Download Chapter 5 – Newton`s Laws of Motion I

Chapter 5: Applying Newton’s Laws
Newton’s 1st Law
The 1st law defines what the “natural” states of motion: rest and constant velocity.
Natural states of motion are and those states are when a = 0. In essence, the first law
sets up (i) the basis to define what accelerated motion is and (ii) which coordinate
systems Newton's laws hold in.
Newton’s 2nd Law: a 

Fnet
 Fx  max
 
m

 Fy  may
Physical Interpretation
Case 1: Equilibrium or Force-free states
From Newton’s 2nd law, there are two ways to be in force-free states:
a  0  rest or constant velocity  Fnet  0
Physically, this means that ALL the forces acting on an object must sum to zero or
balance out:

 Fx  0 (horizontal directions)  Fright  Fleft
Fnet   Fi  0 
 

 Fy  0 (vertical directions)  Fup  Fdown
That is, there must be at least two forces acting in opposite directions relative to
each other. Equilibrium has two possible situations: (i) rest is static equilibrium and (ii)
constant velocity is dynamic equilibrium.
CASE 2: Nonequilibrium Situations
Whenever an object accelerates, that means that there is an unbalanced force and that
the object moves in the direction of the unbalanced force. That is,
a  0 
Fnet  Fhand  f  0 (unbalanced) 
 Fhand  f
Example: A tossed ball is in projectile motion. When the ball reaches the peak of its
trajectory, is the ball in equilibrium or nonequilibrium motion? Nonequilibrium! Why there is an unbalanced force acting on the ball:
a  g  0  Fnet  mg  0 (unbalanced)
Free Fall Motion
By definition, free fall motion is defined when the only force acting on a projectile is the
force of gravity. In chapter 2 and 4 we saw that two objects in free fall (a heavy and a
lighter object) hit the ground at the same time when dropped simultaneously. Do they
hit the ground simultaneously because the same force of gravity acts on both
objects? NO! The force of gravity does not act the same for both masses because Fg
= mg is mass dependent, and both of these objects have different masses. Newton's
2nd law shows that
a
Fnet
m


mg

m
mg  g
m
Because it does not depend on the mass, we say that it is mass independent.
DEMO Use feather and coin in airless tube
VIDEO Show the video of feather and hammer on the moon
Newton’s 3rd Law
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I can’t touch you, without you touching me in return; I can’t nudge this hair without the
chair in turn nudging me back. That is, I can’t exert a force on a body without that
body in turn exerting a force on me. When a CONTACT force occurs  there is a
SINGLE INTERACTION  requires a PAIR of FORCES between two things
NEWTON'S 3rd LAW (Action-Reaction)
For every action, there is an EQUAL & OPPOSITE reaction: F12  F21
Conceptual Examples: Pushing on a Wall, hitting a wall with your hand and
Suppose I walked into a ring with Mike Tyson: who will feel a bigger impact force: my
face or Mike Tyson’s fist?
I am driving my car at night and a bug smashes into my wind shield leaving a big gob.
Who received the great impact force: the car or the smashed bug? Either.
 F
 a  a large mass has a very small acceleration



 F

 a small mass has a very large acceleration

 m
m
a
It is the acceleration (or deceleration) that causes all of the damage (if not death) in
certain situations. That is, it is the acceleration that kills a person in fatal car
accidents.
Interesting Point
Astronauts on the very first trip into space had major problems in turning a screw outside
the space craft. Let me explain. On earth, whenever one turns a door knob there will
always be a force applied by the door knob on your hand in the opposite direction that
you turn it (Newton’s 3rd law). Question: why is it that you don’t move according to this
force? Gravity stops you from rotating. Getting back to the first NASA space flight, one of
the astronauts tried to turn a knob on the outside of the space craft. Because there is not
friction, as he tried to turn the knob, the knob turned him. In fact, the astronaut took one
hour to finally turn the knob and in the process lost 10 lbs of water due to sweating. So
how did they solve this problem for future flights – add foot petals so their feet would lock
in place.
Applying Newton’s 2nd Law
Problem Solving Strategies
Step 0: Is the object in equilibrium or nonequilibrium?

