13.4   Some Examples of Static Equilibrium

Suppose you want to lift a 100-lb rock using a light (but strong) 8-ft rod. You place the support so that it is 2 ft away from the rock and 6 ft from the end where you are pushing. What is the minimum weight you need to put on your end of the rod to hold up the rock?


This is relatively a simple problem involving static equilibrium. However this problem will be solved with the greatest details possible to make the concepts very clear and show you all the techniques of solving problems of this type. (You have more example problems listed at the "end of the section box" at the bottom).

Assume the object is in equilibrium horizontally as shown in the figure and find just the value of F to maintain this equilibrium, but how?

Click on each item below to expand/collapse:

Identify the object on which you are going to apply the conditions of equilibrium. Is it

  • the rock?
  • the support? Or
  • the rod?


The rod, of course. We want to see how the rod (not the rock or the support) is in equilibrium under the action of the all the forces acting on it.

After identifying the object on which you are going to apply the equilibrium condition, draw the free body diagram (FBD) of that object, together with the wisely chosen x-y direction.

Note: You have to draw the whole object while drawing the FBD; you cannot just represent it by a dot as you would for a particle-like object. This (rod) is an extended rigid body and we need to know dimensions to calculate the torque.







The yellow dot shows where the support is and the red dot is the cog (center of gravity) point where the weight of the rod is acting. But this force of gravity on the light rod is so small we neglect it (for this particular example). The reaction force from the support on the rod (N) is shown vertically upward; there can't be any horizontal component as this will result in a net force horizontally (there is no other horizontal force to balance it).

Some selected points in the rod are labeled L (left end), S (support), C (center of gravity) and R (right end) for easy referencing.

Now look at the three (for 2D problems) conditions for equilibrium:

The first of the two translational equilibrium conditions, doesn't really give us anything but the trivial result .The second condition gives the result ; this (alone) cannot give us the value of F since there is another unknown, N, in the equation.

Let us now look at the third condition (rotational equilibrium), which says that the net torque about any point P you choose is zero. Here we are at liberty to choose any point as we like. So which point would you choose? L, S, C, R or any other point not labeled? This is where we need to act wisely. We choose point S to apply the torque equation because this will result in an equation which contains only the one unknown we are interested in:

about point L

about point C

about point S

Note that the torque of N is positive when taken about the point L (N tries to rotate the rod counter-clockwise about a rotation axis passing through that point L) but negative when taken about the point C (N tries to rotate the rod clockwise about a rotation axis passing through that point C)! The magnitude of the torque is very easily calculated as the force x moment arm.

Solving eq. 4, we get the answer F = 33 lb.

You should also note that we can still find F (and N) by solving eq. 3 and eq.1 (or any other combination of two equations) as two independent equations with two unknowns. But, please don't be fooled into thinking that we have 4 independent equations, there are only two independent equations, and the other equations can be derived using those two equations you chose to be the independent equations.


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