Torque and the Second Condiction of Equalibrium
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This text is meant to accompany class discussions. It is not everything there is to know about the basics of torque and the two conditions of equilibrium. It is meant as a prep for class. More detailed notes and examples are given in the class notes, presentations, and demonstrations (click here.)
 
Questions Click for the questions that go with this reading
2 Conditions of Mechanical Equilibrium
When a body is in mechanical equilibrium,
    1. it is not moving in any linear direction AND
    2. it is not rotating..

If both of these conditions are met, then scientists and engineers say, "The body is in mechanical equilibrium." There are two other forms of eqilibrium called static and dynamic equilibrium. Static equilibrium means the body is moving at a constant velocity. Mechanical equilibrium is a subset of static equilibrium. Dynamic equilibrium is where a body is accelerating and moving at constant velocity at the same time. An example of this is a ball that is spinning while not moving linearly. Every spinning object experiences a centripetal acceleration. In the example the object is spinning but not experiencing any translational motion.

 

  • The red car is in mechanical and static equilibrium.
  • The blue car is only in static equilibrium
  • The spinning "eight ball" is in dynamic equalibrium

This unit will only deal with mechanical equilibrium.

 

Putting it All Together

When you solved free body problems, in a previous unit, you were applying one of the conditions of mechanical and dynamic equilibrium by summing up the forces. This is called the first condition of equilibrium. It looked like this...

Sum of the forces example image

The second conidition of equilibrium states that a body is not rotating. This is mathematically stated by saying that, "The sum of torques about any point is zero." This means you have an tool that can be used when solving problems. It looks like this...
Here is a video describing how the math sentence above makes sense.

This YouTube video is here, http://youtu.be/PhOCrlQqI3A

Solution Strategies

 

When do you use the summing up the torques technique?

  • Use summing up the forces and torques when the question asks how far away from a point a force is applied.This includes questions that ask...
      • how far a person can walk on a board or beam.
      • Or where can a weight be placed on beam
      • Or where the lifting force needs to be placed. All these questions involve a force(s) and a distance. -and that's a torque.

General strategy when solving equilibrium problems

  1. Draw an extended free body diagram, "EFBD."
  2. Sum up the forces in the "x" and "y" directions.
    • Try to solve with this information.
  3. Sum up the torques
    • Choose an axis of rotation where you don't know a variable.
  4. Solve

 

Example 1: Using a torques to write a math sentence
  • Question
  • Solution
  • Video Solution

Below is an extended free body diagram for a beam that is held horizontally. Each letter represents a force. Write the math sentence that sums up all the torques.

Example answer:


This is a YouTube video that can be found at, http://goo.gl/9gNQlI

 

Example #2 Beam Question and Solution
 
  • Question
  • Strategy
  • Solution on Paper
  • Video Solution
Question

Strategy

  • Sum up the force in the x direction.
  • Sum up the forces in the y-direction.
  • Sum up the torques about a point on the beam. You can pick ANY point on the beam. Wherever the point is located, all force that point towards or away from the point will not exert a torque about this point. Mathematically this means that the force in toque equaiton will dissappear. Therefore, pick a pivot point where an unknown force points towards or away from. This is one of the secrets to using toque to solve equilibrium problems.
  • Use some or all of the three eqautions you created to solve for an unknown.
Solution

This YouTube video is found at, http://youtu.be/e_T0s-gpjnE

 

Example #3 Balancing
 
  • Question
  • Solution on Paper
  • Video Solution
Solution

This video can be is on YouTube at, http://youtu.be/b7EA_F38hn4

 
Example #4 (The "Ladder Problem")
  • Question
  • Paper Solution
  • Video Solution

A 5.00 m, 200.0 N, long ladder rests on a wall. The ladder’s center of mass is 3.00 m from the bottom of the ladder, along the ladder. There is no friction on the wall, but there a coefficient of friction along the ground of 0.300. How far along the ladder can a 750. N person climb above the ground?

ladder Setup

Solution

 
 

by Tony Wayne ...(If you are a teacher, please feel free to use these resources in your teaching.)