Download Balancing Act: Solving Torque Problems in Physics and more Lecture notes Law in PDF only on Docsity!
EN PHYS 1 – PHYSICS FOR ENGINEERS
Semester AY 20 22 - 2023
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Direction: Go to the link - https://phet.colorado.edu/sims/html/balancing-act/latest/balancing-act_en.html
Choose “Intro” tab. Play around with the simulation for a few minutes to understand/ explore the functionalities of
the different components. To check if the board is balanced at the pivot, remove supports under the board by toggling
the switch at bottom of the screen to right.
Submit your output to this link
https://drive.google.com/drive/folders/13TMvJ0ufac6QgET43CMiLVld12EqIFOJ?usp=share_link
PART 1: BALANCING TORQUES ON A SEESAW
Choose “Balance Lab” tab. Select the following options in right corner of the screen:
Mass Labels, Level, Rulers
For each of the following cases, balance the board at the pivot. Don’t forget to remove the supports under the board.
For each case, calculate total clockwise torque ( Σ τ cw
), total counterclockwise torque ( Σ τ ccw
) and net torque ( Σ τ = Σ
τ cw
). Don’t forget to convert masses to weights and the signs of toques (cw torque is negative & ccw torque
is positive).
NOTE: Show calculation steps. Do not just compute on calculator and write the answer. Instead write the complete
expression showing what numbers you multiplied/added to get the answers.
Name: Paul Jeremy L. Mendoza Rating:
Course/Block: BSEE 1A Date:
April 2 5 , 2023
Activity No. 3 : Balancing Act
Learning Objective: At the end of this activity, the student must be able to:
- Demonstrate through the simulation concept of torque.
- Analyze and calculate a scenario involving torque.
EN PHYS 1 – PHYSICS FOR ENGINEERS
Semester AY 20 22 - 2023
ACTIVITY WORKSHEET
CAMARINES NORTE STATE COLLEGE
F. Pimentel Avenue, Brgy. 2, Daet, Camarines
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Case 1.
5 kg on the right of the pivot, 10 kg on left of the pivot.
Screenshot:
Given:
𝑚
1
= 10 𝑘𝑔; 𝐿
1
= 0. 25 𝑚
𝑚
2
= 15 𝑘𝑔 ; 𝐿
2
= 0. 5 𝑚
Calculations:
𝒄𝒘
𝟏
𝟏
Σ τ cw
2
𝒄𝒘
𝟏
𝟏
Σ τ ccw
2
Σ τ = Σ τ cw
EN PHYS 1 – PHYSICS FOR ENGINEERS
Semester AY 20 22 - 2023
ACTIVITY WORKSHEET
CAMARINES NORTE STATE COLLEGE
F. Pimentel Avenue, Brgy. 2, Daet, Camarines
COLLEGE OF ENGINEERING
Case 3.
5 kg & 20 kg on right of the pivot, 10 kg & 15 kg on left of the pivot.
Screenshot:
Given:
𝑚
1
= 10 𝑘𝑔; 𝐿
1
= 0. 5 𝑚
𝑚
2
= 15 𝑘𝑔 ; 𝐿
2
= 0. 75 𝑚
𝑚
3
= 20 𝑘𝑔; 𝐿
3
= 0. 5 𝑚
𝑚
4
= 5 𝑘𝑔; 𝐿
4
= 1. 25 𝑚
Calculations:
𝒄𝒘
𝟏&𝟐
𝟏&𝟐
Σ τ cw
2
2
𝒄𝒄𝒘
𝟑&𝟒
𝟑&𝟒
Σ τ ccw
2
2
Σ τ = Σ τ cw
EN PHYS 1 – PHYSICS FOR ENGINEERS
Semester AY 20 22 - 2023
ACTIVITY WORKSHEET
CAMARINES NORTE STATE COLLEGE
F. Pimentel Avenue, Brgy. 2, Daet, Camarines
COLLEGE OF ENGINEERING
Case 4.
5 kg, 10 kg & 15 kg on right of the pivot, 20 kg on left of the pivot.
