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MCMASTER – MOHAWK JOINT VENTURE
BACHELOR OF TECHNOLOGY PARTNERSHIP
FOUR-YEAR
UNIVERSITY DEGREE PROGRAMS
EXPERIMENT NO:
TITLE:
Submitted by:
Lab Section:
Partner:
Instructor:
Date lab performed:
Date of submission:
EXPERIMENT NO. 2
FORCE AND ACCELERATION
ENG TECH 1PH3
Page
1
PURPOSE:
To study the acceleration of a free falling body to determine g, the acceleration due to gravity,
and to study the effect of changing the mass and the force on acceleration.
APPARATUS:
Lab cart (Hall's carriage), Spark timing device, Power supply, Spark sensitive paper, Tape, Meter
stick, Weights, Clamps, Thread, Pulleys, Retort stand, Spring balance, Balance.
THEORY:
An object falling in a gravitational field picks up speed as it falls. This implies that its velocity
changes (in this case, increases) with time. The rate of change of velocity with time is called
acceleration. This may be written as:
Δ
v v
−
v
a
= =
0
t
t
where: a
= acceleration (m/s
2
)
v
= final velocity (m/s)
v
0
= initial velocity (m/s)
t = time (s)
For a period of constant acceleration the displacement, x, of a moving object (distance object
travels) is the average velocity times the time period for the motion. The average velocity is the
sum of starting velocity plus final velocity all divided by 2.
⎛
v
+
v
0
⎞
x
=
v
× =
t
⎜
⎟
×
t
⎝
2
⎠
From equation (1) it is possible to manipulate the equation to solve for v.
v
=
a
×
t
+
v
0
(1)
If we substitute this value for v into the equation that defines x, the displacement can be
expressed as:
⎛
v
+
v
0
⎞
x
=
⎜
⎟
×
t
⎝
2
⎠
⎛
a
×
t
+
v
0
+
v
0
⎞
x
=⎜
⎟×
t
⎝
2
⎠
⎛
a
×
t
+
2
v
0
⎞
x
=⎜
⎟×
t
⎝
2
⎠
a
×
t
2
+
2
v
0
×
t
x
=
2
ENG TECH 1PH3
Page
2
x
=
a
×
t
2
+
v
0
×
t
x
=
at
2
+
v
0
t
x
α
t
2
(2)
The displacement is proportional to t
2
, when v
0
is zero.
The acceleration of a free falling object is commonly referred to as the acceleration due to
gravity, g. This value is approximately constant for the earth and is 9.81 m/sec
2
. This acceleration
is a constant value because it depends on the mass and radius of the earth and does not depend on
the mass of the object. A force may be defined as a push or pull that changes or tends to change
the motion of an object. A change in motion is another term for acceleration. Newton's 2
nd
law
describes the relationship between force, mass and acceleration. It is a general description of the
effect an unbalanced force will have on an object, and may be written as:
force = mass
×
acceleration
or, more simply:
f = m
×
a = ma
where:
f = force acting on an object causing it to accelerate (Newtons, N)
m = mass of the object undergoing acceleration (kilograms, kg)
a = acceleration of the object (m/s
2
)
Newton's law of universal gravitation describes a particular force that exists because of mass. It is
an attractive force that pulls together two objects of masses
M
1
and M
2
.
This attraction decreases with the square of the distance between the masses. The universal
gravitational constant, G, compensates for our earth-based metric system. The force of gravity, f
g
is defined by:
G
×
M
1
×
M
2
GM
1
M
2
f
g
=
2
=
2
r
r
where:
f
g
= gravitational force (N)
M
1
= mass of first object (kg)
M
2
= mass of second object (kg)
r
=
distance between objects (m)
G
= universal gravitational constant
=
6.67×10
-11
(N-m
2
/kg
2
)
Now if we consider the gravity between the earth (mass M
e
) and an object of mass m that is
close to the surface of the earth, the distance between the object and the centre of the earth is
essentially the radius of the earth,
r
e
.
