Download Newton`s Second Law of Motion

PHYSICS SEMESTER ONE
UNIT 4
UNIT 4: DYNAMICS AND NEWTON’S LAWS
DYNAMICS
So far we have covered aspects of motion relating displacement, velocity, acceleration and time. We
will now look at what causes the motion. Specifically, we will look at what causes the acceleration to get
items moving, slow items down and change the direction of the velocity.
A force is described as that which causes a change in motion (an acceleration). Force is a vector. The
change in motion of an object has both magnitude and direction.
At the end of the 2D motion section, we covered the concept of frames of reference and how an object
that is moving in one frame of reference can be stationary in another. We can also have accelerating
frames of reference.
When Bob accelerates around a corner in his car, the air freshener, hanging from the
rear view mirror, tilts away from the centre of rotation for the turn. The observer in the car
(Bob) is not moving relative to the air freshener. In his frame, he sees the air freshener swing
away from vertical without anything pushing on it, so he must invent a fictitious force to
explain the fact that the air freshener are not moving (in his frame of reference) yet have the
strange tilt.
An observer on the ground near the car notes that the velocity of the car must change as it
goes around the corner; the car undergoes acceleration (direction change). The tilt of the air
freshener can be explained entirely by the forces required to move it around the corner, so no
fictitious forces are necessary.
Bob’s frame of reference inside of the car is known as a non-inertial reference
frame because he had to invent the fictitious force to explain the tilt of the air freshener. The
reference frame of the observer on the ground is known as an inertial reference frame
because the tilt of the air freshener can be explained due to the motion of the car using known
laws of physics.
Bob’s non-inertial frame of reference is accelerating relative to the inertial
frame of reference of the observer on the ground. Any frame of reference moving at a
constant velocity with respect to an inertial reference frame is also an inertial frame of
reference.
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PHYSICS SEMESTER ONE
UNIT 4
We will define an inertial reference frame as a frame of reference where an object that is not
interacting with other objects (no net force, no fictitious forces) has zero acceleration. We generally
consider anything on the surface of the earth, or moving with a constant velocity relative to the earth as
an inertial reference frame.
Newton’s First Law of Motion (the Law of Inertia)
Under the condition that an object is viewed from an inertial reference frame, an object at rest will
remain at rest and an object in motion will remain in motion with a constant velocity (that is, with a
constant speed in a straight line), unless the object is acted upon by an external force.
The net force on an object, Fnet 
 F  F1  F2  F3  ... , is the sum of all forces on the object.
Consider an ice rink as our inertial reference frame. We will assume that the ice is frictionless so a puck
sliding along the ice surface has no net force on it. The puck will keep sliding with the same speed and
direction (same velocity) because the net force is zero. When the net force on the puck is not zero
(extra force applied to puck using a stick) the velocity will change.
Newton’s Second Law of Motion
Under the condition that the object is viewed from an inertial reference frame, the acceleration of an
object is directly proportional to the net force acting on it, and inversely proportional to its mass.
We will be using Fnet or
 F to denote the net force on an object. The symbol  F indicates that
we are making a sum of all of the forces. According to Newton’s Second Law of Motion
a
F
m
or
a
Fnet
m
This is more commonly written as
 F  ma
or
Fnet  ma
Force is a vector quantity so we can split this equation into separate component equations
 Fx  max ,  Fy  ma y
and
 Fz  maz
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PHYSICS SEMESTER ONE
UNIT 4
Each component equation relates the forces on the object in that direction with the acceleration in that
direction. A net force in the x direction will cause acceleration in the x direction. We will often just use
the x and y directions in 2D. We will skip the vector notation when we are dealing with motion and
acceleration in only one direction.
With the two or three vector component directions, we have two or three equations that have some
variables in common. In a typical 2D problem, there are two unknown variables and the two net force
component equations. We can combine the net force equations to eliminate one of the unknown
variables and solve for the other. We can then substitute the newly found variable in one of the original
equations to find the second unknown variable. More later …
Dimensional Analysis for Force
Define the dimensions of mass, length and time as M, L, and T respectively.
 x  dimensions of the variable x
The dimensions of mass, acceleration and force in Fnet  ma are then
 m  M ,
a 
L
T2
and
 Fnet    m a  M
L
T2
.
The SI unit for force is the newton, 1 N = 1 kg·m/s2.
If an object is released near Earth’s surface it will accelerate at g = 9.8 m/s2 due to gravity so we can say
that the gravitational force (force due to gravity) on 1 kg of mass is mass times the gravitational
acceleration or 9.8 N. In the imperial system, the term pound, lb, is often used to describe both mass
and the force of the same mass. The distinction is often made using “pound force” and “pound mass”,
though, to be correct, the “slug” should be used instead of the “pound mass”.
Example
Bob and his car have a mass of 1.20 × 103 kg. The engine and transmission produce a net horizontal
force of 1.7 × 103 N.
a) Calculate the acceleration of the car.
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PHYSICS SEMESTER ONE
UNIT 4
Here we have no mention of direction so assume that the net force is in a single direction.
Fnet  1.7  103 N, m  1.20  103 kg
Newton’s second law of motion for one dimension is
Fnet  ma
Rearranging gives
a