Screenshot:
Given:
𝑚
1
= 20 𝑘𝑔; 𝐿
1
= 1. 5 𝑚
𝑚
2
= 5 𝑘𝑔 ; 𝐿
2
= 0. 25 𝑚
𝑚
3
= 10 𝑘𝑔; 𝐿
3
= 1 𝑚
𝑚
4
= 15 𝑘𝑔; 𝐿
4
= 1. 25 𝑚
Calculations:
𝒄𝒘
𝟏
𝟏
Σ τ cw
2
𝒄𝒘
𝟏,𝟐&𝟑
𝟏,𝟐&𝟑
Σ τ ccw
2
2
2
Σ τ = Σ τ cw
EN PHYS 1 – PHYSICS FOR ENGINEERS
Semester AY 20 22 - 2023
ACTIVITY WORKSHEET
CAMARINES NORTE STATE COLLEGE
F. Pimentel Avenue, Brgy. 2, Daet, Camarines
COLLEGE OF ENGINEERING
Choose “Balance Lab” tab. Select the following options in right corner of the screen:
Mass Labels, Level, Rulers
In the “Bricks” section on the right of the screen, click on the right arrow. Keep on clicking to the right until you see
“Mystery Objects”.
Find out the mass of each of the Mystery Objects “A”, “B”, “C”, “D”, “E”, “F”, “G” and “H” by balancing each of
these against the bricks of known masses. For each Mystery Object, paste a screenshot of balanced seesaw and show
all calculation steps.
Mystery Object “A”
Screenshot of balanced seesaw:
Given:
𝑚
1
=? ; 𝐿
1
= 0. 25 𝑚
𝑚
2
= 5 𝑘𝑔 ; 𝐿
2
= 1 𝑚
EN PHYS 1 – PHYSICS FOR ENGINEERS
Semester AY 20 22 - 2023
ACTIVITY WORKSHEET
CAMARINES NORTE STATE COLLEGE
F. Pimentel Avenue, Brgy. 2, Daet, Camarines
COLLEGE OF ENGINEERING
Calculations:
𝒄𝒄𝒘
𝟐
𝟐
Σ τ ccw
2
Mystery Object “A” = Σ τ cw
𝒄𝒘
𝒄𝒄𝒘
𝟏
1
Στ
ccw
1
1
- 05 𝑁∗𝑚
(
- 81 𝑚 𝑠
2
⁄ ) ( 0. 25 𝑚)
𝒄𝒘
𝟏
𝟏
Σ τ cw
2
Σ τ = Σ τ cw
Proving
𝒄𝒘
𝒄𝒄𝒘
𝒎
𝟏
𝒈 ∗ 𝑳
𝟏
= 𝒎
𝟐
𝒈 ∗ 𝑳
𝟐
𝒎
𝟏
2
2
1
2
2
1
2
2
EN PHYS 1 – PHYSICS FOR ENGINEERS
Semester AY 20 22 - 2023
ACTIVITY WORKSHEET
CAMARINES NORTE STATE COLLEGE
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COLLEGE OF ENGINEERING
1
- 26 𝑁∗𝑚
( 9. 81 𝑚 𝑠
2
⁄ )( 0. 25 𝑚)
𝒄𝒘
𝟏
𝟏
Σ τ cw
2
Σ τ = Σ τ cw
Proving
𝒄𝒘
𝒄𝒄𝒘
𝒎
𝟏
𝒈 ∗ 𝑳
𝟏
= 𝒎
𝟐
𝒈 ∗ 𝑳
𝟐
𝒎
𝟏
2
⁄ ) ∗
(
- 25 𝑚
)
( 5 𝑘𝑔
) ( 9. 81 𝑚 𝑠
2
⁄ ) ∗
(
- 25 𝑚
)
1
2
2
1
2
2
Mystery Object “C”
Screenshot of balanced seesaw:
EN PHYS 1 – PHYSICS FOR ENGINEERS
Semester AY 20 22 - 2023
ACTIVITY WORKSHEET
CAMARINES NORTE STATE COLLEGE
F. Pimentel Avenue, Brgy. 2, Daet, Camarines
COLLEGE OF ENGINEERING
Given:
𝑚
1
=? ; 𝐿
1
= 0. 25 𝑚
𝑚
2
= 15 𝑘𝑔 ; 𝐿
2
= 0. 25 𝑚
Calculations:
𝒄𝒄𝒘
𝟐
𝟐
Σ τ ccw
2
Mystery Object “C” = Σ τ cw
𝒄𝒘
𝒄𝒄𝒘
𝟏
1
Στ
ccw
1
1
- 79 𝑁∗𝑚
( 9. 81 𝑚 𝑠
2
⁄ )( 0. 25 𝑚)
𝒄𝒘
𝟏
𝟏
Σ τ cw
2
Σ τ = Σ τ cw
Proving
𝒄𝒘
𝒄𝒄𝒘
𝒎
𝟏
𝒈 ∗ 𝑳
𝟏
= 𝒎
