Transcript Physics 121 4. Motion and Force

Physics 121
4. Motion and Force: Dynamics
4.1 Force
4.2 Newton’s First Law of Motion
4.3 Mass
4.4 Newton’s Second Law of Motion
4.5 Newton’s Third Law of Motion
4.6 The Force of Gravity and Weight
4.7 Applications of Newton’s Laws
4.8 Applications involving Friction
4.9 The art and science of problem solving
4.1 Force
A force is a push or a pull
A force is that which causes a body to accelerate
Example 4.1
You see an object moving towards the East
with a constant speed of 3 m/s. What can you
say about the force acting on it?
Solution 4.1
The force (or net force to be precise) must be
zero. The keyword is acceleration. The
acceleration of the object is given to be zero.
(Neither speed nor the direction is changing).
If there is no acceleration then there cannot be
a net force.
So you see, motion without force is possible but
acceleration without force is impossible!
Example 4.2
The term “net force” simply means the
resultant or unbalanced force. (a) What is the
net force acting on the object shown? (b)
Would you expect this object to accelerate?
3 units
7 units
Solution 4.2
(a) The net force is 4 units to the right
(b) This object will accelerate towards the right
3 units
7 units
4.2 Newton’s First Law
In the absence of external forces, an object at
rest remains at rest and an object in motion
continues in motion with a constant velocity
(that is with a constant speed in a straight
line)
Newton’s First Law (in English)
When no force acts on an object, the acceleration
of the object is zero
4.3 Mass
The mass of an object depends on how much
stuff is in the object. The mass of a given object
is the same on Earth, under water, and on the
moon!
The tendency of an object to resist acceleration is
called inertia. The greater the mass of an object
the greater the inertia. In other words, the mass
of an object is a measure of its inertia and viceversa.
4.4 Newton’s Second Law
The acceleration of an object is directly
proportional to the net force acting on it and
inversely proportional to its mass
F = ma
Newton’s Second Law . . . In English
F = ma
Acceleration is greater if the force is greater
and
Acceleration is smaller if the mass is greater
Example 4.3 . . . 0 to 60 in 3 seconds!
Explain two different ways to design a car in order
for it to go from 0 to 60 in 3 seconds.
Solution 4.3 . . . 0 to 60 in 3 seconds!
There is a right way and a wrong way to go
about this and, by golly, I am going to show
you the right way!
1. Design a bigger engine in order to apply a
greater force. The downside of this is that it
will cost you!
2. Take a tin can (in order to reduce the mass)
and paint it red
4.5 Newton’s Third Law
When two objects interact, the force exerted
by object 1 on object 2 is equal in magnitude
but opposite in direction to the force exerted
by object 2 on object 1
F12 = -F21
Newton’s Third Law . . . in English
Action and reaction are equal and opposite
Action: You stand on a skateboard and throw
a ball
Reaction: The ball pushes you back!
Note Action and reaction act on different
objects and so cannot cancel each other.
Example 4.4 . . . Head-on Collision
A big truck slams into a Neon. Which statement is
most nearly correct:
A. Force of truck on neon equals the force of
Neon on truck
B. Force of truck on neon far exceeds the force of
Neon on truck
C. Force of truck on Neon is much less than the
force of Neon on truck
D. Neither experiences any force because the
equal and opposite forces cancel each other out
Solution 4.4 . . . Head-on Collision
A. Force of truck on neon equals the force of Neon
on truck
Note Do not confuse force with “damage”
(acceleration). The same force will cause a
tremendous amount of acceleration on the puny
Neon.
4.6 The Force of Gravity and Weight
The weight of an object is the attractive force of
gravity acting on it
Acceleration due to gravity Near the surface of
the Earth, the attractive force of gravity
accelerates objects at the rate of approximately
9.8 m/s2
Example 4.5 . . . weight of a crate
(a) What are the units of weight?
(b) What is the weight of a 5 kg crate?
Solution 4.5 . . . weight of a crate
(a) Weight is a force (of gravity) so the units would
be newtons.
A newton is a kg m / s2. A force of 1 N will cause a
1 kg mass to accelerate at the rate of 1 m / s2.
(b) F = ma
Fg = mg (“g” stands for “a” due to gravity)
Fg = (5)(9.8)
Fg = 49 N
Example 4.6 . . . Moon walk
Object P weighs the same on the Moon as
object Q does on Earth. Identify the correct
statement:
A. Mass of P is more than the mass of Q
B. Mass of P is less than the mass of Q
C. Mass of P on the Moon is more than its mass
on Earth
D. Weight of P on the Moon is more than the
weight of Q on Earth
Solution 4.6 . . . Moon walk
A. Mass of P is more than the mass of Q. In
fact, mass of P must be 6 times the mass of Q
because gM = 1/6 gE
Note: Mass never changes. Mass is mass is
mass. (Unless you break the object in two!).
Weight changes if the acceleration due to
gravity changes.
4.7 Applications of Newton’s Laws
We will present situations from daily life and
apply Newton’s Laws to answer questions
about forces and motion. This will aid us in
comprehending the material learned so far.
Practice makes perfect!
Example 4.7 . . . The sun also rises!
What is the tension in the string (T) if the mass of
the Van Gogh painting is 2 kg and the price is 37
million dollars?
A. 10 N
B. 15 N
C. 20 N
D. 40 N
T
T
45
45
Solution 4.7 . . . The sun also rises!
Free - body Diagram [FBD]
T sin 45 + T sin 45 = 20 N
T = 14 N
Example 4.8 . . . Slippery when wet
What is “a” in m/s2 if there is no friction?
2 kg
300
Solution 4.8 . . . Slippery when wet
What is “a” in m/s2 if there is no friction?
Free - body Diagram [FBD]
F = ma
mg sin  = ma
a = g sin 
a = 5 m/s2
m
mg sin
mg cos
mg

4.8 Applications involving Friction
Frictional forces resist the motion of objects.
They can arise when one surface rubs against
another, or an object encounters resistance as
it moves through air, water, and other media.
Example 4.9 . . . Calculate 
You apply a 30 N force on a 5 kg crate to push
it horizontally across a factory floor. The
speed of the crate increases at the rate of 1.5
m/s every second. What is the coefficient of
friction?
Solution 4.9 . . . Calculate 
Fnet = ma
F - f = (5)(1.5)
30 - f = 7.5
f = 22.5 N
Free - body Diagram [FBD]
n = 49N
f = 22.5N
F = 30N
f=n
22.5 = () (49)
 = 0.46
mg = 49N
All about friction . . .
Static friction must be overcome in order to start the motion
fs   s n
Kinetic friction must be overcome in order to maintain the
motion.
fk =  k n
n = normal reaction or the force that the surface exerts
perpendicularly on the object
s and k are the respective coefficients of friction and
depend on the nature of the surfaces. s is slightly greater
than k .
Example 4.10 . . . Who wins?
Calculate the acceleration and tension if  = 0.1
3 kg
2 kg
300
Solution 4.10 . . . Who wins?
Apply F = ma to each mass individually
20-T =2a ….(1)
T – 30 sin 30 – 30 cos 30(0.1) =3a …(2)
Now solve (1) and (2) for a and T
3 kg
a = 0.5 m/s2
T = 19 N
2 kg
300
That’s all folks!