Fnet
m
1.7  103 N
The acceleration of the car is 1.4 m/s2.
1.20  10 kg
3
 1.42 m/s 2
b) Calculate the time required to accelerate from rest to 95 km/h.
There is no escape from kinematics problems in first year physics. They show up as part of
most sections. We cover kinematics and dynamics at the start of the course because many of the
other sections deal with forces and motion.
We need to convert the final velocity to m/s.
vi  0,
v f  95
km  1000 m   1 h 
m
,


  26.39
h  1 km   3600 s 
s
t ?
We now know the acceleration, velocities and need to find the time so use
v f  vi  at
t

v f  vi
a
.
26.39 m/s  0
1.42 m/s 2
 18.6 s
It takes Bob 19 s to accelerate from 0 to 95 km/h in his car.
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PHYSICS SEMESTER
ONE
y
F1 = 456 N
UNIT 4
Problem 1: Consider the following diagram showing the
F2 = 642 N
two horizontal forces applied by two friends as they
pull the 1.20 × 103 kg car along an icy road.
a)
Find the total force (in component form).
b)
Calculate the acceleration of the car. State
50.0°
75.0°
x
the answer in magnitude – direction form.
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PHYSICS SEMESTER ONE
UNIT 4
GRAVITATIONAL FORCE AND WEIGHT
When Newton wasn’t playing with forces or inventing calculus, he was, among other things, examining
planetary motion and the relationship between the masses, separation distance and gravitational
forces. He came up with what is now called Newton’s Law of Universal Gravitation.
Newton’s Law of Universal Gravitation
Every particle in the Universe attracts every other particle with a force acting along the line joining the
centres of mass of the two particles. This force is directly proportional to the product of the masses of
the two particles and inversely proportional to the square of the distance between them. If the particles
have masses m1 and m2, and their separation is r, the gravitational force has a magnitude Fg of
Fg 
Gm1m2
r2
where G = 6.67 × 10-11 N·m2/kg2 is the Universal gravitation constant.
This force applies to both particles. The particle with mass m1 experiences a force of Fg towards the
particle with mass m2. The m2 particle experiences a force with the same magnitude Fg towards the m1
particle. The separation distance is measured between the centers of the particles. The term particle
can refer to anything from atoms and subatomic particles to planets, stars and galaxies.
The masses m1 and m2 are often written as M and m when one item M is much bigger than the other
item. M is often used for the mass of the earth or a planet while m is the mass of a small object, like a
fish, on the planet.
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PHYSICS SEMESTER ONE
UNIT 4
We will deal primarily with objects on or near the surface of the earth. The separation distance is the
radius of the earth and forces are in the radial direction (along the radius). The gravitational force on an
object of mass m on the surface of Earth (mass M) is
Fg  mg
where
g
GM
r2
rˆ
is the gravitational field strength or gravitational acceleration. The negative sign in the equation for g
accounts for the fact that the gravitational acceleration direction towards (instead of away from) the
centre of Earth. We will usually use the scalar forms
Fg  mg ,
g
GM
r2
and know that the direction as “down” or the negative y-direction perpendicular to the surface of the
earth. We will assume g  9.8 m/s2 for this course. Those in Canada can check out the Natural
Resources Canada website (http://www.geod.nrcan.gc.ca/products-produits/ggns_e.php) for the values
in their area. The two significant figure value 9.8 m/s2 is good anywhere on Earth’s surface.
The magnitude of the gravitational force on an object is also called the weight of the object. The weight
of mass m is given by the equation
w = mg
The units of weight are the same as those of force, newtons N. Bathroom scales measure your weight,
but report your mass. A 63 kg physics instructor has a weight of 620 N.
How can the scales measure weight and report mass?
When you step on your bathroom scales, the scales mechanisms rotate a disk inside the scales.
When the disk stops, the number under the arrow/line is your mass. Manufactures design the
disk with numbers that are your weight divided by 9.8 m/s2 so the number you see is your mass.
The value of g on the moon is about 1.6 m/s2. The weight of the 63 kg physics instructor on the
moon is


