2
NEWTON’S LAWS OF MOTION
Law
of
Inertia
explains
the
phenomenon where any object in motion will
continue to move with constant magnitude,
unless a net external force acts on it.
What happens when a car abruptly comes to a full stop in
the middle of an expressway? Due to the stop the
speeding SUV behind could not stop in time and crashed
into the front car. Smoke is seen and the rear of the car is
ripped into pieces due to the impact.
(Source: https://stomp.straitstimes.com)
The previous chapter introduces the language
and mathematics of kinematics to describe
motion in one, two or three dimensions. But
what causes objects to move, or why motion
occurs the way they do and similar questions
is answered in the
second division of
mechanics, the dynamics. The concepts
provide the relationships of motion to the
forces that causes it. Such concepts are
clearly stated by Sir Isaac Newton (1642 –
1727), summarized in his famous three laws
of motion. These laws became the
foundation of classical mechanics (also
called Newtonian mechanics).
The laws apply only to macro objects that do not move at extremely high speeds, not as fast
as the speed of light like minute particles do such as electrons. Apart from kinematics
variables two new physical quantities will be introduced here, the force and mass, to analyse
the principles of dynamics.
2.1 Force and Interactions
Force, ⃑ is commonly termed as “a push or a pull” which
may or may not cause motion. It is a quantitative measure of
the interaction between two bodies or between a body and its
environment. It is a vector quantity, thus with magnitude and
direction.
Contact force occurs between two surfaces that are directly in contact with each other.
Common to this are shown below:
Friction and Normal force
Tension force
Friction or frictional force ( ) is an opposing force acting parallel to the surface of contact.
Its direction is always resistive to motion.

Kinetic friction,
– opposing frictional force to motion when one body slides on the
surface of another body.
29
f k  k

Static friction,
relative motion.
–
(2-1)
frictional force between surfaces in contact which are not in
(2-2)
k
where:
s

– the coefficient of kinetic friction.
– the coefficient of static friction.
– normal force perpendicular to the interacting surfaces.
For most surface interfaces, the coefficient of kinetic friction is less than the coefficient of
static friction:
 k < s
where:
– the coefficient of static friction
– the coefficient of kinetic friction
– normal force perpendicular to the
interacting surfaces
(a)
(b)
Figure 2.1: (a) Static Friction, (b)
Kinetic Friction.
Centripetal force,
is a force “seeking the center”, without which circular motion cannot
occur and motion of object will be directed tangent to the curve. It acts to change the
direction of motion of moving object in a circle.

