Skip to content Skip to navigation Skip to collection information
Inside Collection (Book): Accessible Physics Concepts for Blind Students
Summary: This module explains rotational equilibrium in a format that is accessible to blind students.
This module is part of a collection (see http://cnx.org/content/col11294/latest/ ) of modules designed to make physics concepts accessible to blind students. The collection is intended to supplement but not to replace the textbook in an introductory course in high school or college physics.
This module explains rotational equilibrium in a format that is accessible to blind students.
In addition to an Internet connection and a browser, you will need the following tools (as a minimum) to work through the exercises in these modules:
The minimum prerequisites for understanding the material in these modules include:
I recommend that you also study the other lessons in my extensive collection of online programming tutorials. You will find a consolidated index at www.DickBaldwin.com .
This module will consist of a short discussion followed by a lot of hands-on practical problem solving.
Translational and rotational equilibrium
You learned in an earlier module that an object is in translational equilibrium when the net force acting on it is zero. In other words, the vector sum of the forces acting on the object must be zero.
However, you also learned in an earlier module that it is possible for the net force to be zero, while the net torque is not zero. If the net torque is not zero, the object would have non-zero acceleration, and therefore would not be in rotational equilibrium.
Be careful when climbing a ladder
When you are climbing a ladder, for example, it would be highly undesirable for the system that includes you and the ladder to not be in rotational equilibrium.
Conditions for translational and rotational equilibrium
While zero net force is sufficient to ensure translational equilibrium, zero net torque is also required to ensure rotational equilibrium. When both translational and rotational equilibrium is achieved, the object is said to be in a state of static equilibrium.
In order to achieve static equilibrium, the sum of the forces acting on the object must be zero and the sum of the torques about all axes must also be zero.
Choosing an axis of rotation
If an object is not already rotating, it may not be obvious how to determine the axis about which the object is likely to rotate. You need to ensure that the sum of the torques about that axis are zero.
As it turns out, you can choose just about any axis you please when doing the calculations to confirm that the sum of the torques are zero. You touched on this in an earlier module involving the parallel axis theorem. It can be shown that if the net force acting on an object is zero and the net torque about one axis is zero, the net torque about every other parallel axis will also be zero.
The worst case scenario
In the worst case, therefore, you can ensure rotational equilibrium by ensuring that the net torque about three orthogonal axes are zero. In some cases, it will be obvious that you only need to worry about one axis. In some cases, you would need to worry about two or three axes.
A ladder
Before climbing a ladder, for example, you need to ensure rotational equilibrium about
The lack of rotational equilibrium about any of these three axes can result in a nasty fall from a ladder.
Ensuring rotational equilibrium for a ladder
Ladder manufacturers install special "feet" with a high coefficient of friction on ladders to avoid accidents from the first item in the above list. These feet are designed to prevent the ladder from sliding away from the wall that it is leaning on.
They also install wide outriggers on the base of very tall ladders to avoid accidents from the second item in the list. The outriggers are designed to prevent top of the ladder from sliding sideways on the wall.
Tree trimmers that use ladders to trim trees take special precautions to prevent accidents from the third item in the list in those cases where both sides of the top of the ladder are not firmly supported.
Simplification through the choice of an axis
A judicious choice of axis can often greatly simplify the solution of an equilibrium problem by causing one or more forces to go through the axis. This eliminates the effect of those forces from the computation of rotational equilibrium.
I will present several example scenarios involving rotational equilibrium in this section. In several cases, I will ask you to construct a replica of the scenario on your graph board.
All scenarios assume that counter clockwise rotation is the positive direction of rotation. All scenarios also assume that the positive horizontal axis is to the right and the positive vertical axis is up.
Using your graph board, draw a picture of a beam of length L hanging from a rope at its center. Hang a mass M1 from the left end of the beam. Hang a mass M2 from the right end of the beam. Hang a mass M3 half way between M1 and the tie-point of the rope.