w  Fg  mg   63 kg  1.6 m/s2  101 N
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PHYSICS SEMESTER ONE
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2
The scales report this weight divided by 9.8 m/s , or about 10 kg. This is false because mass is a
measure of the amount of matter inside the teacher, which is still 63 kg. Electronic scales do the
division electronically.
Example
Use the above equation for g, along with g = 9.80 m/s2, and the radius of Earth to find the mass of Earth.
Define terms
g = 9.80 m/s2, r = 6.37 × 103 km
Rearrange equation and substitute with g 
GM
. I will leave that to you.
r2
Note that it always a good idea to convert from km to m before you plug the numbers into the formula.
The mass of the earth is 5.96 × 1024 kg.
Check out the gravitational force examples in your textbook.
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PHYSICS SEMESTER ONE
UNIT 4
Reduced to their simplest level, all forces are due to the interaction of two items. When you push on a
wall, the wall pushes on you. Two magnets repel or attract each other. When we walk, we push on the
earth and the earth pushes on us. The gravitational force from planet A on satellite B is the same as that
from satellite B on planet A. In each of these cases we have two objects and two forces, one on each
object. We can actually show that the force from object A on object B is equal in magnitude and
opposite in direction to the force from object B on object A. This is Newton’s Third Law of Motion.
Newton’s Third Law of Motion
For two interacting objects (interacting = exertion forces on each other), the force exerted by object 1
on object 2 is equal in magnitude and opposite in direction to the force exerted by object 2 on object 1.
F12  F21
Example
The gravitational forces exerted by two objects on each other.
Fg12
Fg21
m1
m2
r
The magnitudes of the two gravitational forces are
Fg12 
Gm1m2
r2
and
Fg 21 
Gm2 m1
r2
The magnitudes are identical. The forces are in opposite directions.
Problem 2
Earth is attracted towards that 70 kg man with a force of 686 N. Why don’t we notice the effect of our
gravitational force on Earth?
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PHYSICS SEMESTER ONE
UNIT 4
More Examples
FN = Fdo
An object, subscript o, sitting motionless on a horizontal
desktop, subscript d, exerts force equal to its weight on the
desk Fdo  Fgo . The desk exerts a “normal” force FN  Fod
on the object equal and opposite the force from the object.
Fdo  Fod
Here, the net force on the object is
Fod = Fgo
Fnet , o  Fgo  FN
 Fdo  Fod

