The Physics Philes, lesson 50: Torque It
Holy cow. We’re on lesson 50 of the Physics Philes. I’ve learned so much on this endeavor, and yet, there is so much to go. So onward! This week we will start a discussion of torque.
We know from previous installments that the net force applied to a body gives that body its acceleration. But what if we’re dealing with angular acceleration? In that case, we need to talk about torque, which is the twisting or turning effort of a force. When determining how a force changes the rotational motion of an object, three things are important: the magnitude of the force, the direction of the force, and the point on the body where there force is applied. The quantitative measure of the tendency of a force to cause or change a body’s rational motion is called torque.
Think about loosening a bolt or something with a wrench. If you exert a force at the bottom of the handle, it’s more effective than if you were to exert a force closer to the top. And if you pull or push the parallel to the handle? Well that’s not effective at all.
Before we get deeper into this, we need to define a few things. To do that, I’m going to draw you a diagram, and it’s going to be pretty incomprehensible. Just try to calm down. It will be OK.
Stop hyperventilating! I told you to remain calm! I’ve got you. Let’s take this one step at a time.
Let’s say that the body in this diagram (the square) rotates around an axis perpendicular to point O. This diagram illustrates three forces acting on the body: F1, F2, and F3. Let’s define some terms using F1.
The tendency of F1 to cause the body to rotate around point O depends in part on the magnitude of the force, but it also depends on the perpendicular distance l1, which is the distance between point O and the line of action of the force. The line of action is the line along which the force vector lies. In the diagram, the line of action is indicated in blue. The distance l1 is called the lever arm or the moment arm of force F1 about point O. The twisting effort is directly proportional to F1 and l1, so the torque of the force F1 with respect to O is F1 multiplied by l1. We can do the same thing with the other forces indicated on the diagram. Thus, the equation for torque (represented by τ) is:
Look for a second at F3. It’s line of action is through point O. That means that the value of its lever arm is equal to zero, which means that the torque of F3 with respect to point O is zero. This is what we’d expect.
This is just one way of finding torque. There are actually three methods. You could use the method just discussed. You could also use the angle between two vectors. In that case, the lever arm is equal to rsinΦ, which would make τ = rFsinΦ. Alternatively, you could represent force F as a radial component and a tangential component.
If you’re like me, neither of the two additional methods will make sense right now. Don’t worry. Next week I’ll show you exactly how they work. But right now we have more basics to pin down.
Do you remember several weeks ago when we discussed radians and angles and stuff? For an angle to be positive, we have to measure from the positive x-axis to the terminal side of an angle in the counterclockwise direction. To get a negative angle, we measure from the positive x-axis to the terminal side in the clockwise direction. The same can said for how we measure torque. The force F1 would tend to cause rotation in the counterclockwise direction, which means the value is positive. Force F2 tends to cause rotation in the clockwise direction, which means the value for torque is negative.
Alrighty. That’s it for today. Next week I’ll clarify how to apply the methods for finding torque. I’m excited for it already!
Featured image credit: Flickr