Assume that the system is in static equilibrium.
Find the value of M3 given the following:
Solution:
Choose the tie-point of the rope as the axis of rotation.
The net torque about the axis of rotation is
Tnet = (L/2)*M1 + (L/4)*M3 - (L/2)*M2
For rotational equilibrium, the net torque must be equal to zero.
Rewriting and simplifying yields
L*(M1)/2 + L*(M3)/4 = L*(M2)/2
Dividing both sides by L yields
(M1)/2 + (M3)/4 = (M2)/2
Solving for M3 yields
M3 = 2*M2 - 2*M1
Substituting numeric values yields
M3 = 2*2kg - 2*1kg, or
M3 = 2kg
If the mass of the beam is 2kg, what is the tension in the rope?
Solution:
In order for the system to be in translational equilibrium, the downward forces must be equal to the upward force exerted by the rope.
tension = (M1 + M2 + M3 + Mbeam)*g
Substituting numeric values yields
tension = (1kg + 2kg + 2kg + 2kg)*9.8m/s^2, or
tension = 68.6 newtons
Draw the following picture on your graph board. A vertical wall is on the left. A beam is attached to the wall with a hinge and extends outward horizontally to the right. (Without further support, the beam is free to rotate around the hinge causing the end of the beam to move up and down.)
To support the horizontal orientation of the beam, a cable is attached to the end of the beam and is attached to the wall above the attachment point of the beam. The angle between the beam and the cable at the end of the beam is 30 degrees.
Assume that the system is in static equilibrium.
Assume that
Draw vectors showing all of the forces that are exerted on the beam.
Solution:
The following forces are exerted on the beam:
What is the most judicious location for computing the sum of the torques to establish rotational equilibrium?
Solution:
This is probably a matter of opinion. However, if the torques are computed around the point where the beam attaches to the wall, three of the forces listed above pass through that point and can be ignored when computing the sum of the torques around that point. Therefore, I consider that to be the most judicious point.
Given the above assumptions , what are the magnitudes of the forces labeled V, H, and T required to produce translational and rotational equilibrium?
Solution:
We begin by determining the required value of the tension in the cable, T, to produce rotational equilibrium.
Solve for the value of T
Choose the hinge as the axis of rotation for the reasons given above.
For rotational equilibrium,
L*T*sin(30 degrees) - (L/2)*M*g = 0
Dividing both sides by L yields
T*sin(30 degrees) - M*g/2 = 0
Note that the actual length of the beam is not important for this computation.
Solving for T yields
T = (20kg*(9.8m/s^2)/2 )/sin(30 degrees), or
T = 196 newtons
Therefore, the tension in the cable is a force of 196 newtons directed from the end of the beam toward the wall at an angle of 30 degrees north of west.
Solve for the value of V
Now that we know the tension in the cable, we can compute the values for V and H.
For translational equilibrium in the vertical dimension,
V + T*sin(30 degrees) - M*g = 0, or
V = M*g - T*sin(30 degrees)
Inserting numeric values from above,
V = (20kg*9.8m/s^2) - 196newtons * sin(30 degrees), or
V = 98 newtons
Check the result for V
We can check this result by computing the sum of the torques about the center of the beam.
(L/2)*T*sin(30 degrees) - (L/2)*V = 0
Dividing through by L/2 yields
T*sin(30 degrees) - V = 0
Inserting numeric values yields
196*sin(30 degrees) - 98 = 0
and the values for T and V check.
Solve for the value of H
Knowing the value of T also makes it possible for us to compute the value of H.
For horizontal equilibrium,
H - T*cos(30 degrees) = 0, or
H = T*cos(30 degrees)
Inserting numeric values yields
H = 196newtons*cos(30 degrees), or
H = 169.7 newtons
Draw the following schematic diagram of a crane on your graph board.