 Fdo  Fdo

ΣF = Fgo + n = 0.
0
The net force is zero so the motionless object on the table will remain motionless.
We can use this to equate the magnitudes
FN  Fg
 mg
In this case the normal force on the object is exactly equal to and opposite the force of gravity acting on
the object. The net force is zero so there is no change from the motionless state of the object.
A wrecking ball, subscript b, crashing into the wall of a building, subscript w, exerts a force Fwb on the
wall while the wall exerts an equal and opposite force on the ball Fbw  Fwb
Fbw
Fwb
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PHYSICS SEMESTER ONE
UNIT 4
In these examples, the two forces in the action reaction pair are forces on the two objects involved (one
on each object). This is an important because if the equal and opposite forces were always on the same
object, nothing would move.
Accelerations are often compared to the forces and accelerations due to gravity. An object experiencing
an acceleration of one G-force (also G, g-force and g-load) is accelerating at 9.8 m/s2. An object
accelerating at 5.0 G (or 5.0 g for even more confusion) has an acceleration of
 9.8 m s 2 
a  5.0 G  
 49 m s 2
 1 G 


This is often used for spacecraft, fighter jets and roller coasters. Note the confusion that the term Gforce usually refers to an acceleration and not force. We have to multiply by the noted acceleration
but the mass to get the net force applied to the object. The net force required to accelerate a 65 kg
person at 5.0 G is
 9.8 m s 2 
a  5.0 G  
 49 m s 2
 1 G 