𝟐
𝒈 ∗ 𝑳
𝟐
𝒎
𝟏
2
⁄ ) ∗
(
- 25 𝑚
)
( 15 𝑘𝑔
) ( 9. 81 𝑚 𝑠
2
⁄ ) ∗
(
- 25 𝑚
)
1
2
2
1
2
2
EN PHYS 1 – PHYSICS FOR ENGINEERS
Semester AY 20 22 - 2023
ACTIVITY WORKSHEET
CAMARINES NORTE STATE COLLEGE
F. Pimentel Avenue, Brgy. 2, Daet, Camarines
COLLEGE OF ENGINEERING
Σ τ = Σ τ cw
Proving
𝒄𝒘
𝒄𝒄𝒘
𝒎
𝟏
𝒈 ∗ 𝑳
𝟏
= 𝒎
𝟐
𝒈 ∗ 𝑳
𝟐
𝒎
𝟏
2
2
1
2
2
1
2
2
Mystery Object “E”
Screenshot of balanced seesaw:
Given:
𝑚
1
=? ; 𝐿
1
= 1. 25 𝑚
𝑚
2
= 15 𝑘𝑔 ; 𝐿
2
= 0. 25 𝑚
Calculations:
𝒄𝒄𝒘
𝟐
𝟐
Σ τ ccw
2
EN PHYS 1 – PHYSICS FOR ENGINEERS
Semester AY 20 22 - 2023
ACTIVITY WORKSHEET
CAMARINES NORTE STATE COLLEGE
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Mystery Object “E” = Σ τ cw
𝒄𝒘
𝒄𝒄𝒘
𝟏
1
Στ
ccw
1
1
- 79 𝑁∗𝑚
( 9. 81 𝑚 𝑠
2
⁄ )( 1. 25 𝑚)
𝒄𝒘
𝟏
𝟏
Σ τ cw
2
Σ τ = Σ τ cw
Proving
𝒄𝒘
𝒄𝒄𝒘
𝒎
𝟏
𝒈 ∗ 𝑳
𝟏
= 𝒎
𝟐
𝒈 ∗ 𝑳
𝟐
𝒎
𝟏
2
⁄ ) ∗
(
- 25 𝑚
)
( 15 𝑘𝑔
) ( 9. 81 𝑚 𝑠
2
⁄ ) ∗
(
- 25 𝑚
)
1
2
2
1
2
2
EN PHYS 1 – PHYSICS FOR ENGINEERS
Semester AY 20 22 - 2023
ACTIVITY WORKSHEET
CAMARINES NORTE STATE COLLEGE
F. Pimentel Avenue, Brgy. 2, Daet, Camarines
COLLEGE OF ENGINEERING
𝒄𝒘
𝟏
𝟏
Σ τ cw
2
Σ τ = Σ τ cw
Proving
𝒄𝒘
𝒄𝒄𝒘
𝒎
𝟏
𝒈 ∗ 𝑳
𝟏
= 𝒎
𝟐
𝒈 ∗ 𝑳
𝟐
𝒎
𝟏
2
2
1
2
2
1
2
2
Mystery Object “G”
Screenshot of balanced seesaw:
Given:
𝑚
1
=? ; 𝐿
1
= 0. 25 𝑚
𝑚
2
= 5 𝑘𝑔 ; 𝐿
2
= 1. 25 𝑚
Calculations:
𝒄𝒄𝒘
𝟐
𝟐
EN PHYS 1 – PHYSICS FOR ENGINEERS
Semester AY 20 22 - 2023
ACTIVITY WORKSHEET
CAMARINES NORTE STATE COLLEGE
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COLLEGE OF ENGINEERING
Σ τ ccw
2
Mystery Object “G” = Σ τ cw
𝒄𝒘
𝒄𝒄𝒘
𝟏
1
Στ
ccw
1
1
- 31 𝑁∗𝑚
( 9. 81 𝑚 𝑠
2
⁄ )( 0. 25 𝑚)
𝒄𝒘
𝟏
𝟏
Σ τ cw
2
Σ τ = Σ τ cw
Proving
𝒄𝒘
𝒄𝒄𝒘
𝒎
𝟏
𝒈 ∗ 𝑳
𝟏
= 𝒎
𝟐
𝒈 ∗ 𝑳
𝟐
𝒎
𝟏
2
2
1
2
2
1
2
2
EN PHYS 1 – PHYSICS FOR ENGINEERS
Semester AY 20 22 - 2023
ACTIVITY WORKSHEET
CAMARINES NORTE STATE COLLEGE
F. Pimentel Avenue, Brgy. 2, Daet, Camarines
COLLEGE OF ENGINEERING
1
- 79 𝑁∗𝑚
( 9. 81 𝑚 𝑠
2
⁄ )( 0. 5 𝑚)
𝒄𝒘
𝟏
𝟏
Σ τ cw
2
Σ τ = Σ τ cw
Proving
𝒄𝒘
𝒄𝒄𝒘
𝒎
𝟏
𝒈 ∗ 𝑳
𝟏
= 𝒎
𝟐
𝒈 ∗ 𝑳
𝟐
𝒎
𝟏
2
2
1
2
2
1
2
2
Answers:
Mystery Object “A” = 20 kg
Mystery Object “B” = 5 kg
Mystery Object “C” = 15 kg
Mystery Object “D” = 10 kg
Mystery Object “E” = 3 kg
Mystery Object “F” = 50 kg
Mystery Object “G” = 25 kg
Mystery Object “H” = 7.5 kg
EN PHYS 1 – PHYSICS FOR ENGINEERS
