Friction Lab
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Mechanical Engineering
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Feb 20, 2024
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(1 point)
Title of the Experiment: Friction
Student’s name: Anjana Shyam
Section SLN: A
TA’s Name: Ayush Kumar Singh
Week of the experiment: 4
Objectives: (3 points)
The objective of this lab is to calculate the frictional force in different situations and determine the
coefficients of static and kinetic friction from the calculated frictional force. Calculating the static and
kinetic frictional forces can help determine a distinction between the two forces on objects. Static friction
acts in the opposing direction on objects at rest and has a larger coefficient of friction than kinetic friction
which acts in the opposing direction on objects in motion.
Experimental Data (3 points):
Kinetic Friction on Horizontal Plane
Kinetic Friction on Horizontal Plane Run 2:
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Kinetic friction on an inclined plane 1
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Kinetic Friction on an Inclined Plane 2:
PART 1. Static Friction
1a) Cart on the horizontal track
Mass of the cart,
M
= 0.25 kg + 400 g = 0.65 kg
Mass of the hanger,
m
= 0.05kg
Mass on the hanger,
m
w
= 0.07 kg
Total mass of the hanger
m
h
= 0.12 kg
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1b) Cart on an inclined plane
Mass of the cart,
M
= 0.25 +0.4 = 0.65 kg
Trial
Critical angle (Θ)
1
10.3
2
10.3
3
10.3
PART 2. Kinetic friction
2a) Cart on the horizontal track
Mass of the cart,
M
= 0.25 +0.32= 0.57 kg
Mass of the hanger,
m
= 0.05 kg
Mass on the hanger = 0.01 kg
Total mass of the hanger m
h
= 0.06 kg
Run
acceleration, a
1
-0.02 m/s^2
2
0.13 m/s^2
2b) Cart on an Incline plane.
Run
𝜭
acceleration, a
1
5°
0.02 m/s^2
2
7°
-0.32 m/s^2
Data Analysis (10 points):
PART 1. Static Friction
1a) Cart on the horizontal track.
The experimental value of the coefficient of the static friction
µ
s
calculated as follows:
T =
s max
= μ
s
N
?
T = W
h
= m
h
g
μ
s
F
n
= m
h
g
μ
s
Mg = m
h
g
μ
s
=m
h
/M
μ
s
= 0.184615 ~ 0.18
The discrepancy between the experimentally determined
µ
s
and its actual given value
has been calculated to be:
% discrepancy = |(
= |(0.184-0.18)/0.18| *100
?𝑥??𝑟𝑖???𝑡𝑎? − 𝑡ℎ??𝑟?𝑡𝑖𝑐𝑎?)/𝑡ℎ??𝑟?𝑡𝑖𝑐𝑎?| * 100
= 2.22%
1b) Cart on an Incline plane
The experimental value of the coefficient of static friction
µ
s
is calculated with equation
(4) as follows:
Mgsin(
𝜃
)= μ
s
Mgcos(
𝜃
)
μ
s
= tan(
𝜃
)
tan(
𝜃
) = tan(10.3)=0.18173
μ
s
= 0.18173
Determine the discrepancy between the above-calculated value and the given one to
compare them: 100%* (
|
calc. value– expect. value
|
/ expect. value).
% discrepancy = |(
?𝑥??𝑟𝑖???𝑡𝑎? − 𝑡ℎ??𝑟?𝑡𝑖𝑐𝑎?)/𝑡ℎ??𝑟?𝑡𝑖𝑐𝑎?| * 100
(0.181-0.18)/0.18*100 = 0.56%
PART 2. Kinetic friction
2a) Cart on the horizontal track
For run 1 the coefficient of kinetic friction
μk
is calculated as follows:
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W
h
- f
k
= (M + m
h
) a
W
h
-
𝛍
k
F
N
= (M + m
h
) a
m
h
g -
𝛍
k
F
N
= (M + m
h
) a
m
h
g -
𝛍
k
Mg = (M + m
h
) a
𝛍
k
= (m
h
g - (M + m
h
) a)/Mg
(0.05*9.8 -(0.62)(-0.02))/(0.57*9.8) = 0.0899
For run 2 the coefficient of kinetic friction
μk
is calculated as follows:
𝛍
k
= (m
h
g - (M + m
h
) a)/Mg
(0.06*9.8 -(0.63)(0.13))/(0.57*9.8) = 0.0906
The average values of coefficient of kinetic friction
μk
from both runs is:
(0.08994+0.0906)/2 = 0.09
The discrepancy between the experimental and theoretical values is:
(0.09-0.09)/0.09*100 = 0%
2b) Cart on an Incline plane.