The force of gravity, f
g
, is:
G
×
M
e
×
m
GM
1
M
2
f
g
=
2
=
2
r
e
r
But force is also defined by:
f
=
m
×
a
ENG TECH 1PH3
Page
3
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Since f must equal f
g
, we can equate the equations to get:
Dividing by m we get
G
×
M
×
m
m
×
a
=
2
e
r
e
G
×
M a
=
2
e
r
e
G,
M
e
, the mass of the earth and
r
e
,
the mean radius of the earth, are all constant. The
acceleration caused by gravity (close to the surface of the earth) is also constant and is equal to
9.81 m/s
2
. If objects of different masses fall at different rates, such as a nail and a feather, it is
because the force of air friction acting on the feather is greater than the air friction acting on the
nail.
In a vacuum, the feather and the nail fall at the same rate (see
Figure 2.1).
In this
experiment, the student will determine the acceleration of gravity at the surface of the earth, g,
and measure changes in acceleration as mass and the accelerating force are changed. In order to
measure acceleration the student will use a timing device that marks a moving paper tape at
regular time intervals (see
Figure 2.2
). The distance between the marks should be increasing if
the object is undergoing acceleration because although the time difference between when the
dots are made is the same, the actual distance between the dots is increasing with time as the
tape is be speeding up. In this particular case, the time between dots is l/60
th
of a second.
Figure 2.1
Figure 2.2
1
By measuring the distance from the first dot to consecutive dots we can calculate the velocity at
every newl/60
th
of a second interval. Knowing the change in velocity from a period of l/60
th
of a
second to the next allows the student to calculate the acceleration due to gravity.
Figure 2.3
shows the paper tape connected to a thread which is run through a series of pulleys in order to
allow the mass to fall from a greater height and permit sufficient points on the tape to make the
1 Nakamura Spark Timer, Sargent-Welch, www.sargentwelch.ca.
ENG TECH 1PH3
Page
4
necessary calculations. Different masses can be used in the experiment to show that for any mass,
the acceleration due to gravity is a constant.
Figure 2.3
Figure 2.4
The mass undergoing acceleration, shown in the set-up in
Figure 2.3
, can be varied. The
accelerating force, which is caused by gravity, is directly proportional to this mass. In the set-up
shown in
Figure 2.4,
it is possible to vary the mass of the object undergoing acceleration without
necessarily increasing the accelerating force. With a frictionless (low friction) lab cart tied to the
thread, the mass of the system being accelerated can be increased while the accelerating force
pulling the system at the other end of the thread stays the same. Or this force can be increased
while simultaneously increasing the mass on the cart. Knowing the mass of the entire system, cart
+ falling mass, and the gravity acting on the falling mass, Newton's second law concerning
acceleration can be verified by comparing the ratio
of f/m
with the measured acceleration on the
tape. This acceleration is determined by measuring the increasing space between dots on the tape.
How can the force causing the acceleration be measured? The measure of any unknown force is
best determined by comparing the unknown force to a known force. This is the principle of a
spring scale shown in
Figure 2.5
. The restoring force in the spring acts against the force of
gravity. We commonly refer to the force of gravity as the weight. When the weight is balanced by
the extension of the spring in a hanging scale or the compression of the spring in a bathroom
scale, the calibration of the spring indicates how much force is causing the extension or
compression.
Figure 2.6
shows the internal construction of a typical spring scale. Mass on the
other hand is measured by a balance, which compares a known mass to an unknown mass as
shown in
Figure 2.7.
Note that often mass and weight are used in an interchangeable manner such as using kilogram
for weight instead of the correct unit, which is of course the Newton. The mass of an object is a
constant value and does not change unlike the weight, which does vary as an object moves
further away from the earth.
ENG TECH 1PH3
Page
5
[Recall that weight is inversely proportional to distance squared.]
Figure 2.5
2
Figure 2.6
Figure 2.7
3
PART I: TO DETERMINE THE ACCELERATION DUE TO GRAVITY
PROCEDURE:
1.
Set up the apparatus as shown in
Figure 2.3
.
The tape should be attached to the thread
and the tape should pass between the clapper of the timing device and the sensitive side
of the tape must face upward.
There should be enough length of tape (1.5 m) to travel
the distance of the table.