The crane consists of a boom attached to a horizontal surface with a pin. (The pin acts as a hinge and the boom can rotate around the pin.) The boom extends upwards and to the right at an angle of 30 degrees relative to the surface.
A weight of 6000 newtons is hanging from the end of the boom.
A vertical cable is attached to a winch on an overhead beam and is also attached to the boom 2 meters from the lower end.
Assumptions:
Draw a vector diagram showing the vertical forces being exerted on the boom.
Solution:
Three vertical forces are exerted on the boom:
What is the downward force on the cable?
Let the lower end of the boom be the axis of rotation.
Compute the torques due to the cable and the weight. The sum of those torques must be zero for rotational stability.
2m * F*cos(30 degrees) - 8m*6000 newtons * cos(30 degrees) = 0
Divide both sides of the equation by cos(30 degrees)
2m*F = 8m*6000 newtons, or
F = (8M*6000newtons)/2m, or
F = 24000 newtons
Therefore, the force pulling down on the cable is 24000 newtons.
What is the magnitude and direction of the force on the pin at the bottom of the boom?
For vertical equilibrium, the vertical forces must sum to zero.
V + 24000 newtons - 6000 newtons = 0, or
V = 6000 newtons - 24000 newtons, or
V = -18000 newtons
Therefore, the force on the pin is 18000 newtons down.
A crate is being slid to the right on a horizontal floor by pulling on a rope tied to the right side of the crate.
Width of the crate = w = 4 m
Height of the crate = h = 3 m
Coefficient of kinetic friction = u = 0..5
Attachment point of rope = d = ?
Draw a picture of the crate showing the forces that are being exerted on the crate.
Solution:
The crate is a rectangle with the dimensions given above. There are three forces acting on the crate.
There is a downward force at the bottom center of the crate, which is the weight of the crate. Label it F1.
There is a horizontal force pointing to the left at the bottom of the crate. This is the force of friction. Label it F2.
There is a horizontal force pointing to the right on the right side of the crate that is d units above the floor. This is the force that is pulling the crate to the right. Label it F3.
What is the maximum value for d that allows the crate to slide without tipping over?
Compute the sum of the horizontal forces.
F3 - F2 = 0, or
F3 - m*g*u = 0, or
F3 = m*g*u
Compute the torques about the bottom right corner.
(w/2)*F1 - d*F3 = 0, or
d*F3 = (w/2)*F1, or
d*F3 = (w/2)*m*g
Substitution yields
d*m*g*u = (w/2)*m*g
Simplification yields
d*u = (w/2), or
d = (w/2)/u
Inserting values yields
d = (4/2)/0.5, or
d = 4 m
Since the height of the crate is only 3 m, it is not possible to tip it over by pulling on a rope attached to the right side of the crate.
The boom of a crane is pinned at the intersection of a horizontal floor and a vertical wall. The boom slopes toward the upper right.
A mass hangs from the top end of the boom.
The top end of the boom is connected to the vertical wall by a cable that is stretched horizontally from the wall to the boom.
Length of cable = X = 4 m
Height of cable = Y = 3 m
Mass = M = 2000 kg
The weight of the boom is negligible.
Draw the forces acting on the boom.
Solution:
There are four forces acting on the boom:
A downward force that is attributable to the mass hanging on the top end of the boom. Label this force F1.
A force pointing to the left from the top end of the boom due to the cable. Label this force F2.
A force pointing up at the bottom end of the boom. Label this force F3.
A force pointing to the right at the bottom of the boom. Label this force F4.
Find the tension in the cable.
Solution:
For rotational equilibrium, the sum of the torques about the bottom of the boom must be zero.
F2*Y = F1*X, or
F2*Y = M*g*X, or
F2 = (M*g*X)*/Y
Inserting numeric values yields
F2 = (2000kg*(9.8m/s^2)*4m)/3m
F2 = 26133 newtons
Therefore, the tension in the cable is equal to 26133 newtons
Find the force pushing against the bottom end of the boom along the length of the boom. Also find the angle between that force and the floor.
