**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, F_{net} = SF
= 0, and equilibum of torques, t_{net} =
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 F_{V} and F_{H} 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, **r**_{T},
to **T** is q + 90°. Then
.
The net torque equation is:
Solving for T,
Substituting this expression for T back into the F_{net,y}
equation and simplifying gives us F_{v}.
Substituting T into the F_{net,x} equation and
simplifying gives F_{H}.
As a check on our results, examine T. The horizontal
component of T is ½F. This is equal to F_{H}. 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 F_{V} 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.
- The torque of F is 0.
- The torque of F
_{V} is 0, because it's line of
action passes through the axis of rotation.
Thus, the non-zero torques are those due to T and F_{H}.
The horizontal component of T is the only component that will contribute
to the torque about the chosen axis of rotation. T_{x} and F_{H}
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, T_{x} = F_{H} = ½F.
From this point, T and F_{V} can be found. |