2.
One student will start the timing device while the other student will allow a 100g mass
at the end of the thread to fall into a box.
Insure that no one holds on to the tape and
cause addition friction while it is in motion.
3.
Place the tape on the bench.
Insure the tape is flat and held securely at the ends with
masking tape.
2 Ohaus Pull Type Spring Scale, Ohaus Corporation 2009, www.ohaus.com.
3
Ohaus Model 311 Cent-O-Gram Balance Instruction Manual
, Ohaus Corporation, 2004, pg 2, www.ohaus.com.
ENG TECH 1PH3
Page
6
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4.
Label the first clear dot 1, count 1 dot and label it 2, and so on for at least 16 dots.
The
time interval between successive dots is 1/60
th
of a second.
5.
Use a meter stick to measure the distance between each consecutive dot to the nearest
millimetre.
Measure the distance from dot 1 to dot 2, x
1-2
.
Then measure the distance
from dot 1 to dot 3, x
1-3
.
Continue, until the 15
th
measurement x
1-16
.
Use the
Table 2.1
to record your measurements of the x's in the second column.
6.
Measure the falling mass,
m
, at the end of the thread with a balance and record in the
first column of the bottom row. Measure the weight of the falling mass using the spring
scale, and record in the second column of the bottom row.
Find the ratio of
f
g
/m
and
record in the bottom row
7.
Repeat steps 1 to 6 using a 500 g falling mass at the end of the thread.
Record this data
in
Table 2.2
.
Note:
In the data tables, f
g
in the bottom row of the second column, always refers to the
weight on the end of the string. This is the force that causes acceleration. The mass in
the bottom row of the first column always refers to the total mass of the system
(neglecting the string), the mass on the end of the string, the mass of cart (when used),
and the mass inside the cart (if used).
PART II: TO DEMONSTRATE NEWTON'S SECOND LAW PROCEDURE:
1.
Set up the apparatus as shown in
Figure 2.4
.
Do not place a block on the cart as is shown
in
Figure 2.4
. The tape should be attached to the back of the cart and the thread attached
to the front of the cart.
The tape should pass through the timing device with the correct
side upwards.
2.
One student will start the timing device while the other student will allow a 200 g mass at
the end of the thread to fall into a box and will attempt to catch the cart before it collides
with the retort stand.
Insure there are no obstacles in the carts path.
3.
Place the tape on the bench.
Insure the tape is flat and held securely at the ends with
masking tape.
4.
Label the first clear dot 1, count 3 dots and label it 2, and so on for at least 15 sets.
Record the distances between every 3rd dot (time incremented by 3/60
th
s) and record the
data in
Table 2.3
.
5.
Use a balance to determine the combined mass of the system,
m
sys
, consisting of the cart
and the falling mass and use a spring scale to determine the weight,
f
g
, of the falling mass,
ENG TECH 1PH3
Page
7
m
, at the end of the thread and record these values and the calculated ratio of
f
g
/m
(m =
200 g) in the bottom row of
Table 2.3
.
6.
Repeat the above steps 2 to 5 with the 200 g mass still at the end of the thread but, in
addition, place a block (300 g) on the cart and record the data in
Table 2.4
and determine
the total mass of the cart + block + falling mass on a balance use the same mass on the
end of the thread.
7.
Repeat the procedure but decrease the mass on the end of the thread to 50 g and leave the
300 g mass in the cart.
Record the data in
Table 2.5
.
