Introduction:
If you are unsure how to find the magnitude of a torque using "force times moment arm", or if the use of the cross product of the radius vector and the force to get a torque is difficult for you, you should review Vector Algebra III (the previous minilecture).
The complete form of Newton's First Law covers rotational as well as translational equilibrium. Not only is the body "at rest or in uniform motion", it isn't rotating either. (Rotations, even at constant angular speed, aren't uniform motion since a radial acceleration is present.) If there isn't any acceleration, there can't be any net torque, since it is net torque that causes an acceleration.
Newton's First Law:
In an earlier module, we stated that the first law requires the sum of the forces to be the null vector. But that wasn't complete because rotational equilibrium was not included. The full statement is:
That is, there can't be any tendency for either the translational or rotational motion to change. When the forces lie in a plane (and the torques are perpendicular to the plane), it is easier to work with the scalar portions of these equations:
Since we have three equations, a maximum of three variables can be determined. Always count the variables that you're showing in the free body diagram. If you have more than three, you can't solve the problem (at least, for every variable). Notice that I've shown the sum of the torques about the z-axis = 0? You have to be careful interpreting this. It should actually be about an axis perpendicular to the plane containing the forces (that will usually be the page or screen). So there will be many possible "z-axes". I'll use a letter to indicate the point about which torques are being taken. Examples are the best way to show this. Incidentally, "moment of a force", "moment", and "torque" are used synonymously.
Equilibrium Examples:
Example 1 - A truck
crossing a bridge
Example 2 - A
ladder
Example 3 - A
marquee
Example 4 - A
crane
Example 5 - A "lead-you-by-the-nose" practice
crane problem
Example 6 - Wanna push a refrigerator around?
Center of Gravity:
The "center of gravity" (c.g.) of an object is that point at which, for the purposes of calculation, we can consider all of the weight to act. Consider a closed paper box which has several stones inside. You could find the line along which the c.g. lies by finding the point at which it would balance on your finger. Mathematically, this is really a torque problem. I'm showing just the bottom of the box below because I'm assuming that the box has very little weight to contribute and besides, it's uniformly distributed. To avoid cluttering the diagram, I also didn't show the coordinates for each stone - they are similar to those for stone #1. Your finger, which is supporting the box, is shown by the shaded arrow which has coordinates (xcg, ycg) and a magnitude of F.
First I take torques about the x-axis. Notice that the moment arms for the forces are y-coordinates?
From the sum of the forces in the z-direction (up in this diagram but not shown), we see that:
Substituting this into the equation for ycg and dividing out the gravitational acceleration g, we get:
A similar equation can be obtained for xcg. Compare this with Serway's discussion on pages 210 - 211. Here's an example of how this can be used for an object of uniform density and thickness. Here's another example where masses have to be used.