In a circular motion, the direction of the acceleration vector is directed towards the
center with magnitude of:
and,
(2-3)
Normal force () is exerted on an object by any surface with which it is in contact. It is the
supporting force which is directed perpendicular to the surface area of contact.
Tension force (T) is a pulling force which is transmitted through a string, rope, cable or wire
when it is pulled tight by forces acting from opposite ends. It is directed along the length of
force and pulls equally on the objects on the opposite ends of the string.
Long-range force or force at a distance is any force applied to an object by another body
that is not in direct contact with it. Some examples are weight, electrostatic force or magnetic
force.
30
Weight (w) is the gravitational force that the earth exerts on a body and is always
directed downward (towards the earth).
(2-4)
⃑⃑
The SI unit for weight is Newton.
Mass (m) characterizes the inertial properties of a body (the quantity of a body). It is
measured in terms of kilogram or kg.
Group of forces can also be classified relative to their line of actions. Concurrent forces are
set of two or more forces that pass through a common point. These are forces whose lines of
action intersect at a point at the same time. The principle of equilibrium is also used to
determine the resultant force of non-parallel, non-concurrent systems of forces. Nonconcurrent, non- parallel system of forces have lines of action of forces that do not meet
at one point. Refer to the diagram below:
Concurrent forces
The SI unit of the magnitude of force is kg.m/s2 equivalent to Newton, abbreviated as, N.
Superposition of Forces
When two or more forces acts at a common point (when lines of force are extended)
at the same time, the effect on the body’s motion is the same as if a single force (the
⃑⃗
⃗
⃗
⃗
resultant) were acting equal to the vector sum of the original forces: ∑ ⃗
More generally, any number of forces applied at a point on a body have the same effect as a
single force equal to the vector sum of the forces referred to as Principle of Superposition.
Free-Body Diagram (FBD) is a diagram isolating
the body from surroundings but showing all the
concurrent forces acting on only that object.
i.
With the aid of FBD, forces acting on a body
can be resolve into x and y-components.
ii.
For problems with several bodies involved,
FBD must be drawn separately for each body.
(a)
(b)
Figure 2.2: Examples of FBD. Block on
(a) a level surface with an applied force,
and on (b) an inclined surface.
31
2.2 Newton’s Three Laws of Motion (Sir Isaac Newton, 1687)
1. First Law (Law of Inertia): “An object at rest will remain at rest and an object in motion
will stay in motion with constant magnitude of velocity unless a NET external force acts on
it”.
First Condition of
Equilibrium
(2-5)
Inertia is the tendency of a body to continue moving once set in motion.
Sample Problem
1. A traffic light weighing 100 N hangs from a vertical cable tied to other two cables that are
fastened to a support, as shown in Figure 4.4. Find the tension in each of the three cables.
Solution (Refer to the FBD of the traffic light and the cables.)
The system is at rest, then we will apply First Equilibrium Condition, Fnet=0.
Traffic light: ∑
, gives
Cables: ∑
gives

(eq. 1)
Also, ∑
gives
 (eq.2)
Substituting eq.1 in eq.2:
(
)

Substituting the value for T2 in eq.1: (

)
2. Second Law (Law of Acceleration):”When a net external force, F net , acts on an object of
mass of mass m, the object accelerates in the direction of
”.
32
(2-6)
In component form:
⃑⃑
⃑
⃑⃑
⃑⃑
⃑
⃑
SI unit: kg.m/s2 = Newton, N
Sample Problems
1. A 5.0kg block in the figure is pulled westward across a
frictionless horizontal surface by a 30.0N force applied at
63º angle above the horizontal.
a) What is the block’s acceleration?
b) What is the normal force?
Solution
a) The acceleration can be calculated as:
F
x
 ma x
 Fx  ma
Fx  F cos   30 N cos 63


m
m
5kg
m
a  2.72 2
s
a
FBD
The negative sign indicates that the object is
moving to the left.
b)
F
y
 ma y
Fy    w  0
There is no component of a along
F sin     mg  0
  mg  F sin 
m
  (5kg )(9.8 2 )  (30 N )(sin 63)
s
  22.27 N
(
).
33
2. Figure 4.6 shows block A of mass 2.25 kg that rest on a
horizontal tabletop. It is connected by a horizontal cord
passing over a light, frictionless pulley to a hanging block B
of mass 1.15 kg. Assume that the surface is frictionless.
a) What is the acceleration of the block after the system is
released?
b) What is the tension in the cord as the system is moving?
c) If there exists friction between the two surfaces, what is the minimum coefficient of
static friction between the table and block A if the system remains at rest even if they
are released from rest?
Solution (Refer to the FBD of the load of bricks and the counterweight.)
a) The system will accelerate in the direction of
,
then we will apply Newton’s 2nd Law, Fnet=ma.
Along the axis of motion,
Block A: ∑
𝒎𝒐𝒕𝒊𝒐𝒏
, where
gives
𝒇
(eq.1)

Block B: ∑
(
𝒘𝑨
Equating eqs.1 and 2:

b) The tension in the cord (using Eq.1) is: (
)
(
)(
Equation=g Eq.3 and 4:
)
=
(
(
)
)
. Since the system does not move, we

gives
(Eq.3)
gives


)(
)
. Also, ∑
so,
𝒎𝒐𝒕𝒊𝒐𝒏
)
gives
Block B: ∑
(
)(
c) In the presence of static friction, where
will apply the Newton’s 1st Law.
Block A: ∑
𝒘𝑩
(eq.2)