RESULTS:
Table 2.1
DOT #
TOTAL
DISPLACEMENT
(mm)
TOTAL
TIME
(s)
TOTAL TIME
2
(3600
th
s)
VELOCITY
(m/s)
ACCELERATION
(m/s
2
)
1
x
1-2
= 9.5
1/60
1
v
1
=
2
x
1-3
= 24.5
2/60
4
v
2
=
a
1
=
3
x
1-4
= 43.5
3/60
9
v
3
=
a
2
=
4
x
1-5
= 70.5
4/60
16
v
4
=
a
3
=
5
x
1-6
= 104.0
5/60
25
v
5
=
a
4
=
6
x
1-7
= 143.5
6/60
36
v
6
=
a
5
=
7
x
1-8
= 190.0
7/60
47
v
7
=
a
6
=
8
x
1-9
= 243.5
8/60
64
v
8
=
a
7
=
9
x
1-10
= 301.8
9/60
81
v
9
=
a
8
=
10
x
1-11
= 369.0
10/60
100
v
10
=
a
9
=
11
x
1-12
= 442.0
11/60
121
v
11
=
a
10
=
12
x
1-13
= 513.0
12/60
144
v
12
=
a
11
=
13
x
1-14
= 583.5
13/60
169
v
13
=
a
12
=
14
x
1-15
= 651.0
14/60
196
v
14
=
a
13
=
15
x
1-16
= 719.0
15/60
225
v
15
=
a
14
=
falling mass (m) =
0.10022kg
Average accel
=
m/s
2
weight (f
g
)of
falling mass =
9.8
N
accel =
f
g
/m =
m/s
2
Table 2.2
DOT # DISPLACEMENT
(mm)
TIME
(s)
VELOCITY
(m/s)
ACCELERATION
(m/s
2
)
1
x
1-2
= 12
1/60
v
1
=
2
x
1-3
= 28.5
2/60
v
2
=
a
1
=
3
x
1-4
= 55.5
3/60
v
3
=
a
2
=
4
x
1-5
= 77
4/60
v
4
=
a
3
=
5
x
1-6
= 99.5
5/60
v
5
=
a
4
=
ENG TECH 1PH3
Page
8
6
x
1-7
= 124.5
6/60
v
6
=
a
5
=
7
x
1-8
= 151
7/60
v
7
=
a
6
=
8
x
1-9
= 182.5
8/60
v
8
=
a
7
=
9
x
1-10
= 215
9/60
v
9
=
a
8
=
10
x
1-11
= 250.5
10/60
v
10
=
a
9
=
11
x
1-12
= 270
11/60
v
11
=
a
10
=
12
x
1-13
= 330.5
12/60
v
12
=
a
11
=
13
x
1-14
= 374.5
13/60
v
13
=
a
12
=
14
x
1-15
= 420
14/60
v
14
=
a
13
=
15
x
1-16
= 459.5
15/60
v
15
=
a
14
=
falling mass (m) =
0.203kg
Average accel
=
m/s
2
weight (f
g
) of falling mass =
N
accel =
f
g
/m =
m/s
2
Table 2.3
DOT # DISPLACEMENT
(mm)
TIME
(s)
VELOCITY
(m/s)
ACCELERATION
(m/s
2
)
1
x
1-4
=
21
3/60
v
1
=
2
x
1-7
=
45.4
6/60
v
2
=
a
1
=
3
x
1-10
=
76.5
9/60
v
3
=
a
2
=
4
x
1-13
=
134
12/60
v
4
=
a
3
=
5
x
1-16
=
194
15/60
v
5
=
a
4
=
6
x
1-19
=
273.5
18/60
v
6
=
a
5
=
7
x
1-22
=
368
21/60
v
7
=
a
6
=
8
x
1-25
=
470
24/60
v
8
=
a
7
=
9
x
1-28
=
592
27/60
v
9
=
a
8
=
10
x
1-31
=
768.5
30/60
v
10
=
a
9
=
11
x
1-34
=
921.5
33/60
v
11
=
a
10
=
12
x
1-37
=
1087.5
36/60
v
12
=
a
11
=
13
x
1-40
=
1263.5
39/60
v
13
=
a
12
=
14
x
1-43
=
1450
42/60
v
14
=
a
13
=
15
x
1-46
=
1597
45/60
v
15
=
a
14
=
falling mass + cart (
m
sys
) = 0.303
kg
Average accel
=
m/s
2
weight (f
g
) of falling mass =
N
accel =
f
g
/
m
sys
=
m/s
2
Table 2.4
DOT # DISPLACEMENT
(mm)
TIME
(s)
VELOCITY
(m/s)
ACCELERATION
(m/s
2
)
1
x
1-4
=
16.5
3/60
v
1
=
2
x
1-7
=
37.5
6/60
v
2
=
a
1
=
3
x
1-10
=
59.5
9/60
v
3
=
a
2
=
4
x
1-13
=
91
12/60
v
4
