Can we TORQUE?

You have been using torque since before you were born, so you are an expert on the subject. Torque is for doing the twist. To twist, you apply a force on something in a direction not through the center of rotation, and torque is defined as the force times the distance to the force-line along a perpendicular to the force-line. This is known as the cross product of the distance vector and the force vector, but you might not need to get into that yet. t = F∙d_^ or F_^∙d

What I call the center of rotation is really any point, to tell you the truth. Many times, the choice of that point is obvious and people do not bother to mention it. If you are an auto mechanic and the specs say to tighten up that bolt with a final torque of 20 foot-pounds, you do not ask your boss "Hey, boss-man, it says 20 ft-lb but it doesn't say with respect to what point, so wha do I do?"

"Look for a new job, that's what."

Consider a door, free to rotate on its hinges. In your mind, push on it in various ways and note how effective those forces are to cause rotation of the door. You have known for many years the importance of that perpendicular distance to the hinges and the amount of force applied. Now it is time to formalize it. If the door knob is 3 feet from the hinge-edge of the door, and if you pull or push on the knob with a force of 2 lb perpendicular to the door, then the torque is 6 ft lbs. If your force is not perpendicular, use the perpendicular component.

Consider the door again. The force of gravity acts on the door, and it turns out that this can be treated as one single force acting on the center of gravity, which for a uniform door, is at the geometric center. Does this exert a torque on the door? Now we do need to specify a reference point. Take the lower hinge. Gravity is trying to rotate the door about a horizontal axis through the lower hinge. It does not succeed because there is another torque opposite to the gravity torque, exerted on the upper hinge. If you remove the upper hinge, to keep the door in its proper location you would need to exert a force on the door toward the wall where the hinge was attached. Thus you would be exerting a torque equal and opposite to gravity's torque. In these equilibrium situations we define a positive rotation (clockwise or counter-clockwise, it doesn't matter). Then we have positive and negative torques and their sum is zero. A typical problem would give you the weight of the door and various distances, then it would ask for the forces on the upper hinge. A well-hung door would have the vertical forces distributed approximately equally, so that a 2-hinge door would have half the weight on each.

The picture above shows forces on the door. Imagine the door off its hinges and you are holding it with each hand at a hinge location. To make the door have its usual orientation, your lower hand would need to push on it to the right and upper to the left as shown in the picture.

Then go back to the main page on mechanics.

Other main pages:
fluid/heat/e&m,
vibrations/waves,
quantum
or alphabetical list of topics, definitions, short discussions: index

Comments, questions: email fredrick.gram @tri-c.edu (remove space before @)