  F  ma  0
If equilibrium


If nonequilibrium   F  ma  0
Step 1: Sketch the situation and draw a Free-Body Diagram (FBD).
 isolate the object and reduce to a center-of-mass point
 identify all the forces acting on this point
 choose a coordinate system convenient to you
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Step 2: Breakup all forces into its components along the x- and y-axes, and sum the
components. A force table can be convenient.
Force
x-component
y-component
F1
F1x
F1y
F2
F2x
F2y
F3
F3x
F3y
Fx  F1x  F2x  F3x
Fy  F1y  F2y  F3y
Step 3: Apply Newton’s 2nd law (ΣFx = max, ΣFy = may), pick the acceleration direction
as positive, and solve the two equations for the desired unknown quantities.
Example 5.1
The figures show FBDs for an object of mass m. Write the x- and y-components of
Newton's second law. Write your equations in terms of the magnitudes of forces F1, F2,
... and any angles defined in the diagram.
ma x   Fx 
ma y   Fy 
ma x   Fx 
ma y   Fy 
Example 5.2
The three ropes are tied to a small, very light ring. Two of these ropes are
anchored to walls at right angles with the tensions shown in the figure.
What are the magnitude and direction of the tension T3 in the third rope?
Solution
We are asked to solve for the magnitude and direction (θ) of T3.
Step 0: Is the ring in equilibrium or nonequilibrium?
The massless ring is in static equilibrium, so all the forces must balance out; that is,
forces acting on it must cancel to give a zero net force.

 Fright  Fleft
 Fx  0 

 Fup  Fdown

 Fy  0 
Step 1: Sketch the situation and draw a FBD.
Step 2: Breakup all forces into its components and sum them.
There are two ways to usually solve the problem: the hard way or the easy way.

 T1  T3 cos   T3x
 Fx  T3 cos   T1  0 

 T2  T3 sin   T3y

 Fy  T2  T3 sin   0 
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Step 3: Apply Newton’s 2nd law (ΣFx = 0, ΣFy = 0),solve the two equations for T3 and θ.
I will first solve for the angle θ:
solve for T3
T1  T3 cos  
 T3 
T1
sub this into

 T2  T3 sin  T  T1
3
cos 
cos 
T sin 
 1
 T1 tan 
cos 
Taking the inverse tangent,
T 
 80 
T2  T1 tan  
   tan1  2 
 tan1    58  
 50 
 T1 TT1 50N
80N
2
Next solve for the magnitude, I use one of the equations to solve for T3:
F
x
solve for T3
 0 
T1  T3 cos  
 T3 
T1
50 N

 94 N  T3
cos  T1 50N cos58
This is the usual way that all physics book will show you how to solve Newton 2nd type
problems. However, it is not the efficient way to solve them. Forces are vectors and
therefore, follow vector rules. That is,
2
2
T3 
 T3  T3x
 T3y
 (50)2  (80)2  94 N;
 T3y 
1  80 
θ  tan1 
  tan    58
 50 
 T3x 
Example 5.3
a. A 0.60 kg bullfrog is on a log tilted 30° above horizontal. How large is the normal
force on the log on this bullfrog?
Solution
Is weight a scalar (a number) or is a vector? This example will clearly show you that the
weight is a vector without doubt. How much does the frog actually weight on a flat
surface?
mg m0.60kg   0.60kg  9.81 m/s2  5.9 N  mg


Step 0: Is the frog in equilibrium or nonequilibrium?
The frog is in static equilibrium, so all the forces must balance out; that is, forces acting
on it must cancel to give a zero net force.

 Fright  Fleft
 Fx  0 

 Fup  Fdown

 Fy  0 
Step 1: Sketch the situation and draw a FBD.
Step 2: Breakup all forces into its components and sum them.

 fS  mgcos 60
 Fx  mgcos 60  fS  0 

 N  mgsin 60

 Fy  N  mgsin 60  0 
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Step 3: Apply Newton’s 2nd law (ΣFx = 0, ΣFy = 0), solve the two equations for N.
Solving for the normal force using the y-equations,


N  mgsin60 m0.60kg   0.60kg  9.81 m/s2 cos 60  5.1 N  N
The normal force is supporting only
N
5.1 N