)(
𝑻
)
gives
or (using Eq.2): (
𝑻
𝜼𝑨
(Eq.4)
=
3. A crate of mass 10 kg is released from rest on a frictionless plane inclined 30˚ from the
horizontal. The distance from the front edge of the crate to the bottom of the plane is 5.0
m.
a) Determine the acceleration of the crate after it is released from rest.
b) How long does it take for the crate to reach the bottom?
c) What is its speed just as it gets there?
FBD
34
Solution
a) ∑
∑
(
⁄ )(
)
⁄
b)
√
(
)
c)
(
√
(
(
√
⁄
)
)
)
√ (
(
⁄ )(
)
)
4. A small object with mass m is suspended from a length of 1 m. The object revolves with
constant speed
in a horizontal circle of radius r, as shown in the figure. Find an
expression for .
Solution
Free-Body Diagram (FBD)
From Newton’s first law:
Vertical: ∑
𝑇𝑐𝑜𝑠
𝑇𝑠𝑖𝑛
𝑤
(eq. 1)
Horizontal: ∑
(eq. 2)
Equating the two equations A and B, we can obtain the velocity
From the figure, we can find
. Hence,
√
√
.
.
From the result, the speed v is independent of the mass m of the object.
3.
Newton’s Third Law (Law of Interaction): “For every action
there is always an equal and opposite reaction.”
35
 If object 1 exerts a force on object 2,
, (the action) body 2 will in turn
exert a force on 1,
, equal in magnitude but opposite in direction.
Banked Curves
Banked Curves are curved roadbeds tilted inward, i.e. outer surface is elevated so surface of
the road is inclined.

On a level (horizontal) surface road,
is furnished by , between the tires and the
road.
The maximum speed any vehicle can navigate through the curve is given by the
relationship:
(2-7)
√

As a speeding car turns a curve and if the radius of curvature is small,
would be
large and
may not be enough that may lead to skidding, especially if the road is wet
and slippery.

To reduce the chance of skids, highway curves are often BANKED so roadbed tilts
inward.
(a)
(b)
Figure 2.3: (a) Car turning a curve, (b) The free body diagram.
(Source: http://www.clickandlearn.org)
From the FBD of the vehicle, summation of the forces along the x and y axis shows that
eq. (a),
and
eq. (b)
Dividing eq. (a) by (b) gives:
(2-8)
Sample Problem
1.
A car is rounding a flat, unbanked curve with radius of 150 m.
a) If the coefficient of static friction between the tires of the car and the road is 0.35,
what is the maximum speed at which the driver can take the curve without skidding?
b) At what angle should the curve be bank so that the car travelling with a speed in (a)
can safely navigate the curve in the absence of static friction?
36
Solution (Refer to the FBD of car for the unbanked and banked cases.)
a) The car moves around a circular path under the influence of a centripetal focre.
gives
∑
and
𝜽
𝜼
then

∑
𝜼 𝐜𝐨𝐬 𝜽
𝜼
𝜼 𝐬𝐢𝐧 𝜽
(
)(
)(
) 
𝒇𝒔
𝜽
𝒘

b) ∑
(eq.1)
( ⁄ )
Dividing Eq.1 with Eq.2:

(a)
(eq.2)

