Torque and Equilibrium of Rigid
Bodies
Dr. Komu Thirunavukkuarasu
9 October 2020
PHY 2053 Lab
Contributors:
Shurline Doss: Purpose, data table (I), sample calculations (I), questions
Liliana Giraldo: Experimental Procedures part 1, questions, conclusion part one , and discussion
Noyimot Ibrahim: Experimental Procedures & Diagram (II), Data (II), Sample Calculations(II)
Aria’ Odom: Introduction, Part (II) conclusion
TABLE OF CONTENTS:
Purpose:
2
Introduction:
2
Experimental Procedures and Diagrams:
2
Data:
5
Discussion:
12
Conclusion:
12
Part II
12
Questions:
13
1
Purpose:
The purpose of this experiment is to examine how factors like force, mass, and distance yield and
affect torque value. And to examine how each of these factors are related to one another.
Introduction:
Torque is a measure of how much a force acting on an object causes that object to rotate. The object
rotates on its axis called the pivot point. Before calculating for torque you would first have to
determine the lever arm. The lever arm is the perpendicular distance from the axis of rotation to the
line of action of the force. When mass is present a downward force happens which in turn
counterbalances the actual force of the pivot point. As the two forces are equal the torques must
equal as well. The mass on the left or the right times the distance to the axis of rotation must be the
same. Total counterclockwise torque is the positive rotational direction of the applied force and total
clockwise is the negative direction.
Experimental Procedures and Diagrams:
I.
The objective of Part 1of this experiment is calculating the torque by adding different masses
to either side of the see saw and placing the masses to different placements from their
corresponding sides to create equilibrium throughout the trials.
Trial 1: A 10kg block was placed on the right side of the see-saw at the 1.5 m mark. A 15kg
block was placed on the left side. The 15 kg mass was moved around until the plank was
balanced. The masses and distances were recorded. The left, right, and total torque were
calculated and recorded.
Trial 2 :A 10 kg brick was placed on the plank at 1.5 m to the right side and the 20 kg brick
was placed at 0.75 m to the right side. A 15 kg mass was placed randomly on the left side .
The supports were removed. The 15 kg mass was moved around until the plank was
balanced. The masses and distances were recorded. The left, right, and total torque were
calculated and recorded.
2
Trial 3: The father and son were placed on the see-saw. The father and son were moved
around randomly until the see saw was balanced. The masses and distances were recorded.
The left, right, and total torque were calculated and recorded.
Trial 4: Mystery object F was placed at 1 m to the left side. A 20 kg and the 15 kg bricks at 2
m and 0.5 m were placed on the right side. Once the support was removed, the 20 kg and 15
kg bricks were moved around on their corresponding side without moving the mystery object
F on the left until the plank was balanced. The mass of the mystery object was obtained once
the other masses and distances were calculated. The left, right, and total torque were
calculated and recorded. ( The image below represents the results of trial 4)
3
II.
The objective of Part 2 of this experiment is to investigate the equilibrium conditions of a
case and determine the masses of the mystery items.
The instruction requires students to use a stimulation to place the mystery items (one at a time) on a
see-saw, and balancing them with either bricks or people placed on the right side. For each case,
calculate total clockwise torque (Σ τcw ), total counterclockwise torque (Σ τccw ) and use the
equilibrium condition to calculate the mass of the mystery item.
Below is the diagram for Mystery A:
**The rest of the recorded layouts are displayed under Data (II). **
4
Data:
Part ITrial 1
Left side
right side
Mass
(kg)
Weight
(N)
Distance
(m)
Torque
(N * m)
Mass
(kg)
Weight
(N)
Distance
(m)
Torque
(N * m)
Total
torque
(N * m)
15
147
1
147
10
98
1.5
147
294
Trial 2
Left side
Mass (kg)
Weight (N)
Distance (m)
Torque (N * m)
15
147
2
294
Mass (kg)
Weight (N)
Distance (m)
Torque (N * m)
10
98
1.5
147
20
196
0.75
147
Right side
=294 N * m
Total torque (N * m)= 588 N * m
Trial 3
Left side
right side
Mass
(kg)
Weight
(N)
Distance
(m)
Torque
(N * m)
Mass
(kg)
Weight
(N)
Distance
(m)
Torque
(N * m)
Total
torque
( N * m)
20
196
2
392
80
784
0.5
392
784
5
Trial 4
Left side
Mass (kg)
Weight (N)
Distance (m)
Torque (N * m)
50
490
1
490
Mass (kg)
Weight (N)
Distance (m)
Torque (N * m)
15
147
2
294
20
196
1
196
Right side
= 490 N * m
Total torque= 980 N * m
Part I: Sample Calculation:
Trial 4:
A- right side
Mass 1 (kg): 15 kg
Mass 2 (kg): 20 kg
Weight 1 (N):
W= mg
= (15 kg) (9.8 m/s)
= 147 N
Weight 2 (N):
W= mg
= (20kg) (9.8 m/s)
= 196 N
Distance 1: 2m
Distance 2: 1m
6
Torque 1 (N * m) = (147 N) (2m) = 294 N * m
Torque 2 (N * m)= (196 N) (1m) = 196 N * m
Total torque on right side =490 N *m
B- Finding mass of left side object (object F)
We know the torque of the right side is equal to 490 N * m. Therefore, the torque of the left side is also equal
to 490 N * m. We also know the distance of the object on the left is 1 meter. Knowing these values, we can
calculate the weight of the object on the left, in Newtons. The derived weight will then allow us to calculate
the mass ,in kilograms, by manipulating the formula for weight.