F  ma
 65 kg  49 m s 2
 3185 N
 3200 N
I will leave it to you to show that the net force required to accelerate an object at n G is n times the
weight of the object.
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PHYSICS SEMESTER ONE
UNIT 4
NORMAL FORCE
The normal force ( N or FN ) is a force from solid surface to keep objects from sinking into the surface.
If we look close enough, the normal force is created by the stretching and compressing of the material
with the surface just enough to support the item on the surface (think of a grapefruit on top of a solid
bowl of lime Jell-O). We are not worried about the deformation of the surface, only that it provides just
enough force to keep the object from sinking into the surface.
For a text sitting on a desk, the surface of the desk pushes back on the book just enough to counter the
gravitational force. The normal force is just enough to balance the gravitational force. If the normal
force was any less, the book would fall through the table. If it was any more the book would lift off the
table.
FN
Fg
The normal force gets it name from the fact that its direction is perpendicular to the desk surface.
Normal is another word for perpendicular. (Technically, the force from the text on the table is a normal
force. The gravitational pair of forces is between the text and Earth. We will ignore this fact.)
All normal forces are perpendicular to the surfaces.
All normal forces are exactly the right size to keep the object from sinking into or lifting off the
surface.
Draw the normal and gravitational forces in the following systems (you may have to print this page).
Answer with solution to Problem 3.
surface
surface
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surface
12
PHYSICS SEMESTER ONE
UNIT 4
FRICTION
Friction covers everything from drag in air or water to friction between an object and its solid
surroundings. We will look primarily at friction between two items that are in contact. Chapter 6.4 for
the 120 students describes the friction due to motion in fluids.
Imagine a large boulder shows up on the horizontal road in front of your residence, and you decide to
move it. The boulder is too big to lift and too awkward to roll so you decide to push it. Initially, the
boulder
!
doesn’t move, so you push harder. Eventually, you push with enough force that the boulder moves.
You continue pushing the boulder without letting up because you “know” that it is easier to keep the
boulder sliding than it is to let it stop and then start it moving again.
Before the boulder starts moving, the friction known as the static friction, Fs . There three other forces
acting the boulder: your force F , the gravitational force Fg and the normal force FN from the
driveway. The net force on the boulder is zero because the boulder is not moving.
Fnet  Fg  FN  Fs  F  0
The forces are shown in the following diagram. Force diagrams are incredibly useful for organizing
information when doing problems. We will develop more formal force diagrams, later.
F
FN
FN
F
Fg
Fnet  Fg  FN  Fs  F
0
Fs
Fg
Fs
Your force and friction are in the horizontal (x) direction while the force of gravity and the normal force
are in the vertical (y) direction. In the x-direction
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PHYSICS SEMESTER ONE
UNIT 4
Fnet , x  Fs  F  0
Whether you push with a large or small force (that doesn’t exceed the maximum static friction), the
static friction exactly balances the force F applied by you so the net force remains zero.
Fs   F
Increasing your effort, the force applied by you eventually exceeds the maximum static friction force
and the boulder starts to move. The maximum static friction is proportional to the normal force
between the driveway and the boulder
Fs ,max   s FN
Fs  Fs ,max
where μs is the coefficient of static friction. The coefficient is dimensionless (both forces have the same
units).
Once the applied force exceeds the maximum static friction, the net force is no longer zero and the
boulder starts to slide (grind). The friction for the moving boulder is the kinetic friction, Fk . Kinetic
friction is proportional to the normal force between the surface and the object.
Fk  k FN
where μk is the coefficient of kinetic friction. We will assume that kinetic friction is constant for this
course. The coefficient of static friction is generally larger than the coefficient of kinetic friction
 s  k
so it is harder to get things moving than to keep them moving.
If we plot the coefficient of friction, static or kinetic, versus the applied force we typically get the
following relationship
Ff
Fs ,max   s FN
Fk  k FN
Fs = F
F
static, kinetic if moving
kinetic only
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PHYSICS SEMESTER ONE
UNIT 4
The following diagram shows the forces just after you have managed to start moving the boulder.
Because you haven’t reduced your applied force yet, the net force is not zero.
F
FN
Fnet
F
FN
Fnet  Fg  FN  Fk  F
Fg
0
Fk
Fg
Fk
The coefficients of friction are generally less than one. This makes sense because, if we can push with a
force larger than the normal force, we can lift the object off the surface, eliminating the normal force.
Typical Friction Coefficients
Note: the values shown here are approximate, and are intended for demonstration purposes only.
Check your textbook, on-line, and reference books to find more accurate values between a wider array
of materials.
contact
μs
μk
rubber on concrete
1.05
0.90
waxed wood on wet snow
0.20
0.10
lubricated metal on metal
0.15
0.05
Example
FN
A boulder on a horizontal surface has a mass of 520 kg.
The coefficients of static and kinetic friction are 0.80 and
F
0.60 respectively. Find the maximum static friction and
the kinetic friction.
Fk or Fs
Fg
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PHYSICS SEMESTER ONE
UNIT 4
Define terms:
m  520 kg,
s  0.80,
k  0.60
The normal force and force of gravity are the only vertical forces acting on the stationary boulder.
Fnet , y  Fg  FN  0
We will define the positive y-direction as “up”. This makes the force of gravity in the negative direction.
mg  FN  0
FN  mg