∑
𝒘
(
)
*
(
(b)
gives
(
)
)(
)
+
37
Exercise No. 2
1. A man is dragging a trunk up the loading ramp of a mover's
truck. The ramp has a slope angle of 20.0 o, and the man pulls
upward with a force
whose direction makes an angle of
30.0 with the ramp (Figure 4.9).
a) How large a force
is necessary for the component
parallel to the ramp to be 43.3N?
b) How large will the component
perpendicular to the ramp
then be?
2. A sled is tied to a tree on a frictionless, snow-covered hill, as
shown in the second figure in the right. If the sled weighs 80
N, find the
a) force exerted by the rope on the sled; and the
b) normal force exerted by the hill on the sled.
3. An athlete whose mass is 90.0 kg is performing weight-lifting
exercises. Starting from the rest position, he lifts, with
constant acceleration, a barbell that weighs 490 N. He lifts the
barbell a distance of 0.60 m in 1.6 s. Find the total force that
his feet exert on the ground as he lifts the barbell.
4. In a physics lab experiment, a 6.00kg box is pushed across a
flat table by a horizontal force.
a) If the box is moving at constant speed of 0.350 m/s and the
coefficient of kinetic friction is 0.12, what is the magnitude of the force?
b) What is the magnitude of the force if the box is speeding
up with a constant acceleration of 0.180m/s2?
5. A block of m1=20.0 kg is placed on a plane incline at angle
=37.1 and is connected to a second hanging block with mass
m2 by a cord passing over a small, frictionless pulley. The
coefficient of kinetic friction between the plane and the block is
0.30.
a) Find the mass m2 if it is to descend 12.0 m in the first 3.00 s
after the system is released from rest.
b) Find the mass m2 for which the block m1 accelerate down the
plane at a rate same as the magnitude in (a).
6. A small car with mass of 0.800 kg travels at a constant speed
on the inside of a track that is a vertical circle with radius 5.00
m. If the normal force exerted by the track on the car when it
is at the bottom of the track is 25 N,
a) What is the normal force on the car when it is at the (i)
top, (ii) rightmost and (iii) leftmost point of the track?
b) What is the speed of the car as it travels around the
vertical circle?
Figure 2.6
38
7. A man weighs a fish with a spring scale attached to the ceiling of an elevator. While
the elevator is at rest, he measures a weight of 49.0 N.
a) What weight does the scale read if the elevator accelerates upward at 2.00 m/s 2?
b) What weight does the scale read if the elevator accelerates downward at 2.00
m/s2?
8. A 2500 kg van accelerates down an icy driveway, going from
rest to 30.0 m/s in 6.00 s. During its motion, a 0.100 kg stuffed
toy hangs by a cord from the van’s ceiling. The van’s
acceleration is such that the cord remains perpendicular to the
ceiling. Determine the:
a) angle ; and the
b) tension in the cord.
9. In a “Giant Swing” shown in figure, the seat is connected to two
cables. One is from the top of the central shaft which makes an
angle of 40.0 with the vertical and the other cable of 7.50 m is
horizontal. The seat swings in a horizontal circle at a rate of 25.1
m/s. If the seat weighs 245 N and a 540 N person is sitting in it,
find the tension in each cable.
10. Consider a banked wet curved roadway, where there is a
coefficient of static friction of 0.30 and a coefficient of kinetic
friction of 0.25 between the tires and the roadway. The radius of
the curve is R = 50 m. If the banking angle is
a) What is the maximum speed the automobile can have before
sliding up the banking?
b) What is the minimum speed the automobile can have before
sliding down the banking?
Answers:
a) 69.3 N ; b) 34.6 N
a) 40.0 N ; b) 69.3 N
3. 1,395 N
4. a) 7.06N ; b) 8.14 N
