Download Honors physics Newton`s 3rd Law

Force is part of an interaction
• Force is push or pull
• It is an interaction between
one thing and another.
• Example;
– An apple exacts a force
on someone’s palm (force
of gravity).
– The person’s palm exacts
an upward force of
reaction.
• There are therefore two
forces acting in an
interaction.
Newton’s 3rd law of motion
• Whenever one object exerts a force on a second
object, the second object exerts an equal and opposite
force on the first object.
– One force is called Action Force
– The other is called Reaction Force
• Neither force exist without the other.
• Newton’s 3rd law can be summarized;
– To every action there is an equal opposite reaction.
Action = reaction
• If these two individuals are not in motion, which of the two
is pulling with a bigger force?_______
Pairs of Action and Reaction
Event
Pairs of
objects in an
interaction
Object
exacting
the Action
force
Object
exacting
the
Reaction
force
Moving car
Car tires and
the Road
Tires push Road push
road
tires
backwards forward
Firing a rocket Rocket and
Gas
Falling boulder Earth and the
Boulder
Firing a cannon Cannon and
cannon ball
Hitting a nail
Nail and
with a hammer hammer
Hammer
on nail
Nail on
hammer
Action –reaction pairs do cause motion
•
If action equal reaction, how can motion be
caused by such pairs of forces?
• Explain the question above by showing how
1. Michelangelo's assistant is able to accelerate.
(pg80)
2. How the sled is able to move forward.
Action and reaction Vs acceleration?
 How do equal forces (Action and reaction) produce unequal
acceleration?
 Because force is applied on objects with unequal masses.
a
F
m
a

F
m
The cannon
has a large
mass m,
The cannon
ball has a
small mass m
The action
force F is
equal to
reaction force
F. Therefore;
acceleration
4-6: Weight-The force of Gravity and The
Normal Force
• Consider the forces acting on this
box
F =
N
Since the box is at rest, the
Net force is equal to zero.
Therefore,
∑F= FN-mg= 0
FG=mg
Which means FN=mg
Here, Net force is equal to
weight.
Action reaction forces acts on
different objects
• Why are weight and normal force not
considered action reaction pairs. (pg 81)
When is Normal force not equal to weight?
• Consider a box that is at rest on a table
and another force is acting on it from the
top as shown.
FN2
FN1
m1g
m2g
If M1=10 kg, and M2= 20
kg, calculate FN1 and FN2.
∑F= FN1-m1g= 0
∑F= FN2-m2g-m1g= 0
class work and Homework
Textbook page 97 # 13-21.
Elevator problem
• Dave stands on a
bathroom scale in an
elevator . What is
Dave's mass if the
scale reads 720 N
while the elevator
accelerates upward at
2.2 m/s2
Use the formula
•Net force = apparent weight – actual weight
answer
•
•
•
•
•
•
•
•
•
∑F= FN- mg
Ma = 720 - mg
720 = ma + mg
Ma +mg = 720N
m (a +g) = 720N
m (2.2 + 9.8) = 720N
m (12) = 720
m = 720/12
m= 60 kg
Normal
force
Net Force
(ma)
(Action)
Apparent
Weight
(reaction)
Actual
weight
(mg)
Apparent weight loss
•
a)
b)
A 65 kg woman descends in an elevator that
briefly accelerates at 0.20 g downward.
If she’s standing on a scale, what is it reading?
since ∑F= ma then ∑F= mg-FN
What does the scale read if she descends at a
constant speed of 2 m/s
Example
• A 10 kg bucket is lowered vertically by a rope in
which there is 150 N tension. What is the
acceleration of the bucket and is it going up or
down?
• Solution:
– Choose up to be the positive direction. Write Newton’s
2nd law for the vertical direction, and solve for the
acceleration.
∑F= FN- mg
Class work
• Textbook pg 98 # 7-12, 15-18
Friction (textbook pg 91)
• What are the two types of friction?
Coefficient of Kinetic friction.(μk)
coefficient of Kinetic Friction(  k ) =
Friction Force (Ffr )
Normal Force (FN )
Ffr =  k FN
Example (4-16)
•
A box of mass 10 kg rests on a horizontal
force. The coefficient of static friction is
0.4, and the coefficient of sliding friction
is 0.3.
a)
determine the force of static friction Ffr
b)
if an external force of 10 N is applied, will the
box move?
c)
How much force is required to cause the box to
move?
Example 4-19
• A 10 kg box is pulled along a horizontal surface by
a force Fp of 40 N applied at an angle of 300.
Calculate the acceleration f the box if the
surface has a coefficient of kinetic friction of
0.3.
40 N
300
10 kg
Class work
• Workbook pg 37-38 # 7-11
• Workbook pg 47, # A6-A8
Class work
• Textbook pg 99 # 21-25,