Semester AY 20 22 - 2023
ACTIVITY WORKSHEET
CAMARINES NORTE STATE COLLEGE
F. Pimentel Avenue, Brgy. 2, Daet, Camarines
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Conclusion:
In conclusion, the balance of a seesaw is dependent on the precise equality of the forces acting on both
sides of the beam. Any slight variation in the magnitude or direction of these forces can disrupt the balance and
cause one side to rise or fall. It is essential to ensure that the forces acting on either side of the seesaw are
precisely equal, with the same magnitude and direction, to maintain equilibrium. Additionally, to achieve
balance, an equal force must be exerted on the seesaw, either by a person or an object, to keep it level and
stable. Furthermore, the balance of a seesaw is a fundamental example of the principles of physics and the
importance of maintaining equilibrium in various systems.
The balance of a seesaw also depends on how weight is distributed on either side. If one side has more
weight than the other, it will tip down towards the heavier side. However, the position of the weight on each
side is also important. If a heavier weight is closer to the center of the seesaw, it will require a lighter weight to
balance it if it is placed further away from the center. This is because the weight further away from the center
has greater leverage and exerts a greater force on that side of the seesaw. Achieving balance on a seesaw
requires both equal weight distribution and proper placement of the weights on each side.
In conclusion, the simulation of a seesaw or lab activity is a valuable tool for understanding the
principles of balance and equilibrium in physics. By manipulating the forces, weights, and angles involved, we
can observe how different factors affect the balance of the seesaw and adjust them accordingly to maintain
equilibrium. This simulation or lab activity can be useful in a variety of contexts, from designing physical
structures to understanding the behavior of systems in nature. By using simulations like this, we can better
understand the physical laws that govern the world around us and apply this knowledge to solve real-world
problems.
