lab report 10 - PHY2053L
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Broward College *
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Course
2053L
Subject
Mechanical Engineering
Date
Dec 6, 2023
Type
docx
Pages
4
Uploaded by inuyasha078
Cristian Acuna Vasquez
03/06/2020
PHY2053L
Title :
Balancing Act
Purpose:
The purpose of this simulation is to investigate and learn about the concepts of
balancing torques with regards to rotational equilibrium.
Background Information:
A force’s tendency to produce the rotation of a rigid body is called
the torque.
It is equal to the product of the
force and momentum arm, the perpendicular distance
from the axis of rotation to the force’s line of action. An important basic concept when dealing
with torques is the idea of equilibrium or balance. In general, an object can be acted on by several
torques at the same time. A torque is also a vector quantity which means that it has both a
magnitude (size) and a direction associated with it. If the size and direction of the torques acting
on an object are exactly balanced, then there is no net torque acting on the object and the object is
said to be in equilibrium. Because there is no net torque acting on an object in equilibrium, an
object at rest will stay at rest, and an object in constant angular motion will stay in angular
motion.
Hence, at totational equilibrium , it should be true that
T
net
=
Σ T =I T
left
I – I T
right
I =0
Materials:
computer
PhET software
Procedure:
Using the PhET software. We need to place a variety of object at various positions
on
a seesaw and try to balance by making a torque summation.
The objects have different
masses and shapes.
Results/Calculations/
Analysis
Part I.
Balancing two masses.
Mass 1 (kg)
Position 1
(m)
T
right
=
T
left
Mass 2 (kg)
Position 2 (m)
20
0.5
(20)(9.8)(.5)=(5)(9.8)X
X=2.0
5
2.0
20
1.0
(20)(9.8)(1.0)=(5)(9.8)X
X=2.0
10
2.0
80
0.25
(80)(9.8)(0.25)=(20)(9.8)X
X=1.0
20
1.0
30
0.5
(30)(9.8)(.5)=(10)(9.8)X
X=1.5
10
1.5
10
1.0
(10)(9.8)(1.0)=(20)(9.8)X
X=0.5
20
0.5
5
2.0
(5)(9.8)(2.0)=(20)(9.8)X
X=0.5
20
0.5
15
1.0
(15)(9.8)(1.0)=(60)(9.8)X
X=0.25
60
0.25
Part II
Balancing multiple masses
Mass 1
(kg)
Position
1 (m)
Mass 2
(kg)
Position
2 (m)
T
right
=
T
left
Mass
3 (kg)
Positio
n 3 (m)
20
1.5 m
left
30
0.50 m
left
(20)(9.8)(1.5)+(30)(9.8)(0.5)=
(60)(9.8)X
X=0.75
60
0.75 m
right
20
1.0 m
left
60
2.0 m
left
(20)(9.8)(1.0)+(60)(9.8)(2.0)=
(80)(9.8)X
X=1.75
80
1.75 m
right
80
1.5 m
left
30
0.5 m
right
(80)(9.8)(1.5)=(30)(9.8)(0.5)+
(60)(9.8)X
X=1.75
60
1.75 m
right
60
1.0 m
left
80
0.25 m
right
(60)(9.8)(1.0)=(80)(9.8)(0.25)+
(20)(9.8)X
X=2.00
20
2.00 m
right
Part III
Mystery packages
Know
n
Mass
1 (kg)
Positio
n 1 (m)
left
T
right
=
T
left
Mystery
Package
Positio
n (m)
Right
Mass of
Packag
e (kg)
60
0.5
(60)(9.8)(0.5)= X (9.8)(1.5)
X=20
A
1.5
20
30
1.0
(30)(9.8)(1.0)= X (9.8)(2.0)
X=15
C
2.0
15
30
0.25
(30)(9.8)(0.25)= X (9.8)(0.75)
X=10
D
0.75
10
60
0.25
(60)(9.8)(0.25)= X (9.8)(2.0)
X=7.5
H
2.0
7.5
Know
n
Mass
1 (kg)
Positio
n 1 (m)
Know
n
Mass
2 (kg)
Positio
n 2 (m)
T
right
=
T
left
Mystery
Package
Positio
n (m)
Mass of
Packag
e (kg)
20
1.75 m
left
80
0.5 m
left
(20)(9.8)(1.75) +
(80)(9.8)(0.5) =
X(9.8)(1.5)
X= 50
F
1.5m
right
50
60
0.5 m
left
15
1.75 m
right
(60)(9.8)(0.5) = (15)
(9.8)(1.75) + X(9.8)
(1.25)
X= 3
E
1.25 m
right
3
20
0.25 m
left
5
1.0 m
left
(20)(9.8)(0.25) + (5)
(9.8)(1.0) = X(9.8)
(2.0)
X= 50
B
2.0 m
right
5
Part IV
Balancing Act – Game
Level 1; Best score = 12
Level 2; Best score = 12
Level 3; Best score = 12
Your preview ends here
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Level 4; Best score = 12
Results/Calculations/
Analysis
Error Analysis:
In this lab, there were no errors since it is a simulation.
Conclusion:
In conclusion., I was able to identify the relationship between moment arms, torques, mass, and
how to find unknowns for moments arms and mass.
Torque, also called moment or moment
force, is the tendency of a force to rotate an object about an axis. A system is in equilibrium if it
has no acceleration gained or lost, meaning rotational or translational.
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ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY

Mechanics of Materials (MindTap Course List)
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ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning

Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY