Grade 11 Physics Mechanics _Kinematics and Dynamics_ Unit 3 by stariya

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									                    Unit 3 - Mechanics (Kinematics and Dynamics)
                                                                 Back to Physics 11

KINEMATICS

Lesson 1: Scalars and Vectors

Lesson 2: The Position-Time Graph and Velocity

Lesson 3: Position-Time, Velocity-Time, and Acceleration-Time Graphs


Lesson 4: Constant Velocity and Acceleration: Problem Solving

DYNAMICS

Lesson 5 - Fundamental Forces and Newton’s First and Second Laws

Lesson 6 - Adding Force Vectors by Components

Lesson 7 - Free-body Diagrams

Kinematics

Lesson 1: Scalars and Vectors

Introduction

In this lesson we begin our study of mechanics. This is the branch of physics that
deals with the motion of objects. To begin with, we will study this motion without
considering the forces involved in the motion. To describe the motion, we will need
to set up a reference point and a frame of reference. It is important to define and
use in an appropriate way terms such as position, distance, displacement, speed and
velocity. Terms like distance and displacement, or speed and velocity may mean the
same thing in common conversation, but they have slightly different definitions. To
understand the difference between terms that are similar, it is also important to
understand when direction is important and when it is not. Once these terms are
clearly understood, we will be able to discuss how to graph motion. This is the topic
in the next lesson.

Position and Distance

The branch of physics that describes how objects move without referring to the
forces that cause that motion is called kinematics. To begin with, we will discuss
motion along a straight line only.

Imagine travelling in a car in an easterly direction. From your starting point, you
may travel a distance of 80.0 kilometres (km) in a time of 1.00 hours (h). You then
continue your trip towards the east for another 120.0 km and this second part of
your travel takes a time of 1.50 h. The diagram below can help to visualize the
situation. The car is represented by a dot.




In this example, you are always travelling in an easterly direction. When looking at a
map, this direction is to your right. We choose this direction to be a positive direction
just as in a number line or coordinate system in mathematics, the right direction is
considered to be a positive direction. After the first hour, the car was at a position of
80.0 km or 80.0 km east of the starting point. After the car moved another 120.0 km
to the east, the new position from the starting point was 200.0 km. The distance
travelled by the car is also 200.0 km.

In describing the motion of this car, we need to have a starting point and we need to
have a frame of reference. This starting point, or reference point, is the zero
location in a coordinate system or frame of reference. In the example above, the
reference point was the zero point on the number line. The frame of reference is
the coordinate system used to define motion. The frame of reference is the number
line or the east – west line on a map.

The position of an object is the separation between the object and the reference
point. To properly describe the position of an object, we need to know the distance of
the object from the starting point and the direction from this point. So to describe
the position of the car after first part of the motion of the car, we can say that it is
80.0 km east of where it started.

Distance is simply the separation between two points. It does not require a
direction. So we can say that the distance the car travelled in the first part of the
motion is 80.0 km.

      Distance is a scalar quantity which refers to "how much ground an
       object has covered" during its motion.
      Displacement is a vector quantity which refers to "how far out of
       place an object is"; it is the object's change in position.




Vectors and Scalars

Now suppose that the car we discussed earlier continues on its trip. After arriving at
a position of 200.0 km east of the reference point, it turns around and moves to the
west. It moves an additional 250.0 km to the west in 3.00 h. This new movement is
recorded below.
On a number line, the 250.0 km west would be considered a negative direction. After
completing this motion to the west, the new position of the car is 50.0 km west, or -
50.0 km, of where it began.

The distance the car travelled to the west is 250.0 km. If we want to determine the
total distance the car travelled during the trip, we simply add the distances together.
The distance to the east was 200.0 km and the distance to the west was 250.0 km,
so the total distance was 450.0 km.

We can see from the discussion above that position and distance are different kinds
of quantities. Position requires a direction and it is important to know if a movement
is positive or negative. A quantity, like position, that has both magnitude (size) and
direction is called a vector. Distance does not require a direction. Only the number
of kilometres (in this case) travelled is important. A quantity, like distance, that has
only a magnitude, or size, is called a scalar.

(open the link for a quiz/test)

 http://wps.pearsoned.ca/ca_school_albertaphysics/0,9708,4147218-
content,00.utf8.html

Time. Position, and Displacement

Another scalar is time. The diagram below shows the times when the car was at the
various positions.




Each of these times are instantaneous times. These are the times a clock might read
when the car was at that particular position. The clock reading lasts zero seconds
when the car is at that position. At the next instant, the car is at a different position.
We can thus define an instantaneous time as the time a time piece such as a clock
reads at any particular instant and that lasts zero seconds. The instantaneous
position can be defined as the position of an object at a specific or instantaneous
time.

As the car is moving, it is changing its position. The change in position of an object is
called the displacement of the object. To determine the displacement of an object,
determine the position of the object at one instant in time, x2, and the position at an
earlier time, x1. The change in position, or displacement, ∆x is the difference in
position: ∆x = x2 – x1. Displacement is a vector quantity so it can be positive or
negative. The symbol “∆” is a Greek letter pronounced “delta” and is the symbol
used to represent the “change in” a certain quantity. It is generally calculated as “the
final subtract the initial” value of the quantity. In the example above, the “final”
position was indicated as “x2” and the “initial” as “x1.” Different subscripts may be
used to indicate the terms final and initial. Another way to write these terms could be
“xf” or “xi.” In the same way we have a time interval. The time interval can be
defined as the difference between two clock readings, or in other words, two
instantaneous times: ∆t = t2 – t1.

In the example above, the displacement of the car after the first hour was
∆x = 80.0 km – 0.00 km = 80.0 km. The displacement after 2.50 h was ∆x = 200.0
km – 0.00 km = 200.0 km. For the entire trip, the displacement was ∆x = -50.0 km
– 0.00 km = -50.0 km.

Another way to determine the displacement is to add together the displacements at
various points along the trip. In the first hour the displacement was just 80.0 km. In
the next 1.50 h, the car moved another 120.0 km to the right. This is a positive
displacement so the total displacement up until this point is 80.0 km + 120.0 km =
200.0 km. In the last segment the car moves to the left another 250.0 km. This
represents a displacement of –250.0 km. Therefore the displacement for the entire
trip is 200.0 km – 250.0 km = -50.0 km.

Average Speed

The metric system of measurement was created by French scientists in 1795. It is a
convenient system to use because units of different sizes are related by powers of
ten. An international committee determines the standards of the metric system. The
committee has set up the Système International d’Unités (SI). Three basic units in
the SI system are time, length, and mass. The unit of time is the second (s), the unit
of length is the metre (m), and the unit of mass is the kilogram (kg). A wide variety
of other units, called derived units, are combinations of these base units. One of the
derived units is the unit used for speed and velocity.

We can define the average speed of an object as the distance travelled divided by
the time required to cover the distance.




The equation indicates that the unit for the average speed is the metre per second
(m/s) in SI units. Other units may be used. A common one is the kilometre per hour.
In the example discussed above, at the end of one hour, the car moved a distance of


80.0 km. The average speed was therefore             = 80.0 km/h. After a total
time of 2.50 h had passed, the car was now 200.0 km east of where it started from.

The average speed up until this was point                = 80.0 km/h. The car then
turned around and travelled west. It travelled 250.0 km west in an additional time of
3.00 h. Thus the total distance for the entire trip was 450.0 km for a total time of


5.50 h. The average speed for the whole trip was                = 81.8 km/h. Notice
that in calculating the average speed, the direction of travel was not important. Only
the total distance and the total time were important. Since the direction of travel is
not important, speed is considered to be a scalar, not a vector.

Average Velocity

Speed is a useful idea because it indicates how fast an object is moving. However
speed does not reveal anything about the direction of motion. To describe both how
fast an object moves and the direction of its motion, we need the vector quantity
called average velocity.

We can define the average velocity of an object as the displacement divided by the
elapsed time.



                                                        The SI unit for speed is the
metre per second (m/s), although it can also have other units such as the km/h.
In the example discussed above, at the end of one hour, the car moved a distance of


80.0 km and the average speed was                 = 80.0 km/h. Since the car is
moving to the right, this is also the average velocity. Both speed and velocity are
positive quantities. The 80.0 km distance is also the displacement of the car.

After a total time of 2.50 h had passed, the car moved to a position of 200.0 km east
of where it started from. This is also the displacement of the car. The average speed

and velocity up until this was point                             .

The car then turned around and travelled west. It travelled 250.0 km west in an
additional time of 3.00 h. Thus the total distance for the entire trip was 450.0 km for


a total time of 5.50 h. The average speed for the whole trip was          = 81.8
km/h. However, the total displacement of the car was –50.0 km and the average


velocity for the entire trip was            = -9.09 km/h. Note the difference in both
the magnitude and the direction of velocity as compared to speed for the entire trip.

Exercise

   1. One of the following statements is incorrect.
         a. The car travelled around the track at a constant velocity.
         b. The car travelled around the track at a constant speed.

        Which statement is incorrect and why?

   2. Sound travels at a constant speed of 343 m/s in air.
         a. How much time does it take the sound of thunder to travel at distance
               of 5005 m?
           b. If the sound reflects back to the place where it originated, what is the
                average velocity of the sound?
   3.   Andy Green in the car ThrustSSC established a world record in 1997. The
        speed of the car was 341.1 m/s (1228 km/h). The car was powered by two
        jet engines. It was also the first car to officially exceed the speed of sound. To
        establish such a record, the driver makes two runs through the course, one in
        each direction to nullify wind effects. First the car travelled from left to right a
       distance of 1609 m in a time of 4.740 s. Then the car travlled the same
       distance in the reverse direction in a time of 4.695 s. Determine the average
       velocity in each run.

   4. A tourist being chased by an angry bear is running in a straight line toward
       his car at a speed of 4.0 m/s. The car is a distance d away. The bear is 26 m
       behind the tourist and running at 6.0 m/s. The tourist reaches the car safely.
       What is the maximum possible value for d?




Lesson 2: The Position-Time Graph and Velocity

http://home.nc.rr.com/enloephysics/enloephysics/Kinematics/Page_1x.ht
ml

(visit the above link for interesting applets)

Introduction

In this lesson we extend our knowledge of distance, displacement, speed, and
velocity to a graphical representation of these quantities. We will first discuss how to
draw a position-time graph given a data table showing time and position data. We
will then be able to determine the speed or velocity of an object from a position-time
graph. The average velocity and the instantaneous velocity of an object are
determined in similar ways when a position-time graph is given. In the assignment
work, we will apply this knowledge to a person who is exercising and to a moving
bus.

The Position-Time Graph

Suppose that an object is moving at a constant velocity of 5.0 m/s. Using the

equation         , we can determine the change in position at different times: ∆x =
v∆t. If we choose the reference point to be at zero, then the “change in position” just
becomes the new position of the object. We can create a data table as shown below.
A graph that shows how the position depends on the clock reading, or time, is called
a position-time graph. The graph below represents the data from the table.
Suppose that the object is moving twice as fast, that is at 10.0 m/s. Then the new
data table and the new graph for this motion would be as shown below. In order to
keep the same scale on the graph, only the first two seconds of the motion are
shown.
Velocity and the Position-Time Graph

The velocity of an object can be found from a position-time graph. On
a position-time graph, the displacement is the vertical separation of
the two points: ∆x = x2 – x1. A time interval is the horizontal
separation: ∆t = t2 – t1. By definition, the ratio of displacement to the

time interval is the average velocity:         . From mathematics, the
ratio of the vertical separation of two points on a curve, or rise, to the
horizontal separation of the points, or run, is the slope of a graph.

That is,                   .
                   In the graph to the
                   left, the rise is
                   25.0m and the run
                   is 5.0 s. Therefore
                   the velocity is




                   For this graph, the
                   rise is 20.0 m and
                   the run is 2.0 s. The
                   velocity is




                   The steep slope
                   indicates a higher
                   velocity.


                   In general, velocity
                   is given by the slope
                   of a position-time
                   graph.

                                  .




Average Velocity
 In the graphs above, the velocity of the object was constant over the time periods
indicated. The first graph showed a constant velocity of 5.0 m/s for 5.0 s, and the
second graph showed a constant velocity of 10. m/s for 2.0 s. For these graphs, the
average velocity and the velocity over a short time interval will have the same value
because the velocity is constant.

Now let us look at a situation where the velocity may vary over a time interval. The
graph below shows two different constant velocities over a time interval.


                                                      In the first three seconds,
                                                      the velocity is constant.
                                                      During this time the
                                                      velocity is




                                                      Between the time of 4.0 s
                                                      and 5.0 s, the velocity is
                                                      also constant, but different
                                                      from the first three
                                                      seconds. This second
                                                      velocity is




To determine the average velocity over the full 4.0 s, we can find the slope of the
dashed line shown above. This line has a rise of 25.0 m over a run of 4.0 s. The
average velocity is therefore


           = 6.25 m/s or 6.3 m/s.


Note that the answer is not the average of the velocities, that is                   =
7.5 m/s.

In general, we can say that the average velocity is given by the slope of the line
joining two points on a position-time graph.

To confirm this, we can find the average velocity another way. We determine the
total displacement for the trip and divide by the total time. In this case, the total
displacement over the two intervals is ∆d = 15.0 m + 10.0 m = 25.0 m. The total
time over the two intervals is ∆t = 3.0 s + 1.0 s = 4.0 s. Therefore the average
velocity is


                                .
Instantaneous Velocity

When a person is driving a car, the speedometer will show the speed of the car. If
the speedometer swings from 40 km/h to 60 km/h, the car is speeding up. If the
driver glances at the speedometer and the needle reads 50 km/h, it means that at
that instant, the speed was 50 km/h. In other words, the instantaneous velocity was
50 km/h. That is, if its velocity had been constant, the car would have driven 50 km
in 1 h.

We can use a position-time graph to find instantaneous velocity. When a runner does
a 100 m dash, the first two seconds or so are used to speed up. The position is
changing so that the curve is a smooth line.




Suppose we wanted to find the velocity of the person at 1.00 s. We can start by
finding the average velocity over 2.00 s. After 2.00s, the displacement is 12.5 m.

The slope of the line from 0.00 s to 2.00 s is                        . This shown as
region A above. Now we choose a smaller time interval, the one between 0.5 s and
1.5 s. The displacement is interval is 6.8 m
(8.0 m – 1.2 m) and the time interval is 1.0 s. The slope of the connecting line is
therefore 6.8 m/s. This process could be continued, choosing the time interval to be
smaller and smaller until the two clock readings are almost the same. When this
happens, we can no longer draw a line connecting the two points. We draw a straight
line that is tangent to the curve at this point. The slope of the line that is tangent is
called the instantaneous velocity at that instant. A tangent line is shown above.

The slope of this line is                           and this is the instantaneous
velocity of the runner at 1.00 s.
Exercise


The graph below shows a possible situation for a person who is exercising.




   a. For the segments A to C above, calculate the velocity. (Note: convert each of
        the time intervals into seconds so that the units of velocity are in m/s. 1 h
        has 3600 s.)
   b.   During which of the time intervals, A to C, was the runner moving the fastest?
   c.   During which of the time intervals was the runner returning towards the
        starting point?
   d.   During which of the time intervals was the runner most likely resting.
   e.   What was the average velocity over the whole run?
   f.   For the time interval corresponding to D, determine the instantaneous
        velocity at a position of 500.0 m.




Lesson 3: Position-Time, Velocity-Time, and Acceleration-Time Graphs


Introduction
In this lesson we expand our discussion of graphing in kinematics. We begin with
a graph of constant velocity versus time and how to derive the corresponding
position-time graph. We then define the term “acceleration” and how to
determine acceleration from a velocity-time graph. From a constant
acceleration-time graph, we discuss how to produce a velocity-time graph and a
position-time graph. For the purposes of this course, graphing these graphs will
occur only for the case of constant acceleration.




Constant Velocity


In the previous lesson we discussed the situation of an object moving at a constant
speed of 5.0 m/s. We drew a position time graph for that case. We can also draw a
velocity time graph. Since the speed of the object is constant at 5.0 m/s, the
velocity-time graph is a horizontal line.




We learned in an earlier lesson that the displacement of an object can be calculated
using
∆x = v ∆t. Over a time of 5.0 s, an object moving at a constant velocity of 5.0 m/s
would experience a displacement of ∆x = (5.0 m/s)(5.0 s) = 25 m. If we look at the
velocity-time graph, we can see that the horizontal line creates a rectangle under the
line. The area of a rectangle is given by the width, which is the height in this case,
(5.0 m/s) times the length (5.0 s): area = (width)(length) = (5.0 m/s)(5.0 s) = 25
m. We can make the following general statement.


The displacement in any interval is given by the area under the v-t graph for
that interval.


We can find the area under the v-t curve for smaller time intervals, and generate
data for a position-time graph. For example, for the first second of motion, the area
under the curve is (5.0 m/s) (1.0 s) = 5.0 m. Between 0.0 s and 2.0 s, the area is
(5.0 m/s)(2.0 s) = 10.0 m. We can continue this process until the data table for the
whole motion is made.



Time (s)         1.0 2.0     3.0    4.0     5.0

Position (m) 5.0 10.0 15.0 20.0 25.0




This is exactly the data we started with when we first discussed this motion in Lesson
2, and graphing this data would reproduce the position-time graph above.


Average Acceleration and the Velocity-Time Graph


 Suppose now that the object is not moving at a constant velocity of
 5.0 m/s. This object could be a runner. The runner may change
 velocity going from 5.0 m/s to 7.0 m/s in a time of 1.0 s. The runner
 may also slow down from 7.0 m/s to 5.0 m/s in a time of 1.0 s.
 When the velocity changes, we say the object is accelerating. We can
 define the average acceleration as the rate of change of velocity.




 The SI unit of velocity is the metre per second, m/s and the unit of
time is the second, s. The SI unit of average acceleration is the
metre per second squared, m/s2.


The average acceleration is a vector that points in the same direction
as the change in velocity.


For the above runner who is speeding up, the acceleration is



When this runner slows down, the acceleration is




In the graph below, the velocity is changing uniformly from 0.0 m/s
to 10.0 m/s over a time period of 5.0 s.


                                           From the definition of
                                           average acceleration as
                                           the change in velocity over
                                           an elapsed time,




                                           However, this is equivalent
                                           to the rise divided by the
                                           run and this is the slope of
                                           the line. Earlier we saw
                                           that the slope of a
                                           position-time graph gives
                                                    the rate of change of the
                                                    displacement which is the
                                                    velocity. Now we see that
                                                    the slope of a velocity-
                                                    time graph gives the
                                                    rate of change of
                                                    velocity, that is the
                                                    acceleration.




Positive Acceleration and the Velocity-Time Graph


Earlier we saw that the acceleration of the runner was a constant value of 2.0 m/s 2.
A graph of this acceleration on an acceleration-time graph would simply be a
horizontal line.


We have already seen that by determining the area under a velocity-time graph, we
can create a position-time graph. We can also find the area under an acceleration-
time graph to create a velocity-time graph.


In the first second the acceleration-time graph below, the area under the a-t graph is
(2.0 m/s2)(1.0 s) = 2.0 m/s. This means the velocity has increased by 2.0 m/s from
the initial value. Between 1.0 s and 2.0 s, the area is also 2.0 m/s, so the velocity
has increased by another 2.0 m/s, making the total increase in velocity 4.0 m/s from
the initial time. We can continue in a similar manner to generate the data for the
velocity-time graph shown below.



Time (s)           0.0 1.0 2.0 3.0 4.0 5.0

Velocity (m/s) 0.0 2.0 4.0 6.0 8.0 10.0
Graphing this data will reproduce the solid line of the velocity-time graph seen
below. If the initial velocity is not 0.0 m/s, then the starting point of the line will
begin at the initial velocity. For example, if the initial velocity is 2.0 m/s, the graph
of velocity-time will begin there, and then increase by 2.0 m/s for every second. This
is shown as the dashed line. It has the same slope as the original line.




The Position-Time Graph and Constant Positive Acceleration


In the previous section, we drew a velocity-time graph from a constant acceleration-
time graph. Now we will draw the corresponding position-time graph. To do so, we
determine the area under the velocity-time curve at various times and create a data
table showing time and position.




Between 0.0 s and 1.0 s, the area is      (1.0s)(2.0m/s) = 1.0 m.


Between 0.0 s and 2.0 s, the total area is     (2.0s)(4.0m/s) = 4.0 m.
(Or we could determine the area between 1.0 and 2.0s which is 3.0 m and then add
this to the 1.0 m to obtain 4.0 m.)
We can continue in a similar way until we produce the data table below.



Time (s)        0.0 1.0 2.0 3.0 4.0            5.0

Position (m) 0.0 1.0 4.0 9.0 16.0 25.0




We can see that the position-time graph is gradually sloping up and to the right. This
is a typical curve for a position-time graph when the acceleration is constant. In
general the shapes of the a-t, v-t, and d-t graphs for constant acceleration are as
shown below.
The Position-Time Graph and Constant Negative Acceleration


We have seen what the velocity-time and the position-time graphs look like for
constant positive acceleration. We will now examine the case of constant negative
acceleration. The graph below shows a constant negative acceleration of –2.00 m/s2.
We could create a data table for the velocity-time graph by finding the area between
the horizontal axis and the acceleration line. This would result in negative areas and
the velocity-time graph would be sloping downwards and to the right. The solid line
shows the velocity assuming the initial velocity is zero. The dashed line shows the
velocity if the initial velocity is –2.0 m/s.
We could also determine the area under the velocity-time curve to produce a data
table for a position-time graph. These areas would also be negative. The shape of
the resulting velocity time graph is shown below.




Exercise
The following velocity-time graph shows a possible motion of a bicycle.
   a. Produce a data table showing time and position. Then graph the data. The
       time portions of the table have been filled in.



       Time (s)         0.0 2.0 5.0 7.0 9.0 11.0 12.0

       Position (m)




   b. Produce a data table showing time and acceleration. Then graph the data. The
       time portions of the table have been filled in.



       Time(s)                   0 to 2 2 to 5 5 to 9 9 to 11 11 to 12

       Acceleration (m/s2)


Lab: Uniformly Accelerated motion and a Modern Galileo Experiment




Lesson 4: Constant Velocity and Acceleration: Problem Solving


In this final lesson, we emphasize problem solving in kinematics. We begin by
studying a velocity-time graph of constant acceleration when the initial and final
velocities are not zero. From this study, we derive a useful equation for determining
the displacement of an object. We then use three illustrative examples to show three
other equations can be also be used to solve problems involving the motion of
objects that are accelerating at a constant rate.


Displacement and the Velocity-Time Graph


In lesson 1 we learned that the average velocity is defined as the displacement
divided by the elapsed time.
                                    = or

In lesson 2 we learned that if we are given a position-time graph, that the average
velocity can be calculated as the slope of a position-time graph over a time interval.
The instantaneous velocity for a curving position-time graph is the slope of the
tangent line at a given time. In lesson 3 we found that the displacement of an object
over a time interval could be found by determining the area under a velocity-time
curve. The following graph is a velocity-time graph for constant acceleration, but the
initial velocity does not begin at zero.


                                                   To determine the area under
                                                   the line, we could first
                                                   calculate the area of the
                                                   triangular region.




                                                   We could then calculate the
                                                   area for the rectangular
                                                   region.




The total area is 180. m + 120. m = 300. m.


This is equivalent to ∆x = v1t +   (v2 - v1t) =   (v1 + v2t). The term   (v1 + v2t) is
the average velocity. This makes sense since we learned earlier that ∆x = vave ∆t.
Another way to determine the total area is to find the area of the trapeziod formed


by the line above. The area of a trapezoid is found by Area =    (b1 + b2)h where b1
and b2 represent the two parallel lines formed by the trapezoid, that is v1 and v2 ,
and h is the time interval, ∆t between these two velocities. Since this area
represents the displacement, then ∆x =     (v1 + v2) ∆ t. For the graph above, ∆x


  (10.0m/s + 40.0m/s)(12.00s) = 300. m.




Equations for Average Velocity and Average Acceleration

In lesson 1, we defined average velocity as the displacement divided by the elapsed
time.




In lesson 3, we defined average acceleration as the change in velocity divided by the
elapsed time.




In lesson 4, we learned that the average velocity could also be written as



                                              vave = 1/2(v1 + v2)




We will now look at how these three equations can be used to solve problems
involving the motion of objects.

Example 1

An object moving at 3.0 m/s accelerates for 4.0 s with a uniform acceleration of 2.0
m/s2. Find the displacement of the object.

In solving the problem, it is useful to state the information given in symbolic form.
v1 = 3.0 m/s
t = 4.0 s
a = 2.0 m/s2

Then state the unknown in symbolic form. We are looking for the displacement.

∆x = ?



To solve for the displacement, we can use          . But in order to determine the
average velocity, we must first determine the new velocity after the objcet has


accelerated for the given time. We do this using                   . Rearranging this
equation gives a∆t = v2 - v1, and then v2 = v1 + a∆t.

The velocity after a time of 4.0 s is v2 = v1 + a∆t = 3.0 m/s + (2.0 m/s2)(4.0 s) =
11 m/s.



The average velocity is vave =    (v1 + v2) =   (3.0m/s + 11m/s) = 7.0 m/s.

The displacement is therefore ∆x = vave ∆t = (7.0 m/s)(4.0 s) = 28 m.




Example 2

A car travelling at 10.0 m/s accelerates at a rate of 3.0 m/s2 to a speed of 25.0 m/s.
What is the displacement of the car during the acceleration?

The initial conditions are

v1 = 10.0 m/s
a = 3.00 m/s2
v2 = 25.0 m/s

We are looking for the displacement, ∆x.

In order to find the displacement, we will first need the time. Since we are given the
acceleration, we can determine the time.




The average velocity is



vave =   (v1 + v2)     vave =    (10.0m/s + 25.0m/s)= 17.5 m/s
The displacement of the car is

∆d = vave∆t = (17.5 m/s)(5.00 s) = 87.5 m.




Example 3

 If an object is accelerated at 5.00 m/s2 and starts from rest, what is its velocity
after 20.0 m?

The initial conditions are

a = 5.00 m/s2
v1 = 0.00 m/s
∆d = 20.0 m

We are looking for the second velocity v2.

In this example, we will emphasize an approach that derives two equations for the
time interval, and then makes the expressions equal to each other.

The first equation for the time interval comes from the relationship between time,
displacement, and average velocity.




The second equation for the time interval comes from the relationship between time,
acceleration, and the difference in velocities.




We can now make these two expressions of time equal to each other.




At this point, we can insert the numerical values for the variables and then solve for
the second velocity.
Exercise

   1. A sports car is moving initially at 10.0 m/s. It then undergoes a uniform
      acceleration of 1.20 m/s2.

           a. What is the new velocity after a time of 30.0 s?
           b. What is the displacement of the car?




   2. A speedboat has an acceleration of 2.0 m/s2. The initial velocity of the boat is
      6.0 m/s and the final velocity is 22.0 m/s. What is the displacement of the
      boat?


   3. A spacecraft is initially at rest with respect to a space station when it fires its
      rockets. These rockets make it accelerate at a rate of 10.0 m/s 2. What is the
      new velocity of the spacecraft after is has moved a distance of 2.15 x 10 5 m?
      (In solving this question, derive two separate equations for time and then
      make the expressions equal to each other to solve for the second velocity.)


   4. The following graph shows a car changing in velocity over time.




      Determine the displacement of the car.




DYNAMICS
In kinematics (above), we began our study of mechanics by reviewing the principles
of kinematics, a description of motion. In this dynamics, we complete the topic of
mechanics with a discussion of dynamics, the second branch of mechanics. In
dynamics we move beyond the description of motion to examining the “why” of
motion. This means understanding the forces that act on objects.

Lesson 5 - Fundamental Forces and Newton’s First and Second Laws

Introduction

We begin our study of dynamics with a discussion of the four fundamental forces that
hold our universe together. We then begin our discussion of Newton’s laws of motion
starting with the first law which is a nonmathematical law describing whether or not
an object will change its motion when forces act on it. Newton’s second law describes
in a mathematical way the acceleration of a mass when an unbalanced force acts on
the mass. We then show how Newton’s second law and our study of kinematics
combine to allow us to solve problems involving the motion of objects.

The Four Fundamental Forces

Contemporary theoretical physics maintains that in the universe as it exists today,
there are four fundamental forces:

      Gravitational Force
      Electromagnetic Force
      Strong nuclear Force
      Weak nuclear Force

We can define a force as a push or a pull. We can consider a force to be an agent of
change. When an apple falls, or a grasshopper leaps, or a supernova explodes, or a
neutron decays, these very different events unfold because of the involvement of the
four fundamental forces.

Gravitational Force

The gravitational force is an attraction between two objects because of their mass.
It keeps you, the atmosphere, and the seas fixed to the surface of the planet.
Gravity is the weakest of all the interactions, and it is also the least selective
because it acts between all particles. Because its range is unlimited and it is only
attractive, gravity rules the cosmos on a grand scale. It holds the earth in orbit
around the sun and keeps the sun locked in our galaxy of a hundred thousand million
stars. It reaches across the thousands of millions of galaxies that constitute the
universe.

Electromagnetic Force

The electromagnetic force is the force that charged particles exert on each other.
It binds together the atoms and molecules that make up the matter that we see
around us. It can be either attractive or repulsive. It can cause two magnets to
attract or repel, or two charges to attract or repel. It produces contact forces such as
that found between a fist and a punching bag, or a hamburger and teeth. Other
contact forces such as friction, the pull or push of a spring are considered to be
electromagnetic forces. Friction is the bonding at the molecular level of two surfaces
in contact. The movement of a spring is caused by the electromagnetic forces acting
between the atoms of a spring. The electromagnetic force is much stronger than the
gravitational force and its range is also unlimited.

Strong Nuclear Force

The strong nuclear force is an attractive force that holds protons and neutrons in
the nucleus of an atom. It is an extremely strong force necessary to overcome the
repulsive force between two protons and only acts at short ranges of about 10 -15
metres. This is only a few times the diameter of a proton. This tiny range accounts
for this force not coming directly into our normal experience, and as a result, was
not discovered until the 20th century. Without it, familiar matter, from planets to
puppies, would all disintegrate into a fine subatomic dust.

Weak Nuclear Force

The weak nuclear force is a million times fainter than the strong force and a
hundred times shorter in range. But it is much stronger than the gravitational force.
It can cause a proton to change into a neutron and is responsible for the slow
radioactive decay of certain elements.

Our understanding of which forces are fundamental is continually evolving. For
instance, in the 1860s and 1870s James Clerk Maxwell showed that the electric and
magnetic forces could be explained as manifestations of a single electromagnetic
force. In the 1970s, Sheldon Glashow, Abdus Salam and Steven Weinberg presented
the theory that explains how the electromagnetic and the weak nuclear force are
related to each other in a force called the electroweak force.
Newton’s First Law of Motion

During the seventeenth century, Isaac Newton, starting with the work of Galileo,
developed three important laws that deal with force and mass. Collectively they are
called “Newton’s laws of motion” and they provide the basis for understanding the
effect that forces have on an object.

To gain some insight into Newton’s first law, think about the game of ice hockey. If
the player does not hit a stationary puck, it will remain at rest on the ice. After the
puck is struck, however, it coasts on its own across the ice, slowing down slightly
because of friction. Since the ice is very slippery, there is only a relatively small
amount of friction to slow down the puck. In fact, if it were possible to remove all
friction and wind resistance, and if the rink were infinitely large, the puck would
coast forever in a straight line at a constant speed. Left on its own, the puck would
not lose any of the velocity imparted to it when it was struck. This the essence of
Newton’s First Law of Motion. We can state this law as follows.

An object continues in a state of rest or in a state of motion at a constant
speed along a straight line, unless compelled to change that state by a net
force.

The first law uses the phrase “net force.” Often, several forces act simultaneously on
a body, and the net force is the vector sum of all the forces. We will study later how
to add together such forces. Individual forces matter only to the extent that they
contribute to the total force. For instance, if friction and other opposing forces were
absent, a car could travel forever at 50 km/h in a straight line, without using any gas
after it had come up to a speed. In reality, gas is needed, but only so that the engine
can produce the necessary force to cancel opposing forces such as friction. The
cancellation ensures that there is no net force to change the state of motion of the
car.

There are several implications about Newton’s First Law that must be understood.

      An external force is required to change the velocity of an object. Internal
       forces have no effect on the object’s motion. For example, the driver pushing
       on the dashboard does not cause the car’s velocity to change.
      The external forces must be unbalanced. Two equal opposing forces acting on
       an object will not change its velocity. For an object’s velocity to change, the
       vector sum of the applied forces must be different than zero.
      Objects remain at rest unless acted upon by an external unbalanced force.
      Moving objects continue to move in a straight line at a constant speed unless
       acted upon by an external unbalanced force.

The ability of an object to resist changes in its state of motion is a fundamental
property of all matter. Newton called this property “inertia.” Because of inertia, a
moving object tends to remain at rest or in motion along a straight line at a constant
speed. Newton’s First Law can be referred to as the law of inertia.
Newton’s Second Law of Motion

The diagram below shows three situations. The same force is applied to three
different masses on a horizontal frictionless surface. In the first situation, a force F is
applied to a mass m, and the mass accelerates at a value of a. In the second
situation, the same force F is applied to twice the mass, 2m. This results is an
                1
                  a
acceleration of 2 . In the third situation, the force F is applied to three times the
                                           1
                                             a
mass, 3m, resulting in an acceleration of 3 .




From this series of drawings, we can conclude that when the force is constant, that
                                                                       1
                                                                  a
the acceleration of the mass is inversely related to the mass:         m.

The diagram below shows three more situations. This time the mass is held constant.
The force is varied and the resulting acceleration is determined.




From this series of drawings we can see that the acceleration is directly proportional
for the force applied: a a F.

By combining the two relationships, the relationship between acceleration, force, and
                              F
                         a
mass can be written as        m . To produce an equation, a proportionality constant, k,
is inserted, so that

      F
ak
      m or kF = ma. In SI units, mass is measured in kilograms, and acceleration in
metres per second. The SI unit of force that is used in physics is the force needed to
     cause a mass of 1 kg to have an acceleration of 1 m/s2. This amount of force is
     called the Newton (N). If we substitute these values into the equation, we get
                                      kg  m / s2
                                 k 1
     k(1N )  (1kg )(1m/ s2 ) or          N       . Using these units for F, m, and a, the value
     of k will always be 1. Thus, Newton’s Second Law of Motion can be
     mathematically stated as F = ma. Force is a vector, and the direction of the
     acceleration is in the same direction as the force.

     Newton’s Second Law of Motion and Kinematics Equations

     In the previous unit on kinematics, we learned that we could use the following three
     equations in solving problems dealing with motion.

                        x2  x1 x                     v2  v1 v
                                                                                          v  v 
              v ave                         aave                                     1
                        t2  t1 t                     t2  t1  t           v ave 
                                                                                        2 1 2

     We can now combine the use of these three equations with Newton’s Second Law, F
     = ma, in solving problems.