  520 kg  9.8 m/s 2

 5096 N
The formula for the maximum static friction is
Fs ,max   s FN
  0.80  5096 N 
 4080 N
The formula for the kinetic friction is
Fk ,max  k FN
  0.60  5096 N 
 3060 N
The maximum static force is 4100 N and the kinetic friction is 3100 N.
What is the direction of the friction?
The direction of the static friction is opposite the applied force.
The direction of the kinetic friction is opposite the direction of motion.
Friction can be both good and bad.
tire: road friction – good
car: air friction, drive train friction – bad
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PHYSICS SEMESTER ONE
UNIT 4
Check out http://phet.colorado.edu/new/simulations/sims.php?sim=Forces_in_1_Dimension. This
applet shows all of the forces acting on an object with motion in the horizontal direction. Try setting the
applied force to a maximum, letting the object accelerate, and then reverse the direction of the applied
force so the object comes close to but does not hit the wall. Also note that the friction force is just big
enough to counter the applied force until the object starts moving. When the object is moving, the
friction is the constant kinetic friction.
Also play with the ramp http://phet.colorado.edu/new/simulations/sims.php?sim=The_Ramp. When
you start the applet, the parallel force graph is the only graph shown. This shows the components of the
force in the direction parallel to the ramp (forces perpendicular to the ramp are not shown). If you vary
the ramp angle, you will see that the size of the different components varies with angle (the component
of the gravitational force is
Fgx  mg sin  .
The force of kinetic friction is
F fx   FN
  Fg cos 
  mg cos 
Both of these forces are functions of the ramp angle θ. If the angle is too low, the friction will be static,
and will only be large enough to counter the applied and gravitational forces.
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PHYSICS SEMESTER ONE
UNIT 4
EQUILIBRIUM
A system where the net force is zero is said to be in equilibrium. Most of the problems you will do in
this course will involve objects in equilibrium. An object is in equilibrium when its acceleration is zero.
Under equilibrium conditions:
 F  ma  0
The sum of the forces on the object is zero.
The sum of the forces in the x-direction is zero.
 Fx  max  0
The sum of the forces in the y-direction is zero.
 Fy  ma y  0
Can a moving object be in equilibrium?
Yes, as long as the acceleration is zero.
As noted, equilibrium shows up often in dynamics question.
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PHYSICS SEMESTER ONE
UNIT 4
FREE-BODY DIAGRAMS
When solving problems with several forces on a single object, it is often useful to use a free-body
diagram (FBD) to help visualize all of the forces acting on the object. FBDs help you organise your
problem solving process.
Check out the following PDF file on FBDs.
Creating FBDs
Select an Object
-
usually there is only one or two, and the question will often indicate the objects
-
in complex systems, it helps to sketch the whole system before drawing individual object FBDs
Draw All Forces Acting on the Object
-
this is all forces noted in the question plus any additional forces that you can assume are
present (interaction forces, force of gravity, friction if coefficient noted …)
-
gravitational force and normal force
-
usually you will be told whether to include friction
Choose a Convenient Set of Directions
-
this is BIG, it can be the difference between a simple and difficult problem
-
Can you use the regular y-vertical, x-horizontal directions?
-
Is the motion in an obvious direction? Choose this as the either the positive x- or y-direction,
and choose the other direction perpendicular to the motion.
-
Do most forces lie along one of two perpendicular directions? choosing the directions can save
work when calculating the vector components
Using FBDs
Check for hints about the net force on the object
-
Is the object accelerating?
-
Is the object is stationary or moving at a constant velocity in one or more directions? In this
case, Newton’s 1st Law tells us the net force in those directions is zero.
If you are asked to find the acceleration, you will probably need to use Newton’s 2nd Law. In most
cases, you will need to break the forces into components and use Newton’s Laws to solve for unknown
variables.
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PHYSICS SEMESTER ONE
UNIT 4
Example 1
Tossing a ball into the air.
If you toss a 0.10 kg ball into the air with a force applied by your hand is 5.0 N, what is the acceleration
of the ball?
F
+y
draw FBD, pick directions (only one in this case)
Is there acceleration?
– yes, so looking to find the net force and then acceleration
Fg
Define Terms:


m  0.10 kg, Fg  mg   0.10 kg  9.8 m/s2 ˆj  0.98 N ˆj, F  5.0 N ˆj
According to Newton’s Second Law the acceleration is proportional to the net force and inversely
proportional to the mass. We need to find the net force in order to find the acceleration.
Fnet  F  Fg