=
a
3
=
5
x
1-16
=
124
15/60
v
5
=
a
4
=
ENG TECH 1PH3
Page
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6
x
1-19
=
162.5
18/60
v
6
=
a
5
=
7
x
1-22
=
210
21/60
v
7
=
a
6
=
8
x
1-25
=
261
24/60
v
8
=
a
7
=
9
x
1-28
=
316.5
27/60
v
9
=
a
8
=
10
x
1-31
=
380
30/60
v
10
=
a
9
=
11
x
1-34
=
447.5
33/60
v
11
=
a
10
=
12
x
1-37
=
520
36/60
v
12
=
a
11
=
13
x
1-40
=
593
39/60
v
13
=
a
12
=
14
x
1-43
=
665
42/60
v
14
=
a
13
=
15
x
1-46
=
735
45/60
v
15
=
a
14
=
falling mass + cart
+
mass on cart (m
sys
) =
0.602 kg
Average accel
=
m/s
2
weight (f
g
) of falling mass =
N
accel =
f
g
/
m
sys
=
m/s
2
Table 2.5
DOT # DISPLACEMENT
(mm)
TIME
(s)
VELOCITY
(m/s)
ACCELERATION
(m/s
2
)
1
x
1-4
=
16
3/60
v
1
=
2
x
1-7
=
34
6/60
v
2
=
a
1
=
3
x
1-10
=
53
9/60
v
3
=
a
2
=
4
x
1-13
=
74.5
12/60
v
4
=
a
3
=
5
x
1-16
=
97.5
15/60
v
5
=
a
4
=
6
x
1-19
=
120.5
18/60
v
6
=
a
5
=
7
x
1-22
=
147
21/60
v
7
=
a
6
=
8
x
1-25
=
173.8
24/60
v
8
=
a
7
=
9
x
1-28
=
203.5
27/60
v
9
=
a
8
=
10
x
1-31
=
233
30/60
v
10
=
a
9
=
11
x
1-34
=
264.5
33/60
v
11
=
a
10
=
12
x
1-37
=
298.5
36/60
v
12
=
a
11
=
13
x
1-40
=
333.5
39/60
v
13
=
a
12
=
14
x
1-43
=
369
42/60
v
14
=
a
13
=
15
x
1-46
=
407.5
45/60
v
15
=
a
14
=
falling mass + cart
+ mass on cart (m
sys
) =
0.553
kg
Average accel
=
m/s
2
weight (f
g
) of falling mass =
N
accel =
f
g
/
m
sys
=
m/s
2
CALCULATIONS
:
ENG TECH 1PH3
Page
10
1.
Plot a graph of the total displacement (column 2 in
Table 2.1)
vs. the total time (column 3 in
Table 2.1)
as follows with:
ordinate 1 = x
1-2
and abscissa 1 =
1/60 s ordinate 2 = x
1-3
and
abscissa 2 = 2/60 s ordinate 3
=
x
1-4
and abscissa
3 = 3/60 s
…
until the ordinate 15 = x
1-16
is plotted.
Use a scale that gives a full-page graph. Label this graph "Graph d vs. t". Prepare a similar
graph of d vs. t
2
(values in column 4 of
Table 2.1)
and label this graph "Graph d vs. t
2
".
2.
Compute the speed, v, at each successive point starting at dot 1 by dividing
twice
of x
1-2
by
l/60
th
second and so on until v
15
is calculated.
Remember to convert the distances between dots in mm to metres.
2
Δ
x
v
=
Δ
t
2
x
1 2
−
v
1
=
2
x
1 3
−
v
2
=
…
2
x
1 16
−
v
15
=
Record these values in column 5 of the Table 2.1.
3.
Prepare an average speed vs. time curve with
ordinate 1 = v
1
and abscissa 1 =
1/60 s ordinate 2 = v
2
and abscissa
2 = 2/60 s
ordinate 3 = v
3
and
abscissa 3 = 3/60 s
…
until all of the velocities in column 5 have been plotted.
