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For a rigid body to be in a complete state of equilibrium it must first be in a state of
where the sum of all of the forces equals zero.
Then, we must also place it in a state of
where the sum of all of the torques equals zero. For horizontal beams in the plane of the page, the torques will either result in a clockwise (cw) rotation or a counterclockwise (ccw) rotation.
An important item to note is that
the location of the pivot point is absolutely arbitrary
- that is, you can calculate your torques about any convenient point. To simplify your work, it is recommended that you choose a point where the torques of as many of the forces as possible will be zeroed out - that is, a common point through which their "lines of action" will pass. As you become more familiar with these problems, you will develop a sense of the best locations to use.
We will now illustrate the steps you should follow to solve problems dealing with rotational equilibrium by examining an assortment of beam and force configurations.
first type of problem
involves horizontal beams and vertical forces.
Refer to the following information for the next eight questions.
In the following diagram, develop an expression for the mass of the fish and the tension in the rope based on the mass of the blue block, M
, and the mass of the purple block, M
First we must set up the equations for static equilibrium:
Develop expressions for these two conditions.
Now we will begin to develop the beam's equation for rotational equilibrium. Using the point where the rope attaches to the rod as our pivot point, what are the moment arms for each of the forces?
Which forces will produce a clockwise torque? Which ones produce a counter-clockwise torque?
Write expressions for each of the non-zero torques.
Now write the equation balancing the torques.
Solve your equation for torque for the weight of the fish.
What is the mass of the fish?
What is the tension in the supporting rope?
This CP workbook page
will provide you with further practice calculating torques produced on horizontal beams by vertical forces.
Now let's examine
a second type of problem
in which the beam remains horizontal but the forces have a mixture of diagonal, horizontal, and vertical orientations.
Suppose we have a uniform, horizontal beam and we are asked to calculate the tension in its cable.
Since the beam is a rigid body, we must satisfy all three of the following conditions if we wish to place it into a state of complete equilibrium:
Before constructing any of these equations, we must first take components of the tension in the cable and remember that there is (1) a horizontal force pushing outward on the beam and (2) a vertical force supporting the beam where it contacts the wall.
If the pivot is placed at the hinge where the boom meets the wall, H, V, and T cos 30º have no moment arms since they pass through the pivot. Listed below are our three equations. We will not simplify them any further since we do not know a specific value for the beam's mass.
H = T cos 30º
V + T sin 30º = Mg (where Mg represents the weight of the beam)
T sin30º(L) = Mg(½L)
A third type of problem
involves diagonal beams as well as diagonal, horizontal, and vertical forces. In this type of problem you may be required to take components not only of the applied forces but also of the beam in order to determine the moments arms of each respective force.
For example, examine the following problem where you are asked to find the tensions in the two ropes that are connected to the crane's 500-N boom as well as the horizontal and vertical components of the force acting on the hinge.
The first step in this solution is to resolve any diagonal forces into their components. This is illustrated in the following diagram.
Step two is to determine the moment arm for each force. We will be using the hinge as our pivot point.
As you can see in the above diagram, when a force is oriented horizontally, its moment arm is oriented vertically.
the horizontal component, T
cos 37º, has a vertical moment arm equal to L sin 53º
the vertical component, T
sin 37º, has a horizontal moment arm equal to L cos 53º
the vertical weight, mg, has a horizontal moment arm equal to ½L cos 53º
the vertical tension, T
, has a horizontal moment arm equal to L cos 53º
the horizontal and vertical components of the hinge, H and V, do not have any moment arms since their lines of action pass through our pivot point
The equations that satisfy our three conditions become:
H = T
V = mg + T
sin 37º + T
mg(½ L cos 53º) + T
sin 37º(L cos 53º) + T
(L cos 53º) = T
cos 37º(L sin 53º)
Substituting in mg = 500 N and T
= 4000 N give us that T
= 9100 N. Knowing the value of T
will now allow you to calculate the value of the vertical component of the hinge, V, in the equation for
and the value of the horizontal component of the hinge, H, in the equation for
A Physical Pendulum, The Parallel Axis Theorem and A Bit of Calculus
Conservation of Momentum in Two-Dimensions
Density of an Unknown Fluid
Mass of a Paper Clip
Moment of Inertia of a Bicycle Wheel
A Chart of Common Moments of Inertia
A Further Look at Angular Momentum
Center of Mass
Centripetal Acceleration and Angular Motion
Discrete Masses: Center of Mass and Moment of Inertia
Introduction to Angular Momentum
Rolling and Slipping
Rotational Dynamics: Pivoting Rods
Rotational Dynamics: Pulleys
Rotational Dynamics: Rolling Spheres/Cylinders
Rotational Kinetic Energy
Thin Rods: Center of Mass
Thin Rods: Moment of Inertia
Torque: An Introduction
The Baton Twirler
The See-Saw Scene
Center of Gravity
Torque: Cams and Spools
Center of Gravity
Center of Gravity vs Torque
Moments of Inertia and Angular Momentum
Practice: Uniform Circular Motion
Rotational Kinetic Energy
Torque: Rotational Equilibrium Problems
Basic Torque Problems
Center of Mass (Discrete Collections)
Moment of Inertia (Discrete Collections)
Rotational Kinematics #2
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Catharine H. Colwell
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