 
 5.0 N ˆj  0.98 N ˆj

 4.02 N ˆj
The acceleration is then
a

Fnet
m
4.02 N ˆj
0.10 kg
 40.2 m/s 2 ˆj
The acceleration is 40. m/s2 up.
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PHYSICS SEMESTER ONE
UNIT 4
Problem 3
Find the acceleration of a 1600 kg car that is being pushed horizontally with a force of 5200 N to the
right, assuming that the coefficient of kinetic friction is 0.25.
Is there acceleration?
Draw FBD, define directions
Define terms:
m  1600 kg, F  5200 N ˆi,
k  0.25,
a ?
Can you find the acceleration using only one of the component equations?
If not, can you use the other component equation to help find it?
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PHYSICS SEMESTER ONE
UNIT 4
TENSION
Tension is another common force that shows up in dynamics problems. Tension is another form of force
used to describe the force from/on a rope. In this part of the course, we assume that the tensions on
both ends of the rope are equal in magnitude, pulling towards the other end of the rope.
Consider a tug-of-war with a very
light rope where the rope is not moving.
We can ignore the force of gravity on
the light rope, and then assuming we have
the force FL on the left end and FR on the right end, the FBD for the rope, looks like this
FL
FR
The rope is not moving so the net force is zero.
Fnet  FL  FR  0
FL  FR ,
the magnitudes FL  FR
We can now apply Newton’s third law and see that the tension on each end is equal and opposite to the
force on it.
TL  FL
TR  FR
TL
TR
The tensions on the ends are then equal in magnitude and opposite in directions.
TL  TR  T
We often write the magnitudes as T at both ends. This is true even if the rope goes over a light,
frictionless pulley. The case where the pulley is not light and frictionless will be covered later when we
look at rotational motion.
T
T
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PHYSICS SEMESTER ONE
UNIT 4
Finally, there are several cases where we only have to consider one end of the rope. In these cases, the
tension is just another force. Instead of stating that the applied for is 5 N, we have a rope attached with
a tension of 5 N.
Problem 4
A boat is secured to a lakeside pier with two horizontal ropes. A wind is blowing off shore. The tension
in the ropes are T1 = 48 N [16° N of E] and T2 = 48N [16° S of E]. Assuming that the net horizontal force
on the boat is zero, what is the force of the wind on the boat?
There is no movement or tension in the vertical direction so we will only look at the horizontal plane.
Draw FBD, draw directions
Terms defined in question.
Is the object accelerating?
What is the net force on the boat?
The sums of the force components in both directions must equal _______ .
Break the tension forces into components
T1E 
T2E 
T1N 
T2N 
Use the net force to find the components of the wind force Fw .
The force of the wind on the boat is
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PHYSICS SEMESTER ONE
UNIT 4
Problem 5
Imagine, instead of pushing the car in Problem 3, you are tow the same car (mass 1600 kg, coefficient of
friction 0.25) with a tow line that makes an angle of 37° with the horizontal. The tension on the tow line
is 5200 N. What is the acceleration of the car?
Problem 6
A child pulls a sled up a snow-covered hill at a constant velocity with a force parallel to the hillside. If
the sled has a mass of 9.5 kg, the hill slope is 12°, and the coefficient of friction between the sled and
the snow is 0.20, what is the force applied by the child?
Problem 7
Sleds A and B are connected by a horizontal rope, with A in front of B. Sled A is pulled forward with a
horizontal rope with a tension of magnitude 29.0 N. The masses of sleds A and B are 6.7 kg and 5.6 kg,
respectively. The magnitudes of kinetic friction on A and B are 9.0 N and 8.0 N respectively.
B
A
TAB
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Tpull = 29.0 N i
24
PHYSICS SEMESTER ONE
UNIT 4
Problem 8
+y
A loonie (mL = 6.99 g) and a dime (mD = 2.09 g) are attached to the
ends of a thread. The thread lies over a smooth horizontal bar.
Initially, the coins are held motionless. When released, the coins
+y
+y
start to move. The friction between the thread and bar is negligible
(zero).
a) Find the acceleration of the coins.
b) Find the tension in the thread.
*** There are as many different examples of dynamics problems are there are problems needing forces.
Check out other examples and problems in your textbook, especially those with springs, pulleys, inclines
and blocks that are side-by-side. ***
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PHYSICS SEMESTER ONE
UNIT 4
Example
Highway Curve Angles
What angle do you bank a curve in order to
minimize the chance of sliding off?
Each highway curve is designed for a particular speed. If driven at the specified speed, a car on a curve
will only under go centripetal acceleration around the curve. The car will not accelerate sideways up or
down the curve even under icy conditions.
Consider a 1600 kg car moving at a constant speed around an ice covered curve with a radius of
curvature of 150 m. What angle is necessary for the bank to ensure that a car moving at 85 km/h will
not slide up or down the curve?
The forces and accelerations are: the normal force, the gravitational
force and the centripetal acceleration (ice = no friction).
+y
+x
n
Draw FBD
θ
Important: The normal force is always perpendicular to the surface.
mac
Choose Directions: This is very important too. The direction of the
centripetal acceleration is horizontally to the left. Any vertical
Fg
acceleration will cause the car to slide up or down the bank.
Choose regular x-horizontal and y-vertical directions as shown. This way, only the normal
force needs to be split into components.
Define terms:
m = 1600 kg, v = 85 km/h = 23.6 m/s, r = 150 m
The forces on the car are
Fnet  Fg  FN  mac
In the x-direction
Fnet , x  FNx  mac
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PHYSICS SEMESTER ONE
UNIT 4
The centripetal acceleration is
ac 
v2
r
2
23.6 m/s 


