|Solving Static Equilibrium Problems
See Sections 13.1-4 of your text for related material.
A static equilibrium problem is one in which both the linear and angular acceleration of the system is 0. Solving such a problem requires the application of both equilibrium of forces, Fnet = SF = 0, and equilibum of torques, tnet = St = 0. All of the characteristics of the these two problem types must be applied to solve a static equilibrium problem. In addition, heed the following as a way to simplify solutions:
Select the axis of rotation so that the torques of one or more forces will be 0. This is done by selecting the axis of rotation at the point from which a force acts or through which a force acts. The moment arm of the force will be zero; therefore, the torque due to the force will be 0.
Problem: A rigid, vertical rod of negligible mass is connected to the floor by an axle through its lower end. The rod also has a wire connected between its top end and the floor. If a horizontal force F is applied at the midpoint of the rod, find a) the tension in the wire, and b) the horizontal and vertical components of force exerted by the bolt on the rod.
Let's look at the net force problem first. The situation is diagramed to the right. Unlike net force problems that we've done in the past, note that the forces are drawn from their points of application on the rod. This will be important when we do the net torque part of the problem later.
The net force equations are:
We'll come back to these equations later when we need them. Let's look at the net torque problem next. We need to select an axis of rotation. If we select it at the point where FV and FH act, then those forces will contribute 0 torque. With that choice, the moment arms of F and T are shown to the right.
The position vector of F coincides with its moment arm, which equals L/2.
Note that the angle from the position vector, rT, to T is q + 90°. Then
The net torque equation is:
Solving for T,
Substituting this expression for T back into the Fnet,y equation and simplifying gives us Fv.
Substituting T into the Fnet,x equation and simplifying gives FH.
As a check on our results, examine T. The horizontal component of T is ½F. This is equal to FH. That makes sense, because equal horizontal forces of ½F on each end of the rod would not only balance F but would also keep the rod from rotating. As another check, note that the downward component of the tension force is Fsinq /(2cosq) = ½Ftanq. This is equal in magnitude to FV as is necessary for the vertical forces to be balanced.
Let's look at the equilibrium of torques problem using a different axis of rotation to show that the results are the same. Let the axis of rotation be the point of application of force, F. We won't redraw the diagram, but note the following.
Thus, the non-zero torques are those due to T and FH. The horizontal component of T is the only component that will contribute to the torque about the chosen axis of rotation. Tx and FH have equal moment arms of L/2 about that axis. Since they must produce equal and opposite torques, they must have equal magnitudes. Since their sum must equal F, Tx = FH = ½F. From this point, T and FV can be found.