Torque: 490 N * m
Distance: 1m
490 𝑁 ∗𝑁
= 490 N
1𝑁
𝑁 490 𝑁 490 𝑁𝑁 ∗ 𝑁/𝑁
Mass = =
=
= 50 kg
Weight (N)=
𝑁 9.8 𝑁/𝑁
9.8 𝑁/𝑁
7
Part II-
Below are the masses of mystery items A-E by placing the Mystery Items balanced on the scale.
**The calculations are in the next section**
Mystery A:
By establishing static equilibrium, we can observe that the mass of the static mystery object A is 20 kg.
Mystery B:
By establishing static equilibrium, we can observe that the mass of the static mystery object B is 5 kg.
Mystery C:
By establishing static equilibrium, we can observe that the mass of the static mystery object C is 15 kg.
8
Mystery D:
By establishing static equilibrium, we can observe that the mass of the static mystery object B is 10 kg.
Mystery E:
We have established the static equilibrium by placing the brick of mass m= 5 kg at distance d= 3 m
and mystery object of mass mh = ? at distance dh = 5 m.
9
Part II. Sample Calculation:
Formulas Used : 𝜏⃗𝜏⃗ = 𝜏⃗𝜏⃗ × 𝜏⃗𝜏⃗
and mgd = mhgdh
Mystery A-
Στcw = (20kg) (9.8m/s^2) (8m) = 1568
Στccw = (? kg) (9.8m/s^2) (-8m)
Στ =Στcw+Στccw= 1568 + ?
1568= (x) (9.8m/s^2) (-8m)
1568= (x)(78.4)
MA= 20 kg
Mystery B-
Στcw = (5kg) (9.8m/s^2) (8m) = 392
Στccw = (? kg) (9.8m/s^2) (8m)
Στ =Στcw+Στccw= 392 + ?
392= (x) (9.8m/s^2) (8m)
392= (x)(78.4)
MB= 5 kg
Mystery C-
Στcw = (15kg) (9.8m/s^2) (8m) = 1176
Στccw = (? kg) (9.8m/s^2) (-8m)
Στ =Στcw+Στccw= 1176 + ?
1176= (x) (9.8m/s^2) (-8m)
1176= (x)(78.4)
MC= 15 kg
10
Mystery D-
Στcw = (10kg) (9.8m/s^2) (8 m) = 784
Στccw = (? kg) (9.8m/s^2) (-8m)
Στ =Στcw+Στccw= 784 + ?
784= (x) (9.8m/s^2) (8m)
784= (x)(78.4)
MD= 10 kg
Mystery ESince, there is static equilibrium, the torques on the both sides are equal:
mdg = mhgdh
Στcw = (5 kg) (9.8m/s^2) (3 m) = 147
Στccw = (? kg) (9.8m/s^2) ( 5 m)
Στ =Στcw+Στccw= 147 + ?
147= (x) (9.8m/s^2) (5 m)
147 = (x)(49)
ME= 3.55 kg
11
Discussion:
Part one of the lab simulation, the weights that were placed on the left times the distance
to the axis of rotation had equal the weights that were placed on the right times its distance to
the axis of Rotation. During the simulation, a plank that is pivoted at its center sits on two
supports representing the Equilibrium. For us to keep a rigid body in rotational equilibrium,
the sum of the torques acting on it had to be zero. Maintaining total equilibrium is the goal in this
experiment, both translational and rotational equilibrium must be maintained. Therefore, there
must be no net forces or torques acting on the body. The weight was determined by acquiring the
given masses times the gravity constant in the units of Newtons. Once it was found, the torque was then
calculated by getting the weight times the distance (m) in corresponsales of each mass. Once the
torque was determined on the left and right side, the total torque is obtained. To find the mass
of mystery object F in trial 4, We know the torque of the right side is equal to 490 N * m. Therefore,
the torque of the left side is also equal to 490 N * m. We know the distance of the object on the left
is 1 meter. We can calculate the weight of the object on the left, in Newtons. The derived weight will then
allow us to calculate the mass ,in kilograms, by manipulating the formula for weight. Making a new formula
of mass equals weight (N) divided by the gravity constant (9.81 m/s)
Conclusion:
Part I:
In Part I of the experiment the weight was determined by getting the given masses and the gravity
constant. Once the weight was found, it was used to then determine the torque. This part of the
experiment allowed us to view the weights on the left times the distance to the axis of rotation must
equal the weights on the right times its distance to the axis of rotation.
Part II
In part 2 of the lab we were able to determine the masses of the mystery objects while also
calculating total counterclockwise torque and total clockwise torque. In calculating for torque during
the trials we found that each left and right side were equal to each other. The derived weight allowed
us to convert the masses to find our weights.
12
Questions:
1. Does the equilibrium condition (equation 1) depend upon the number of forces and
thereby sources of torque on either side of the fulcrum?
Yes, it does depend on the number of forces and the sources of the torque on either side of
the fulcrum.
2. What are the conditions necessary for translational equilibrium?
The net force should be equal to zero
3. What conditions are necessary for rotational equilibrium?
The net torque should be equal to 0
4. For the see-saw, what provides the upward component of force needed to keep
everything in translational equilibrium
The fulcrum provides the upward motion.
5. In the figure below a plank is pivoted at its right end as shown. Each of the marks
represents 1 m (i.e. the length of the plank is 9 m). A rope attached to the hook pulls up
on the plank while a 65 kg mass sits on the plank as shown. What is the reading of the
scale?
W=
65 𝑁 9.81𝑁6
8
= 478.2 N
13