5. a) 30.6 kg ; b) 8.97 kg
6. a) (i) 9.32 N (ii) 17.2 N (iii) 17.2 N
b) 10.4 m/s
7. a) 59 N ; b) 39 N
8. a) 30.7o
b) 0.843 N
9. a) 1,025 N
b) 6,088 N
10. a) 20.9 m/s
b) 8.46 m/s
1.
2.
39
40
2
NEWTON’S LAWS OF MOTION
PROBLEM SET
Name: ____________________________________ Section: _________________________
Instructor: _________________________________ Date Submitted: __________________
1. In the figure opposite, the weight w is 60.0 N.
a) What is the tension in the diagonal string?
b) Find the magnitudes of the horizontal forces
⃗ and ⃗ that must be applied to hold the system in
the position shown.
2. Two blocks with masses 4.00 kg and 8.00 kg are connected by a string and slide down a
30.00 inclined plane. The coefficient of kinetic friction between the
4.00 kg block and the plane is 0.25; that between the 8.00-kg block
and the plane is 0.35
a) Calculate the acceleration of each block.
b) Calculate the tension in the string.
c) What happens if the positions of the blocks are reversed, so the
4.00-kg block is above the 8.00-kg block?
41
3.
Blocks A, B, and C are placed as in the figure and connected by ropes of negligible mass.
Both A and B weigh 25.0 N each, and the coefficient of kinetic friction between each block
and the surface is 0.35. Block C descends with constant velocity.
a) Draw two separate free-body diagrams showing the forces acting on A and on B.
b) Find the tension in the rope connecting blocks A and B.
c) What is the weight of block C?
d) If the rope connecting A and B were cut, what would be the acceleration of C?
4. Block B, with mass 5.00 kg, rests on block A, with mass 8.00 kg,
which in turn is on a horizontal tabletop. There is no friction
between block A and the tabletop, but the coefficient of static
friction between block A and block B is 0.750. A light string
attached to block A passes over a frictionless, massless pulley, and
block C is suspended from the other end of the string. What is the
largest mass that block C can have so that blocks A and B still
slide together when the system is released from rest?
42
5. Two identical 15.0-kg balls, each 25.0 cm in diameter, are suspended by two
35.0-cm wires. The entire apparatus is supported by a single 18.0-cm wire,
and the surfaces of the balls are perfectly smooth.
a) Find the tension in each of the three wires.
b) How hard does each ball push on the other one?
43
44
LABORATORY ACTIVITY NO. 2
Newton’s Law of Acceleration: Friction
I. Objectives:
1. To determine the acceleration of an object experimentally and compare the
acceleration predicted by Newton’s Second Law of Motion:
.
2. To show how the acceleration of an object is dependent on force and mass.
3. To verify Newton’s Second Law of Motion,
.
4. To determine the coefficient of static friction between two surfaces
II. Introduction:
Newton’s second law of motion states that, “If a net external force is applied on a
body, the body accelerates”. The acceleration of the object is directly proportional to the net
force acting on the body and inversely to its mass. Mathematically:
⃗
⃗
∑⃗
(2.1)
⃗
The direction of the acceleration is the same as the direction of the net force.
In this experiment, the cart will be released from rest and allowed to accelerate over a
distance, . Using a stopwatch, you will determine how long it takes, on the average, for the
cart to move through a distance, . This experimental value of the cart’s acceleration can be
determined from equation 2.3.
d  vo t 
1 2
at ;
2
where vo  0
(2.2)
(Experimental Value)
(2.3)
so that ,
a
2d
t2
Assuming that the tabletop is truly horizontal (i.e. level), Newton’s Second Law
determines theoretically the acceleration of this system from equation 2.4:
a
where:
Fnet
M
or
m
a
M