 86% of the frog's weight
mg 5.9 N
The answer is less than the weight of the frog. Does that mean that the "rest of the
weight of 0.8 N" lies along the incline? NO! Because the weight is a vector force and not
just a simple number. The weight of the frog is distriubted along two directions, the xand y-axis. The other part of the frog's weight is held up by the static friction force (or
with weight along the y-direction) and is given by
Wx  mgcos60 m0.60kg   0.60kg  9.81 m/s2 cos60  2.9 N  Wy


Interpret the solution: Because on the incline plane, the weight gets distributed along
two axes such that
mg  Wx2  Wy2  5.12  2.92  5.9 N .
The force on the plane acting on the object is less than if the object was on flat ground –
the object’s weight is NOT fully supported by the incline. Clearly, the weight has true
vector behavior!
b. A 4000 kg truck is parked on a 15° slope. How big is the friction force on the truck?
Solution
How much does the truck actually weight on a flat surface?
mg m4000kg   4000 kg  9.81 m/s2  39,240 N  mg


Step 0: Is the frog in equilibrium or nonequilibrium?
The truck is in static equilibrium, so all the forces must balance out;

 Fright  Fleft
 Fx  0 

 Fup  Fdown

 Fy  0 
Step 1: Sketch the situation and draw a FBD.
Step 2: Breakup all forces into its components and sum them.

 fS  mgcos 75
 Fx  mgcos 75  fS  0 

 N  mgsin75

 Fy  N  mgsin75  0 
nd
Step 3: Apply Newton’s 2 law (ΣFx = 0, ΣFy = 0), solve the two equations for fS.
Solving for the frictional force using the x-equations,


fS  mgcos75 m4000kg   4000 kg  9.81 m/s2 cos75  10156 N  10000 N  fS
The frictional force is supporting only
N 10156 N

 26% of the truck's weight
mg 39240 N
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Example 5.4
The figure shows two 1.00 kg blocks connected by a rope. Assume the rope is
massless. The entire assembly is accelerated upward at 3.00 m/s2 by force F.
a. What is F?
b. What is the tension of rope?
Solution
a. The two block-system is accelerating upward at a  3.0 m/s2. What force is
accelerating this system? The net force ΣFy.
Step 0: nonequilibrium ⟶ ΣFy = ma
Step 1, 2, 3: Draw a FBD, sum the forces and apply Newton’s second law:
F
y
 F  (mA  mB )g  (mA  mB )a 
F  (mA  mB )(a  g)  25.6 N  F
b. In order to determine the tension, I have to focus on the individual block's. If I
focus on block A, I find that
Step 0: nonequilibrium ⟶ ΣFy = ma
Step 1, 2, 3: Draw a FBD, sum the forces and apply Newton’s second law:
F
Ay
 F  mAg  T  mAa 
 T  F  mA (a  g)  25.6 N  (1 kg)  (3  9.81)m/s2  12.8 N  T
or I can do the same to Block B:
F
By
 T  mBg  mBa 
 T  mB (a  g)  (1 kg)  (3.00  9.81) m/s2  12.8 N  T
FRICTION
There are two types of frictional forces – static and kinetic friction. Symbolically they are
written as
fS  Static frictional force & fk  kinetic frictional force
DEMO Produce a friction curve using a heavy block with a spring scale
Consider a block that is being pulled to my left. When I first apply a small force, the block
does not move – this is because the force of friction is balancing out the force of my
hand. As I apply a larger force, the block still does not move, indicating that the “static
frictional force” increases in strength with the applied force of my hand force. In
other words, the static frictional force is not a constant force. As I continue to apply
an increasing force eventually the static force reaches a certain maximum value and
then the force of my hand over comes it. This is the so-called breakaway point when
the block starts to moves. At this point, friction changes its nature and instead of a
varying force, the frictional force is roughly constant in nature and less than the maximal
value.
Interpretation – this agrees with our intuition. Physically, it is harder to start moving an
object since fS increases with the pushing force. After maxing out f S, the object moves. It
then becomes easier to move the object since fk < fS.
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From a microscopic viewpoint, irregularities in the surfaces cause friction. This can be
seen from polished stainless steel at the micro level.
Properties of Friction
1. Frictional forces always oppose the direction of motion.
2. Experimental facts about Friction
frictional force  Normal force 
  lots of constraints (temp, area, speed,...)
f N