150 m
 3.72 m/s 2
The x-component of the normal force is then
FNx  mac

 1600 kg  3.72 m/s 2

 5950 N
In the y-direction, there is no acceleration so the net force is zero (no movement up or down the ramp)
Fnet , y  Fg  FNy  0
Rearranging gives
FNy   Fg
   mg 
mg to account for direction

= 1600 kg  9.8 m/s 2

 15680 N
The normal force is perpendicular to the bank so we can use the angle of the normal
θ
force relative to horizontal to find the angle between the surface of the road and
n
horizontal.
tan  
FNx
FNy
  tan 1
 tan 1
θ
FNx
FNy
5950
15680
 20.8
The highway must be banked at an angle of 21° to avoid sliding up or down
at 85 km/h under icy conditions.
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PHYSICS SEMESTER ONE
UNIT 4
SUMMARY:
Newton’s Laws of Motion and Universal Gravitation
Newton 1:
v  constant if Fnet  0
Newton 2:
Fnet  ma , Fnet , x  max , Fnet , y  ma y , Fnet , z  ma z
Newton 3:
F12  F21
Gravity: Fg 
Gm1m2
r2
g
gravitational force
GM
r2
gravitational acceleration
The gravitational force and acceleration are radial, directed towards
the centre of the other object.
Friction
Static:
Fs  Fs ,max  s FN
Kinetic: Fk  k FN
Equilibrium
The sum of the forces on the object is zero.
 F  ma  0
The sum of the forces in the x-direction is zero.
 Fx  max  0
The sum of the forces in the y-direction is zero.
 Fy  ma y  0
Free Body Diagrams
Draw all forces acting on an object.
Choose the coordinate directions (x and y axes) to simplify calculations.
Tension
-
The tension is always from the end of the rope in the direction along the rope.
-
Most ropes are considered ideal in that have no mass. Ideal ropes they don’t stretch so the ends
of the rope move with the same velocity and acceleration.
-
If the rope changes direction, you can modify your coordinate systems (one set for each end of
the rope) so both ends have the same velocity and acceleration vectors.
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PHYSICS SEMESTER ONE
UNIT 4
-
A rope that goes over and ideal pulley (no mass, no moment of inertia) or a frictionless surface
has the same tension magnitude at both ends.
NANSLO Physics Core Units and Laboratory Experiments
by the North American Network of Science Labs Online,
a collaboration between WICHE, CCCS, and BCcampus
is licensed under a Creative Commons Attribution 3.0 Unported License;
based on a work at rwsl.nic.bc.ca.
Funded by a grant from EDUCAUSE through the Next Generation Learning Challenges.
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29