g

(Theoretical Value)
(
)
(2.4)
M – is the total mass of the dynamics cart and its load
m–
is the hanging mass
One application of Newton’s 2nd law of motion is friction which is a force that resists the
motion of an object that is in contact with another object or material. The coefficient of friction
45
between two surfaces is a number that determines how much force is required to move an
object that is held back by friction when the two surfaces are pressed together.
Static friction
is related to the normal force,
by the coefficient of static friction,
:
(2.5)
Kinetic friction is the force involved when one body slides over the other surface of another
body while static friction is encountered when two surfaces in contact are not in relative
motion.
(2.6)
In the first method, a block is placed on an inclined plane, and gradually increase the angle of
inclination. At first, static friction keeps the block stuck to the incline. When the angle of
inclination increases, the downhill force exceeds the maximum static friction then the block
breaks loose and starts sliding. The angle at which this happen is termed as the angle of
repose, θ.
From figure 2.1,
where:
, when
The component of gravity pulling the object along the
ramp is given as:
Figure 2.1
(2.7)
where
, is the force required to break loose the block.
From equation 2–6:
and from Figure 2.1:
Equating the two equations gives:
(2.8)
Likewise, from Figure 2.1,
Thus,
(2.9)
46
Another way to supply a frictional force is to
use a weight hanging on a string attached to
the block, as shown in Figure 2.2
Figure 2.2
As we increase the “hanging mass”,
, the tension in the string increases. The static friction
between the block and the table also increases until it reaches its limit where the block breaks
loose. At this point the equality holds in equation (7.1) so that
(2.10)
III. Materials and Apparatus:
Dynamics track and cart, pulley, set of masses, weight hangers, platform balance,
stopwatch, string, paper clips, wooden block, wooden flat board, meter stick
IV. Procedure:
Part A. Acceleration due to varying force applied on the body while its mass is held
constant
Figure 2.3
1.
2.
3.
4.
5.
6.
Measure the mass of the cart using the platform balance.
Set up the pulley, cart, and a bumper of some sort to prevent the cart from hitting the
pulley at the end of its run.
Carefully level the dynamics track until the cart has no particular tendency to drift or
accelerate in either direction along its run.
Put a loop in one end of the string and place this loop over the spring-release trigger on
the Dynamics Cart. Drape the string over the pulley. Adjust the pulley so the string is
level.
Adjust the length of the string so that the longest arrangement of masses that you intend
to use will not hit the floor before the cart has reached the end of its run. Put a loop at
this end of the string.
Hang enough paper clips onto the dangling loop in the string until the cart will just
continue to move without apparent acceleration when slightly pushed. This small added
mass will compensate for friction in the system and will be ignored in the following
calculations. The paper clips should remain attached to the loop throughout the
experiment.
47
7.
Hang a 10-g mass to the end of the hanging loop and pull the cart back to a clearly
marked starting point. Determine the distance
that the cart will move from the starting
point to the bumper block and record this distance at the top of Table 2.1.
NOTE: The total mass of the system will remain constant throughout the experiment.
8.
Practice releasing the cart, being careful not to give it any push as you do so. The best
way to do this is to press your finger into the table in front of the cart, thereby blocking
its movement. Quickly pull your finger away in the direction that the cart wants to move.
At the instant you pull your finger away, start your stopwatch. Stop your stopwatch at
the instant the cart arrives at the bumper. (To eliminate reaction time errors, have
the person who releases the cart to also do the timing.)
9.
Determine the average time for the cart to move through the distance ( ) having been
released from rest. Record in table 6.1 the average of the three time trials in which you
have the most confidence. Repeat for all of the masses given in the data table.
10. Determine the mass M of the cart and record at the top of Table 2.1.
11. Complete Table 2.1 using your data collected and the equations given in the Introduction.
Part B. Acceleration due to constant force applied on the body while it’s mass varies.
1.
2.
3.
4.
Using the same set-up in Part I, replace 80 g on the weight hanger with 10 g.
Record this on top of table 2.2. Its weight serves as the force for the entire experiment.
Place 50 grams on the bed of the dynamics cart and repeat procedure 9 in Part I.
Complete Table 2.2.
Part C. Friction on Inclined Surface