Plot the average speed vs. time curve on a separate graph. Label this graph
"Graph 2.1"
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4.
From equation (1) in the theory section acceleration can be determined by:
Δ
v
a
=
Δ
t
v
2
−
v
1
a
1
=
v
3
−
v
2
a
2
=
…
v
15
−
v
14
a
14
=
Every Δv is divided by 1/60 s. Calculate the values for a
1
to a
14
and record in column 6
(Table
2.1)
or column 5
(Tables 2.2
to
2.5
). Calculate the average acceleration and record it in the first
row of
Table 2.6
. Record the ratio of f
g
/m (which appears in the bottom row of
Table 2.1)
in
Table 2.6
in the second row which is labelled f
g
/m. 5. Repeat the above steps 2 to 4 for
Tables
2.2
to
2.5
noting that there are fewer columns in T
ables 2.2
to
2.5.
So calculations from step 2
will be recorded in column 4
of Tables 2.2
to
2.5
and calculations from step 4 will be recorded in
column 5. Also note that for
Tables 2.3
to
2.5,
calculations are for every third dot and time
intervals are for every 3/60
th
seconds.
Label the graph (requested in step 3) for
Table 2.2 as
Graph 2.2,
and continue labelling the graphs for
Tables 2.3
to
2.5
accordingly. Again note that
the abscissa values for graphs for
Tables 2.3
to
2.5
are every 3/60
th
seconds.
It is not necessary to repeat step 1 which involves the total displacement vs. time and
displacement vs. time
2
for Tables 2.2 to 2.5
DISCUSSION:
Theory states that the acceleration of a falling object should be a constant. This acceleration is
the slope of the line of the velocity vs. time graph. When you plot a graph with measured data
from an experiment, there will always be errors in the measurements. The empirical data
collected may appear to be scattered when plotted on the graph paper. This makes it very difficult
ENG TECH 1PH3
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to choose the best straight line to connect the dots in order to measure the slope. Which straight
line best fits the data?
This problem can be easily solved using Microsoft Excel. The method is called linear regression
using the method of least squares (your instructor will explain this method if you are not familiar
with it). If the relationship between the ordinate (v) and abscissa values (t) is a straight line it is
said to be linear. For a period of constant acceleration:
v
α
t
or the velocity is proportional to time,
and:
v
=
a
×
t
when the initial velocity is zero.
Clearly, acceleration 'a' is the slope of the line:
v
a
=
t
The generic form of a straight line is:
y=m
×
x + b
where 'm' is the slope of the line and 'b' is the y-intercept.
How reliable is this result? There is a value known as the correlation coefficient (r). When this
value is close to ± 1, we can be confident that there is a strong linear relationship between the
data points and that value of the slope is reliable. This technique is valid only for linear
relationships. Velocity vs. time is a linear relationship as the data illustrates in
Graph 2.la.
In the
case of displacement vs. time
(Graph 2.1
) the relationship does not appear to be linear.
In the first row of
Table 2.6
, calculate the average value of the acceleration for each
of Tables
2.1
to
2.5
. In the second row for
Table 2.6
record the values of the ratio f
g
/m from each of the
five tables.
From the data for velocity and time for each of the five tables, determine the slope of
the line that best fits the data using the method described above and from that state the
acceleration. Record the values in the
Table 2.6,
in the row labelled slope. Also determine r, the
correlation coefficient. Record this in the bottom row of
Table 2.6.
Table 2.6
GRAPH
VELOCITY vs.. TIME
N.1
N.2
N.3
N.4
N.5
acceleration:
average value from
table
f
g
/m: from tables
bottom
row
slope: by linear
regression
r: correlation
coefficient
ENG TECH 1PH3
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DISCUSSION QUESTIONS
:
1)
Does the mass of a free falling body appear to be a contributory factor to the acceleration?
What is the acceleration in m/s
2
? If it is less than the accepted value, explain why.
2)
What is the effect of increasing the force on the acceleration of the cart?
3)
What is the effect on acceleration of increasing the mass of the cart without increasing the
force that is causing the acceleration?
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