fr = 69 N

84cos35o

= 68.8184sin35o

= 48.81

0 N

0 -mg

- Fr 0

X Y

84cos35o

= 69.6384sin35o

= 48.18

0 N

0 -mg

- Fr 0

X Y

A coach pulls a 35 kg bag of soccer balls across the field with a force of 84 N directed at an angle of 35o above

the horizontal. If the bag travels at a constant velocity, (a) how strong are the resistive forces from the field on the bag and (b) how strong is the normal force on

the bag?Fy = may

48.75 + N - mg = 0

N = 294.82 N

48.18 + N – (35)(9.8) = 0

48.18 + N - 343 = 0

N = 290 N

N

mg

Fr = ?

84 N

35o

Example: A child holds a sled at rest on a frictionless snow-covered hill. If the sled weighs 100. N, find the force the child must exert on the rope and the force the hill exerts on the sled.

40

40

W=100N

Tn

40

W=100N

TnX Y

0 N

T 0

- 100sin40o

= - 64.28- 100cos40o

= - 76.60

Example: A child holds a sled at rest on a frictionless snow-covered hill. If the sled weighs 100. N, find the force the child must exert on the rope and the force the hill exerts on the sled.

Fx = max

T – 64.28 = 0

T = 64.28 = 64.3 N

Fy = may

N – 76.60 = 0

N = 76.60 = 76.6 N

Example: A traffic light weighing 100. N hangs from a cable tied to two other cables fastened to a support. The upper cables make angles of 37 and 53 with the horizontal. Find the tension in each of the three cables.

53T2

T3

37T1

T3

W

T1 T2

T3=W

37 53

Example: A traffic light weighing 100. N hangs from a cable tied to two other cables fastened to a support. The upper cables make angles of 37 and 53 with the horizontal. Find the tension in each of the three cables.

T3

W

T1 T2

T3=W

37 53

X y

T1

T2

T3

-T1cos37

T1sin37

T2cos53 T2sin53

0 -100 N

Example: A traffic light weighing 100. N hangs from a cable tied to two other cables fastened to a support. The upper cables make angles of 37 and 53 with the horizontal. Find the tension in each of the three cables.

x y

T1

T2

T3

-T1cos37 T1sin37

T2cos53T2sin53

0 -100 N

Fx = 0 = T2cos53-T1cos37T2cos53 = T1cos37

T2 = T1(cos37/cos53)

Fy = 0 = T1sin37+T2sin53-100

100 = T1sin37+1.33T1sin53

100 = T1(.602)+T1(1.06)

T1= 60.2 N

T2 = 1.33T1

Example: A traffic light weighing 100. N hangs from a cable tied to two other cables fastened to a support. The upper cables make angles of 37 and 53 with the horizontal. Find the tension in each of the three cables.

T3 = W = 100N

T1= 60.2 N

T2 = 1.33T1

T2 = 1.33(60 N)T2= 80.1 N

T3= 100 N = 1.00 x 102 N

Forces of Friction

The term friction refers to the resistive forces that arise to oppose the motion of a body past another with which it is in contact.

Forces of Friction

Kinetic friction (sliding friction) is the frictional resistance a body in motion experiences.

Static friction is the frictional resistance a stationary body must overcome in order to be set in motion.

Ff=μN

Where μ = coefficient of friction N = normal force

The magnitude of μ depends on the nature of the surfaces in contactNOTE: friction does NOT depend on the area of contact!

Example: You need to move a box of books into your dormitory room. To do so, you attach a rope to the box and pull on it with a force of 90.0 N at an angle of 30.0.

The box of books has a mass of 20.0 kg, and the coefficient of friction between the bottom of the box and the hallway surface is 0.5. Find the acceleration of the box.

W=mg

n

30

T=90 N

Fkf

X Y

90.0sin30o

=90.0cos30o

=0

0

0

N

-mg

- Ff

Example: You need to move a box of books into your dormitory room. To do so, you attach a rope to the box and pull on it with a force of 90.0 N at an angle of 30.0. The box of books has a mass of 20.0 kg, and the coefficient of friction between the bottom of the box and the hallway surface is 0.500. Find the acceleration of the box.