     For example, an object of mass 2.00 kg is moving at a constant speed of 4.00 m/s.
     It then experiences a force of 10.0 N for a time of 5.00 s. What is the displacement
     of the object?

     The information given in the problem is

     v1 = 4.00 m/s              m = 2.00 kg               F = 10.0 N             t = 5.00 s

     We are looking for the displacement of the object: Dx.

     First we can calculate the acceleration of the object.

          F  10.0N
     a
          m 2.00kg = 5.00 m/s2

     We can now use this value for acceleration to determine the new velocity of the
     mass.

     v2 = v1 + aDt = 4.00 m/s + (5.00 m/s2)(5.00 s) = 29.0 m/s


                                                    4.00m / s  29.0m/ s
                                    1             1
                             v ave  (v1  v2 )
     The average velocity is        2           = 2                        = 16.5 m/s

     Now we can determine the displacement.

     x  vavet   = (16.5 m/s)(5.00 s) = 82.5 m

     Exercise
1. For each of the following situations, which of the four fundamental forces is
acting?

a) In radioactivity, a proton may change into a neutron.

b) An electron will repel another electron.

c) The protons in a nucleus are held together so that they do not push each other
apart.

d) An attractive force exist between two neutrons that are not charged, but are
attracted because they have mass.

2. A person with a black belt in karate has a fist that has a mass of 0.70 kg. Starting
from rest, this fist attains a velocity of 8.0 m/s in 0.15 s. What is the average force
applied to the fist to achieve this level of performance?

3. A bicycle has a mass of 13.1 kg, and the rider has a mass of 81.7 kg. The rider is
pumping hard, so that a horizontal net force of 9.78 N accelerates them. What is the
acceleration?

4. A catapult on an aircraft carrier is capable of accelerating a plane from 0.0 to 56.0
m/s in a distance of 80.0 m. What is the average force that the catapult exerts on a
13 300 kg jet?




Lesson 6 - Adding Force Vectors by Components

Introduction

Newton’s Second Law is a vector relationship. It states that an object accelerates in
the direction of the net force acting on it. If there are two or more forces acting on
the object, we must be able to find the sum of these forces to obtain the “net” force.
Since forces are vectors, we must learn how to add vectors. In this lesson we will
review how to add vectors using what is known as the “component” method. The
component method involves the use of trigonometry. We begin with a basic review of
trigonometry. This will be followed by how to find the vector and scalar components
of a vector, and this will eventually lead to a method of adding vectors by using their
components.

Trigonometry: The Basic Trig Functions

Before we can learn how to add force vectors using a trigonometric approach, it is
necessary to quickly review the basics of trigonometry. There are three basic
trigonometric functions that are often used to solve physics problems. These
functions are the sine, the cosine, and the tangent of an angle q (the Greek letter
theta). They are abbreviated as sinq, cosq, and tanqrespectively. Look at the
triangle below.




                       h: hypotenuse
                                                  ho: length of side opposite the angle q
                                              o
                             q           90

           ha: length of side adjacent to the angle q




Looking at the diagram above, the three trigonometric functions can be defined as




                  ho                        ha                        ho
        sin                      cos                     tan 
                  h                         h                         ha

h = length of hypotenuse of a right triangle

ho = length of side opposite the angle q

ha = length of side adjacent to the angle q

The sine, cosine, and tangent of an angle are numbers with no units. These are
scalars, not vectors

Trigonometry: The Inverse Trig Functions and The Pythagorean Theorem

There are other situations where two sides of a right triangle may be known but the
value of q is not known, In such situations, the concept of inverse trigonometric
 functions is useful. The following equations for the inverse of sine, cosine, and
tangent are given in terms of the symbols used in the original drawing of the right
triangle.




                   ho                   ha                    
                                                                       h 
          sin1                 cos1                 tan1 o 
                    h                    h                    ha 

The symbol “-1” means the inverse trigonometric function. It does not mean “take
the reciprocal.”
For example, if the opposite side of a triangle is 2.0 m long and the adjacent side is
10.0 m long, then the angle involved is


                                                2.0 m
                               q
                           10.0 m




                    2.0m 
           tan1 
                   10.0m 
                              = 11.3o or 11o.




Finally, if two sides of a triangle are known, and it is necessary to find the third side,
then you can use the Pythagorean Theorem.




        h2 = ho2 + ha2




For the triangle above, the length of the hypotenuse would be




                           2                    2
         h =     (2.0 m)           + (10.0 m)       = 10.2 m or 10. m.


Vector Components of a Vector

In earlier lessons we have seen examples of vectors: displacement, velocity,
acceleration, and force. For example, suppose a force vector has a magnitude of
20.0 N at an angle of 60.0o north of east of your starting point. We can call this
force vector F. We could draw a right triangle as shown below so that one side of
the triangle point east and the other north. By using the cosine function we could
determine that the adjacent side is a vector 10.0 N east and the opposite side is a
vector 17.3 N north. We can call the 10.0 N E force vector the x component of the
vector F. Similarly, we can call the 17.3 m N displacement vector the y component of
the vector F.
F: 20.0 N

    o
60.0 N of E




                                               Fy: 17.3 N, N




Fx: 10.0 N, E




In physics, understanding vector components is very important. You can see that the
vector sum of the components is equal to the original vector.




        Fx + Fy = F




There are two characteristics about vector components that are very important.

1. When the components of a vector are added using vector addition, they have the
same meaning as the original vector. As you will see later, in doing calculations it
may be convenient to use the components of a vector instead of the vector itself.

2. The vectors Fx and Fy are not just any two vectors that add together to give the
vector F. They are perpendicular vectors. Because they are perpendicular, the trig
functions discussed earlier can be used.

We can write a formal definition of vector components as follows.




In two dimensions, the vector components of any vector F are two perpendicular
vectors Fx and Fy that are parallel to the x and y axes respectively, and add
together as vectors so that

F = Fx + Fy.
Scalar Components of a Vector

When working with vectors, it is often easier to work with what are known as “scalar
components.” Vector components are properly written in bold face such as Fx +
Fy since they have magnitude and direction as do all vectors. Scalars are not vectors,
so scalar components can be written as Fx and Fy.




The scalar component Fx has a magnitude that is equal to Fx. It is given a positive
sign if Fx points along the + x axis, or to the right. It is given a negative sign if
Fx points along the - x axis, or to the left.




Similarly, the scalar component Fy has a magnitude that is equal to Fy. It is given a
positive sign if Fy points along the + y axis, or up. It is given a negative sign if
Fy points along the - y axis, or down.




The following table compares vector and scalar components for the vector shown
below. The following table compares vector and scalar components for the vector
shown below.




                            h: 20.0 m

                            60.0o N of W

y: 17.3 m N




                     x: 20.0 m W




        Vector Components               Scalar Components




        hx = 20.0 m W (-x axis)                  hx = -20.0 m

        hy = 17.3 m N (+y axis)                  hy = +17.3 m
As we work through the remainder of this course, when we use the term
“component,” this will always mean the scalar component unless otherwise stated.

Resolving A Vector Into Components

When the magnitude and the direction of a vector are known, it is possible to find
the components of the vector using trigonometry. The process of finding these
components is called “resolving a vector into its components.” When we do this, we
are always looking at the perpendicular components that we can visualize as the two
sides of the right triangle formed by the vector. We call these components the x
component and the y component.