For horizontal equilibrium
F4 = F2 = 26133 newtons
For vertical equilibrium
F3 = F1 = M*g = 2000kg*9.8m/s^2, or
F3 = 19600 newtons
The magnitude of the force = sqrt(F4^2 + F3^2), or
Magnitude = sqrt(26133^2 + 19600^2), or
Magnitude = 32666 newtons
Therefore, the magnitude of the force is 32666 newtons.
The angle of the force is
angle = atan(19600 newtons/26133 newtons) in degrees
angle = 36.9 degrees north of east
A ladder is sitting on a horizontal floor and leaning against a smooth wall. A bucket of paint is hanging from a rung 80-percent up from the bottom of the ladder. The parameters are as follows:
Draw a vector diagram of the forces acting on the ladder.
Solution:
There are five forces acting on the ladder:
The ladder is in static equilibrium.
The coefficient of static friction between the ladder and the wall is zero.
What is the minimum possible value for the coefficient of static friction between the ladder and the floor.
F1 = ?
F2 = m*g = 2.5kg * 9.8m/s^2 = 24.5 newtons
F3 = M*g = 12kg*9.8m/s^2 = 117.6 newtons
F4 = F5*u
For vertical equilibrium,
F5 = F2 + F3 = (2.5kg + 12kg)*9.8m/s^2 = 142 newtons
For rotational equilibrium, the sum of the torques about the bottom of the ladder must be zero.
Compute the horizontal distance to the line of action of the weight of the ladder.
X1 = (L/2)*cos(50 degrees), or
X1 = ((5/2)m)*cos(50 degrees), or
X1 = 1.61m
Compute the horizontal distance to the line of action of the weight of the bucket of paint.
X2 = 0.8*5m*cos(50 degrees), or
X2 = 2.57m
Compute the vertical distance to the line of action of the horizontal force at the top of the ladder.
Y1 = 5m*sin(50 degrees), or
Y1 = 3.83m
Compute sum of the torques about the bottom of the ladder.
Y1*F1 - X2*F2 - X1*F3 = 0, or
F1 = (X2*F2 + X1*F3)/Y1, or
F1 = (2.57m*24.5 newtons+ 1.61m*117.6 newtons)/3.83m
F1 = 65.87 newtons
For horizontal equilibrium,
F4 = 65.87 newtons
By substitution,
F4 = F5*u = 65.87 newtons
u = (65.87 newtons)/F5, or
u = (65.87 newtons)/142 newtons, or
u = 0.46
While the coefficient of static friction could be higher than this and still achieve static equilibrium, if the coefficient were any lower, the ladder would slide away from the wall.
I encourage you to repeat the computations that I have presented in this lesson to confirm that you get the same results. Experiment with the scenarios, making changes, and observing the results of your changes. Make certain that you can explain why your changes behave as they do.
I will publish a module containing consolidated links to resources on my Connexions web page and will update and add to the list as additional modules in this collection are published.
This section contains a variety of miscellaneous information.
Financial : Although the Connexions site makes it possible for you to download a PDF file for this module at no charge, and also makes it possible for you to purchase a pre-printed version of the PDF file, you should be aware that some of the HTML elements in this module may not translate well into PDF.
I also want you to know that I receive no financial compensation from the Connexions website even if you purchase the PDF version of the module.
Affiliation : I am a professor of Computer Information Technology at Austin Community College in Austin, TX.
-end-
More about this module: Metadata | Downloads | Version History
This work is licensed by Richard Baldwin under a Creative Commons Attribution License (CC-BY 3.0), and is an Open Educational Resource.
Last edited by Richard Baldwin on Oct 8, 2012 1:33 pm -0500.
"Blind students should not be excluded from physics courses because of inaccessible textbooks. The modules in this collection present physics concepts in a format that blind students can read […]"