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 uniform circular motion. It is meant as a  prep for class.

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Introduction

The beam holding up the boxes are in static equilibrium. This is because they meet the two conditions of static equilibrium.

  1. All the forces horizontally and vertically add up to zero, in other words the beam is not accelerating horizontally or vertically.
  2. The beam is not experiencing any angular acceleration. For us, this will simplify to the fact that the beam is not rotating. (We will not examine constant angular velocity when it is not equal to zero.)

The concept of, "torque," is used to describe the tendency for a body to rotate. Torque is used to mathematically set up the 2nd condition of equilibrium.

Torque

To spin or rotate a car's tire. A force must be applied at some distance away from the axis the tire is spinning around. This force must be pointing tangent to the rotation's direction. At least part of the force must be pointing tangent to the rotation.

the center of the rotation where the axle is located is called the "pivot" or the "pivot point." This ability of a force to cause a rotation is called a moment. The distance between where the force is applied and the axis of rotation is called a moment arm. To physicists a torque is another name for a moment and is the preferred term.

Torque is defined as the product of distance times the perpendicular component of force about a point in space.

Where is the symbol for torque. It looks like a capital " T " with a tail to the right, but is lower case in size. The unit of torque is a "N•m." A, "N•m", is pronounced as if it were one word, "Newtonmeter." d is the symbol for the distance between the axis and the point where the force is applied. (Also called the moment arm.) stands for the force that is 90° to the moment arm's vector.

Torques cause the tendency to rotate.

To calcualte the value of a torque, multiply the magnitude of the perpendicular force times the distance between the pivot point and where the force is applied.

Any force that

 

Example

This wrench is applying a torque that rotates the screw. The torque here is
τ =(0.25m)(100N) = 25Nm
The torque here is
τ=(100Ncos27)(0.25m)=22.4Nm

 

In the picture on the far right, above, right, the component of force parallel to the displacement is found using trigonometry. It is (100N)cos27°. This value times the distance of 0.25m gives the torque.

When the force is pointing at the pivot point, You cannot make a right triangle using the force as the hypotenuse and having one side perpendicular to the moment arm. Therefore, when the force points towards or away from the pivot point, no torque is generated.

Take this 5 question fill-in-the-blank, Quia, reading quiz on the content of this page. When you get to the web page, it will ask you for your name, you can skip this and continue on to the questions.
Below is a second quiz about torques.

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