Suppose you want to find the components of a 5.0 N vector at an angle of 60.0 o N of
E.




                                                  a

                                    5.0 N             y component



                                              o
                                       60.0

                                    x component




The y component of this vector is the side opposite to the angle. We therefore use
the sine function to find its length.

sin q = ho/h or in this case sin 60.0o = y/5.0              y component = (5.0)(sin
60.0o) = 4.3 N




The x component is found using the cosine function.

cos q = ha/h or cos 60.0o = x/5.0                          x component = (5.0)(cos
60.0o) = 2.5 N
Another method is to use the other acute angle in the triangle. This is the angle
a shown above. Since the angles of a right triangle add up to 90o, the value of this
angle is 90.0 - 60.0 = 30.0o. Using this angle, the components as shown above can
be calculated as follows.




y component = (5.0)(cos 30.0o) = 4.3 N                   x component = (5.0)(sin
30.0o) = 2.5 N




You can see that the values of the components are the same. Either acute angle can
be used to determine the components of a vector. Choose whatever angle is
convenient in the question.

A good way to check if your answers are correct is to use the Pythagorean theorem.
This is because the vector components and the original vector form a right triangle.
For this question, the hypotenuse should be

                 2                2
h =    (4.3 N)        + (2.5 N)       = 5.0 N.




Vectors with Zero Components and Vectors That Are Equal

It is possible for vectors to have one of the components equal to zero. In the case of
a vector that is vertical, there is no x component. In this case, the y component of
the vector is itself equal to the whole vector.




                     Vector equivalent to 1.0 N north.

                     y component of vector is +1.0 N.

                     x component of vector is 0.0 N.




For a vector which is pointing in a horizontal direction, there is no y component. The
x component of the vector is itself equal to the whole vector.

                     Vector equivalent to 1.0 N west.

                     y component of vector is 0.0 N

                     x component of vector is -1.0 N.
 For a vector to be zero, both the x component and the y component of a vector
must be zero. So if a vector A = 0, then Ax = 0 and Ay = 0. This may seem trivial,
but in later studies, we will find that the sum of the forces acting on an object may
be zero, so there is no net force. The object is said to be at equilibrium.

For two vectors to be equal, they must have the same magnitude and direction. Two
vectors A and B are equal only if the components of the vector are also equal, that is
Ax = Bx and Ay = By.


Vector Addition Using Components: The Method

Adding vectors using components is a very accurate way of adding vectors. To see
how this is done, look at the vectors A and B below. The x and y components of
these vectors are also shown.
                                                                 Vector B
                   Vector A
                                                                     Bx

              2.00 N
                              Ay                         o
                                                   40.0 S of E                   By
                       Ax
                                                                 4.00 N
               o
          20.0 N of E




To obtain the resultant sum, we would add together the components of the vector.
The sum of the x components would give the x component of the resultant vector:
Ax + Bx = Cx. Also, the sum of the y components would give the y component of the
resultant vector: Ay + By = Cy.




                                                                     Cx

                                   Ax               Bx

                                           q

                                                                     By     Cy

                                               C

                                                                      Ay
Using scalar notation, we can write

        Ax + Bx = Cx               and    Ay + By = Cy




Since Cx and Cy form the sides of a right triangle, it is possible to determine the
length of the hypotenuse using the Pythagorean theorem:

                     2
        C =     Cx       + Cy 2




It is also possible to determine the value of the angle q using

                        y
                        C     
         = tan   -1
                       
                       C    
                              
                        x   




For the diagram above, this is the angle between the resultant and the x axis.




Vector Addition Using Components: The Calculation

Now let’s add together the two vectors A and B described above.


                                                                       Vector B
                          Vector A
                                                                           Bx
                  2.00 N
                                         Ay
                                                             o
                                   Ax                    40.0 S of E              By

                  o                                                    4.00 N
              20.0 N of E




First find the x components and the y components of each vector. Then add them
together, as shown in the table below.
Vector         x component                                                      y component




A                  Ax = (2.00 N) cos 20.0o = 1.88 N                           Ay = (2.00 N) sin 20.0o
= 0.68 N




B              Bx = (4.00 N) cos 40.0o = 3.06 N                               By = -(4.00 N) sin
    o
40.0 = -2.57 N




C                  Cx = Ax + Bx = 4.94 N                                       Cy = -1.89 N




To determine the length of the resultant, use




                    2
         C =   Cx         + Cy 2                     2                 2
                                     =    (4.94 N)       + (-1.89 N)       = 5.29 N




The angle q is found using




                         y
                         C              1.89 N 
          = tan                tan1 
                   -1
                        C    
                         x    =       4.94 N 
                                                      = 20.9o S of E.




Notice that to determine the value of the angle, only positive numbers are used in
the inverse tangent function. Once the value of the angle is determined, its direction
in reference to the points of a compass is stated. In the example above, the positive
x component and the negative y component place the vector below the x axis on the
positive side of the axis. Thus the direction is south of east.


Reasoning Strategy for Adding Vectors Using Components

The following is a summary of the reasoning strategy you should use to add vectors
using the component method.
1. First you need to determine the x and y components for each of the vectors you
are adding together. Always make sure to use positive or negative signs on the
components to show if they are pointing along the positive or negative axes.

2. Add together algebraically all of the x components to obtain the total x component
of the resultant. Also add together the y components to determine the total y
component of the resultant.

3. Use the Pythagorean theorem to determine the size or magnitude of the resultant.
Do this using the x components and the y components of the resultant.

4. Use one of the inverse functions to find the angle that the resultant makes with a
horizontal or vertical reference line. You could use either the inverse sine, the
inverse cosine, or the inverse tangent since all sides of the triangle are known. But it
is probably best to use the inverse tangent function once the x component and the y
component of the resultant have been determined.




There are three basic trigonometric functions that are often used to solve physics
problems.



                       h: hypotenuse
                                                 ho: length of side opposite the angle q
                                             o
                            q           90

           ha: length of side adjacent to the angle q




The three trigonometric functions can be defined as


                  ho                        ha                       ho
        sin                      cos                    tan 
                  h                         h                        ha



h = length of hypotenuse of a right triangle

ho = length of side opposite the angle q

ha = length of side adjacent to the angle q




The inverse trigonometric functions are
                  ho                 ha                h 
          sin1               cos1             tan1 o 
                   h                  h                ha 



The Pythagorean Theorem can be used to calculate the sides of a right triangle.




        h2 = ho2 + ha2




In two dimensions, the vector components of any vector F are two perpendicular
vectors Fx and Fy that are parallel to the x and y axes respectively, and add
together as vectors so that

F = Fx + Fy.




There are two characteristics about vector components that are very important.




1. When the components of a vector are added using vector addition, they have the
same meaning as the original vector. As you will see later, in doing calculations it
may be convenient to use the components of a vector instead of the vector itself.

2. The vectors Fx and Fy are not just any two vectors that add together to give the
vector F. They are perpendicular vectors. Because they are perpendicular, the trig
functions discussed earlier can be used.

The scalar component Fx has a magnitude that is equal to Fx. It is given a positive
sign if Fx points along the + x axis, or to the right. It is given a negative sign if
Fx points along the - x axis, or to the left.