The frictional force is proportional to the normal force with a proportionality constant μ is
called the coefficient of friction:
S  coefficient of static friction
k  coefficient of kinetic friction
fStatic  SN  variable force
fkinetic  kN  constant force
fS (max)  SN  constant force
Materials
Coefficient of
static friction, S
Coefficient of
kinetic friction, k
Steel on steel
0.74
0.57
Copper on steel
0.53
0.36
Copper on cast iron
1.05
0.29
Copper on glass
0.68
0.53
glass on glass
0.94
0.40
Rubber on concrete (dry)
1.00
0.80
Rubber on concrete (wet)
0.30
0.25
Units: [] = 1 (unitless)
Example 5.5
Bonnie and Clyde are sliding a 300 kg bank safe across the floor to their getaway car.
The safe slides with a constant speed if Clyde pushes from behind with 385 N of force
while Bonnie pulls forward with a rope with 350 N of force. What is the safe's coefficient
of kinetic friction on the bank floor?
Solution
Since it is sliding with constant velocity, it is in dynamic equilibrium and the kinetic friction
opposes the motion by pointing to the left. Setting up our notation:
FBonnie  FB  350 N, FClyde  FC  385 N, mSafe  mS  300 kg
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Step 0: equilibrium ⟶ ΣF = 0
Step 1, 2, 3: Draw a FBD, sum the forces and apply Newton’s second law:

 Fx  FB  FC  fk fk kN  FB  FC  kN  0


 Fy  N  mSg  0
To solve for the coefficient of friction, we need to focus solving for the normal force in the
y-equations and substitute it into the x-equations to obtain μk.
FB  FC  fk  0 
 fk  FB  FC  350 N  385 N  735 N
N  mSg  0

 N  mSg  (300 kg)(9.8 m/s2 )  2940 N
Then, for kinetic friction
fk  kN 
 k 
fk
735 N

 0.25  k
N 2940 N
Elevator Problems and Weightlessness
Worked Example
An elevator with a weight of 27.8 kN is given an upward acceleration of 1.22 m/s2 by a
cable. (a) Calculate the tension in the cable. (b) What is the tension when the elevator is
decelerating at the rate of 1.22 m/s2 moving downward?
Solution
a. The mass of the elevator is Fg/g =2837 kg and the acceleration is 1.22 m/s2. It is
moving upwards to that is the positive direction. A FBD and Newton’s 2nd law sets up
the equation to determine the tension:
Tup  mg  ma 
 Tup  m  g  a   3.13  104 N  Tup
b. The only difference from part (a) is now the velocity points downward and therefore,
defines positive as downward; a FBD and 2nd law sets up to give
mg  Tdown  ma 
 Tdown  m  g  a   2.43  104 N  Tdown
One can see that the tension in the cable changes, depending on the direction of the
acceleration.
Reinterpret the solution
While in an elevator at rest or moving at constant velocity, the weight scale reading
would read exactly the same as the normal force of the elevator:
Fnet  0 
 Fscale  N  mg  850N
That is, if you are in a elevator at rest or moving with constant velocity, the weight scale
would read your weight as if you where standing on the ground. I would "feel" normal
and weight 850N. If now you are in an accelerating elevator, there is a net force acting
on you and according to our example above, we should feel "heavier" since the cable
had a higher tension; a higher cable tension implies there is a higher normal force. That
is, my weight increases so
Fnet  0 
 Nup  m(g  a)  (850  105)N  955 N (215 lbs)
It is the net force that causes the "ma" term to appear, which in turn causes the "weight"
increase in this accelerating frame. Very important: the term "ma" is NOT a force acting
on you but is the result of the net force causes this term - acceleration and force are not
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the same thing. Nonetheless you feel the physical sensation of this net force. If now the
elevator decelerates, the normal force of the elevator acting on you will causes my
"weight" to change according to
Fnet  0 
 Ndown  m(g  a)  (850  105)N  745 N (167 lbs)
The physical sensation now is that I feel "light."
Question: what would the scale read if the elevator was in free-fall? ZERO!
Because the scale reads zero one says that you are weightless.
Questions: Is a force acting on you? YES - gravity
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