W=mg

n

30

T=90 N

Fkf

Fy = 0 = n + 90sin30-(20)(9.8)

Fkf = μn = (0.5)(151 N)

n = 151 N

Fx = max = 90cos30-75.5

ax = +0.122 m/s2

Fkf = 75.5 N (to the left)

(20)(ax) = 90cos30-75.5

Example: A hockey puck is given an initial speed of 20.0 m/s on a frozen pond. The puck remains on the ice and slides 120. m before coming to rest. Determine the coeficient of friction between the puck and the ice.

W=mg

nd=120 m

Fkf

Example: A hockey puck is given an initial speed of 20.0 m/s on a frozen pond. The puck remains on the ice and slides 120. m before coming to rest. Determine the coeficient of friction between the puck and the ice.

W=mg

nd=120 m

Fkf

vf2 = vi

2 + 2ad

a = -1.67 m/s2

vi = 20 m/svf = 0 m/s

d = 120 m

0 = (20)2 + 2(a)(120)

Example: A hockey puck is given an initial speed of 20.0 m/s on a frozen pond. The puck remains on the ice and slides 120. m before coming to rest. Determine the coeficient of friction between the puck and the ice.

W=mg

n

d=120 m

Fkf

Fy = 0 = n - W

Thus, Fkf = μn =μmg

n = W = mgFx = max = -Fkf

μ = 0.170

Fkf = μmg = 9.8μm

(m)(-1.67) = -(9.8)(μ)(m)

Example: A 4.00-kg object is connected to a 7.00-kg object by a light string that passes over a frictionless pulley. The coefficient of sliding friction between the 4.00-kg object and the surface is 0.300. Find the acceleration of the two objects and the tension in the string.

4 kg

7 kg

m2g

N

m1

g

T

T

4.00 kg object:Fx = ma = T - Fkf

Fy = 0 = n – mg

4a = T - Fkf

n = 39.2 N

n = (4.00 kg)(9.8m/s2)

Fkf

Example: A 4.00-kg object is connected to a 7.00-kg object by a light string that passes over a frictionless pulley. The coefficient of sliding friction between the 4.00-kg object and the surface is 0.300. Find the acceleration of the two objects and the tension in the string.

4 kg

7 kg

m2g

N

m1

g

T

T

Fkfa = 5.16 m/s2

Example: A 4.00-kg object is connected to a 7.00-kg object by a light string that passes over a frictionless pulley. The coefficient of sliding friction between the 4.00-kg object and the surface is 0.300. Find the acceleration of the two objects and the tension in the string.

4 kg

7 kg

m2g

N

m1

g

T

T

FkfT = 32.5 N

a = 5.16 m/s2

Multiple Objects Hooked Together

• Treat it as a whole system together.

• Add all the masses up and use that as the total mass.

• Then just look at the forces on the outside.

Example: Three boxes are tied together by light strings as shown. They are then pulled along a horizontal floor with a force of 150 N directed parallel to the floor. If each box has a mass of 10.0 kg and experiences a frictional force of 25 N, what is the total acceleration of the system?

10.0 kg 10.0 kg 10.0 kg 150 N

25 N25 N25 N

F = mtotala

150 – 25 – 25 - 25 = (30.0)a

a = 2.5 m/s2

Air resistance & Terminal Velocity

Air resistance or Drag

Air resistance is similar to friction in that it opposes the direction of motion.

Air resistance is the resistance of the air that the object is falling through. When the object is moving slowly, it is not

moving through much air so the air resistance is small.

The faster the object moves, the more air it travels through and the larger the air resistance is.

Weight (mg)

Air Resistance

Weight (mg)

Air Resistance

When the forces are balanced, acceleration

stops.

Terminal velocity

weight

Air Resistance

The forces are balanced. That is, the vector sum is equal to zero.

Free Fall Motion

About Forces

Common misconceptions about forces

When a ball has been thrown, the force of the hand that threw it remains on it for awhile. NO! The force of the hand is a contact

force; therefore, once contact is broken, the force is no longer exerted.

Common misconceptions about forces

Even if no force acts on a moving object, it will eventually stop.

Weight and mass are two names for the same thing.

Air does not exert a force. NO. Air exerts a huge force, but because it

is balanced on all sides, it usually exerts no net force unless an object is moving.