For run 1 the coefficient of kinetic friction
μk
is calculated as follows:
W
h
-
𝛍
k
F
N
= (M + m
h
) a
m
h
g -
𝛍
k
Mgcosθ = (M + m
h
) a
-Mgsinθ +
𝛍
k
Mgcosθ = Ma
𝛍
k
= (gsinθ+ a)/gcosθ
(9.8*sin(5) + 0.02)/9.8cos(5) = 0.0895
For run 2 the coefficient of kinetic friction
μk
is calculated as follows:
𝛍
k
= (gsinθ+ a)/gcosθ
(9.8*sin(7) + (-0.32))/9.8cos(7) = 0.0899
The average values of coefficient of kinetic friction
μk
from both runs is:
(0.0895 + 0.0899)/2 = 0.09
The discrepancy between the experimental and theoretical values is:
(0.09-0.09)/0.09 * 100 = 0%
Results (3 points)
Friction Force
Lab Section
Experimental
Value
Theoretical
Value
Discrepancy
𝞵
s
1a
0.184
0.18
2.22%
𝞵
s
1b
0.182
0.18
0.56%
𝞵
k
2a
0.09
0.09
0%
𝞵
k
2b
0.09
0.09
0%
Discussion and Conclusion (10 points):
The experiment involved a mass-cart system to determine the coefficient of kinetic friction and
static friction on a horizontal plane and inclined plane. The equations for the coefficients of static
and kinetic friction determined by free body diagrams of the hanging mass and the cart were
used to determine the experimental coefficients of kinetic and static friction and theoretical
coefficients were given in the lab manual. The coefficient of static friction varies depending on
the force applied to resist motion. The coefficient of static friction has a maximum value
corresponding to a maximum static frictional force that the applied force in the opposite direction
has to exceed in order for the object to begin moving, which is why the coefficient of static
friction is always larger than the coefficient of kinetic friction. In the experiment, the coefficient
of static friction for the cart on a horizontal plane was determined by the theoretical equation for
㎲
which was determined to be the ratio of the hanging mass to the mass of the cart. The
theoretical
㎲
was calculated to be 0.184 and the theoretical coefficient of static friction was 0.18
resulting in a discrepancy of 2.22%. The coefficient of static friction on an inclined plane was
determined by three trials to calculate the critical angle(the angle of the inclined plane that would
allow the cart to begin motion). Using the calculated critical value 10.3° and the equation for
㎲
determined by the free body diagram on the cart(
) the coefficient of static friction
㎲
=
𝑡𝑎?(𝛉)
was determined to be 0.182 resulting in a discrepancy of 0.56%. The coefficient of kinetic
friction on a horizontal plane was calculated from the quadratic fit performed on the position
time graph of the cart’s motion and the linear fit performed on the velocity time graph of the
cart’s motion which was used to calculate the acceleration of the mass-cart system. The
acceleration for Run 1 was -0.02 m/s^2 and the acceleration for Run 2 was 0.13 m/s^2, and the
equation for
𝛍
k
was the weight of the cart divided by the difference between the weight of the
hanging mass and net force of the mass cart system. The average of the coefficients of kinetic
friction for both runs (0.0899 and 0.0906 respectively) was 0.09, which resulted in an
experimental discrepancy of 0% against the theoretical value 0.09. The accelerations for the cart
on an inclined plane of 5° and 7° were 0.02 m/s^2 and -0.32 m/s^2 respectively. The equation for
𝛍
k
using free body diagrams and Newton’s Second Law was (gsin( ) + acceleration of the
𝛉
cart)/gcos( ), resulting in a coefficient of 0.0895 for Run 1 and 0.0899 for Run 2. The average of
𝛉
the two coefficients was 0.09 which resulted in an experimental discrepancy of 0%. The
equations of the coefficients of static and kinetic friction determined by the free body diagrams
of the cart- block system and the cart on an inclined plane were confirmed by this experiment,
and the fact that the coefficient of static friction is always larger than the coefficient of kinetic
friction was confirmed by this experiment(0.18>0.09).
In conclusion, the objectives of this experiment were achieved. The principles of friction force
were confirmed through Newton’s Second Law, and the frictional force to resist motion is always
greater than the frictional force when the object is already in motion.
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(Follow the general directions given in the Syllabus where it is describing how the Discussion
and Conclusion section should be composed. In particular explain how the values of the
coefficients of the kinetic and static friction compare to each other; what does the frictional force
depend on, explain the nature of the frictional forces, also explain the reasons for discrepancies
between your experimental and theoretical results).
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100
150
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F(t)
Force_Crate
Known values:
Mass of Block
68 kg
TT
Hs μk
0
0.63 0.53 26°
Problem Statement:
A crate is initially at rest on a sloped surface, when a force is applied: F(t)
where t is in seconds.
Answers:
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Find the magnitude of the friction force at that time:
317.771 N
The direction of the friction force at that time is: down the slope
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169.756 S
240+0.2 N,
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Express your answer to three significant figures and include the appropriate units.
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