Similarly, the scalar component Fy has a magnitude that is equal to Fy. It is given a
positive sign if Fy points along the + y axis, or up. It is given a negative sign if
Fy points along the - y axis, or down.

The process of finding the components of a vector is called “resolving a vector into
its components.” When we do this, we are always looking at the perpendicular
components that we can visualize as the two sides of the right triangle formed by the
vector. We call these components the x component and the y component.
It is possible for vectors to have one of the components equal to zero. In the case of
a vector that is vertical, there is no x component. In this case, the y component of
the vector is itself equal to the whole vector.

For a vector which is pointing in a horizontal direction, there is no y component. The
x component of the vector is itself equal to the whole vector.

For a vector to be zero, both the x component and the y component of a vector must
be zero.

For two vectors to be equal, they must have the same magnitude and direction. Two
vectors A and B are equal only if the components of the vector are also equal, that is
Ax = Bx and Ay = By.

The following is a summary of the reasoning strategy you should use to add vectors
using the component method.

1. First you need to determine the x and y components for each of the vectors you
are adding together. Always make sure to use positive or negative signs on the
components to show if they are pointing along the positive or negative axes.

2. Add together algebraically all of the x components to obtain the total x component
of the resultant. Also add together the y components to determine the total y
component of the resultant.

3. Use the Pythagorean theorem to determine the size or magnitude of the resultant.
Do this using the x components and the y components of the resultant.

4. Use one of the inverse functions to find the angle that the resultant makes with a
horizontal or vertical reference line. You could use either the inverse sine, the
inverse cosine, or the inverse tangent since all sides of the triangle are known. But it
is probably best to use the inverse tangent function once the x component and the y
component of the resultant have been determined.

Exercise

1. Newton’s second law indicates that when a net force acts on an object, it must
accelerate. Does this mean that when two or more forces are applied to an object
simultaneously, it must accelerate? Explain.

2. Two children are pulling on a wagon of mass 50.0 kg. One child is pulling to the
right with a force of 200.0 N while another child is pulling to the left with a force of
magnitude 150.0 N. What is the net force and what is the magnitude and direction of
the acceleration of the wagon?

3. A person is stranded on a raft. The mass of the raft and the person is 1500 kg. By
paddling, the man exerts an average force of 25 N in a direction due east (the +x
direction). The wind also exerts a force on the raft. The force of the wind has a
magnitude of 15 N and points north. Determine the magnitude and direction of the
acceleration of the raft and person.
4. Two children are trying to pull a 20.0 kg toboggan out of a deep snow drift that
provides an opposing force of 8.0 N. They are using ropes attached to the toboggan
and parallel to the ground. The forces exerted by the children and the direction in
which they pull on the ropes are shown below. Determine the net force and the
magnitude and acceleration of the mass.




                                             20.0 N



                  8.0 N          30.0o

                                     20.0o

                                          15.0 N




Lesson 7 - Free-body Diagrams

Introduction

When we do physics problems involving several forces acting on an object, it is
important to understand exactly which forces are acting and in what directions they
are acting. To help in this understanding, it is helpful to draw diagrams called “free-
body diagrams.” The focus of this lesson will be to learn how to draw these
diagrams, and later to use this skill in problem solving.

Free-Body Diagrams Defined and Strategy for Drawing

A free-body diagram is a diagram that represents an object and all the forces
acting on it.

In drawing a free-body diagram, the following strategy should be used.

1. Draw a sketch of the object completely removed from its physical surroundings.
Often the force vectors are drawn from a dot at the centre of the object.

2. Draw vectors representing the forces acting on the object. The vectors must show
the direction of the force. It is also useful to approximate the magnitude of the force,
so the length of the vector is important. But in first drawing the vectors, it is often
not clear what the length should be so an approximate length can be shown.

3. Do not include forces exerted on the surroundings by the object.

4. When finding the net force acting on a object from a free-body diagram, use the
rules for vector addition.
We will now examine several situations in which free-body diagrams will be drawn.

Free-Body Diagram for Falling Objects

1. Free fall in a vacuum.

In this case, there is only one force vector. That is the force due to gravity. The force
vector is drawn straight down.

The force of gravity arises because of a gravitational attraction between the earth
and the ball. The earth and the ball are not actually in contact. This is an example of
a “non-contact force.”

The non-contact force can be defined as a force between two objects that arises
when the objects are not in contact with each other. Other examples of non-contact
forces are the electrical forces and the magnetic forces to be studied later.



                                  Fg




2. Free fall with air friction.

The force vector representing gravity is still present. There is also now the vector
presenting air friction pointing straight up. Friction is an example of a “contact
force.” A contact force is defined as a force that arises when two objects are in
contact. One example of a contact force is the force of friction. Friction can be
defined as a force opposing the relative motion of two objects that are in contact. In
this case, friction arises because of air pushing against the ball as it falls.




                                  Ff



                                  Fg




Free-Body Diagram for Objects on a Flat Surface
3. An object resting on top of a table.

When an object such as a book rests on a tabletop, the table pushes back on the
book with an equal and opposite force (Newton’s third law). The force that the table
exerts on the book is referred to as a “normal force.” A normal force is a force that
acts perpendicular to the surface of contact. In this situation, the book is at rest. This
is an example of an equilibrium situation.




                                             FN




                                             Fg



4. An object on a frictionless table being pulled to the right.

When the book mentioned earlier is pulled to the right, the gravitational force and
the normal force are still acting. The pull to the right is an applied force (F a).




                                            FN



                                                   Fa



                                            Fg




5. An object pulled to the right on a surface with friction. Assume that the pull to the
right is larger than the force of friction.
For the book described above, the force of friction is opposite to the applied force,
and is therefore to the left. This friction vector is shorter than the applied force
vector.
                                              FN



                                  Ff                      Fa



                                               Fg




6. An object pulled to the right with an applied force at an angle to the object.

When the applied force is at an angle above the horizontal surface, the normal force
is decreased. This effect will be discussed further later.




                                          FN
                                                         Fa
                                  Ff



                                               Fg




Free-Body Diagram for Objects on an Inclined Surface

7. An object on a frictionless incline.

In this case, there are only two forces acting on the object. Gravity acts on the
object downwards. The normal force is perpendicular to the surface of contact. Note
that we continue our convention of drawing the normal forces from one point, in this
case the centre of the box.




                                                    FN




                                                Fg
8. An object on an incline with friction.

The forces in the previous diagram are still at work. If the object is sliding down the
incline, then the force of friction is opposite to the direction of motion. The force of
friction is therefore up the incline and parallel to it.




                                                 FN

                                     Ff




                                               Fg
Exercise

1. A steel ball is falling through water. Assume that the gravitational force is larger
that the frictional force of the water.

2. A rectangular box is pulled to the left with a horizontal force. There is friction
between the box and the flat surface. The force to the left is larger than the frictional
force.

3. Redraw the situation for the box in question two. But now the applied force
pushing down and to the right on the top left corner of the box. This is like pushing a
lawn mower or a shopping cart.

4. An inclined surface is slanting down and to the right. An object is sliding down the
incline. There is friction between the object and the surface of the incline.

								
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