1. Place the wooden block on the dynamics track at one end and slowly raise such end until
the block starts to slide down.
Note: For consistent results use the same face of the block and start the block from the
same place each time.
2.
3.
4.
5.
6.
Measure the height A and base B of the inclination.
Repeat four times, for a total of five trials to complete Table 2.3.
Calculate the coefficient of friction between the surfaces using Eq. 2-9.
Replace the dynamics track with a wooden flat board.
Repeat steps 1–5. Record the results in Table 2.4.
Part D: Horizontal Surface
1.
2.
3.
4.
5.
6.
Measure and record the mass of the wooden block,
, and record in Table 2.5.
Lay the dynamic track flat on the table.
Attach one end of the string to the block and the other end to the mass hanger.
Drape the mass hanger over the pulley, and place the block’s wider side on the board.
Slowly add set of masses to the weight hanger until the block starts to slide.
Add 100 g on top of the block and repeat step 5. Make three trials.
Note: Remove all of the mass from the hanger after each trial, and start adding it in the
next trial.
7. Do the same procedure for 200, 300, and 400 g respectively on top of the block. Complete
Table 2.5.
48
8. Determine the average mass needed to start the block to move and calculate the weight
of the mass. This gives the maximum force of static friction, i.e.
9. Calculate the normal force between the block and the table, which in this situation is just
the weight of the block,
10. Finally, calculate the coefficient of static friction using equation 2.10
Note: You should get approximately the same value for
in each trial. Get the average.
Calculate the percent difference for
by comparing the values you got in Part III and
IV.
11. Replace the dynamics track with wooden board and repeat steps 1-10. Record results in
Table 2.6
49
50
Date Performed:
Date Received:
University of Science and Technology
of Southern Philippines
Department of Physics
Content (65%)
Performance (15%)
Neatness (10%)
Promptness (10%)
School Year 20__ to 20__
First/Second/Summer/ Semester
Score/Rating:
Name:_____________________________________________ID No: ___________________
Section Code: _____________________________ Course & Year: _____________________
Group No: ___________Instructor’s Name & Signature: _____________________________
Leader: _____________________
Members:
___________________ ____________________ ______________________
___________________ ____________________ ______________________
___________________ ____________________ ______________________
LABORATORY ACTIVITY NO. 2
Newton’s Law of Acceleration: Friction
I. Data and Results :
Part A. Acceleration due to varying force applied on the body while its mass is held
constant.
Table 2.1
Varying Force
d=_______________
mass, m
(g)
F = mg
(N)
M =_____________
t1
time, t
(s)
t2
t3
( )
( )
% Error
tave
10
20
30
40
50
60
70
80
51
Part B. Acceleration due to constant force applied on the body while it’s mass
varies.
Table 2.2
Varying Mass
d=______________
Mass
MT
on the cart
(g)
50
100
150
200
250
300
350
400
t1
Time
t2
t3
m=_______________
% Error
tave
Part C. Friction on Inclined Surface
Table 2.3
Aluminum Board
Trial
A (cm)
B (cm)
µs (Coefficient of Static
friction)
1
2
3
4
5
Average µs
Table 2.4
Wooden Flat Board
Trial
A (cm)
B (cm)
µs (Coefficient of Static
friction)
1
2
3
4
5
Average µs
52
Part D. Friction on Horizontal Surface
Table 2.5
Aluminum Board
(
)
Mass
added
to
(kg)
Total
mass
(
)
(kg)
Hanging mass,
(kg)
Trial 1 Trial 2 Trial 3 Average
(
)
% Diff
(
)
% Diff
0.100
0.200
.300
0.400
Average value of
Table 2.6
Wooden Flat Board
(
)
Mass
added
to
(kg)
Total
mass
(
)
(kg)
Hanging mass,
(kg)
Trial 1 Trial 2 Trial 3 Average
0.100
0.200
.300
0.400
Average value of
II. Computations:
53
III. Discussion of Results:
IV. Conclusion :
V. Answers to Questions and Problems:
1. Cite some possible errors that affected your experimental value of acceleration.
2. Suggest possible procedure/s to lessen the percentage error of the cart’s acceleration.
3. Based on the results of your experiment, give the relationship between
a) acceleration and force and
b) acceleration and mass.
54
4. Comparing the results of µs of wooden board and aluminium board in part 1, which has
greater µs? Why?
5. What is the relationship of the angle of repose in part 1 to the coefficient of static friction?
6. In case the block did not move at constant velocity in part 1, is the friction force equal to
the x-component of the force of gravity? Why or why not?
7. What is the relationship between the weights of the block in part 2 to the friction force?
8. Is µs from part 1 the same with µs in part 2 for wooden board and dynamic (aluminium)
track respectively? Why or why not? Explain the possible errors of this experiment?
9. How will you improve the procedure of this experiment to lessen the percent difference?
55
56