Free-Body Diagrams and Equilibrium

104 > Review the Knowledge You Need to Score High
Step I: Free-body diagram.
We'll use point A as die fulcrum to start with. Why? In a static equilbrium situation, since die
bridge isn't actually rotating, any point on die bridge could serve as a fulcrum. But we have two
unknown forces here, die forces of die supports A and B. We choose die location of one of these
supports as die fulcrum, because now that support provides zero torque—die distance from die
fulcrum becomes zero! Now all we have to do is solve for die force of support B.
The diagram below isn't a true "free-body diagram," because it includes both distance
and forces, but it is useful for a torque problem. Bob's weight acts downward right where
he stands. The bridges weight is taken into account with a force vector acting at the bridge's
center of mass; that is, 40 m to the right of pillar A.
80 m
20 m
Step 2: Vector components.
We don't have to worry about vector components here. (We would have if the forces had
not acted perpendicular to the bridge. Such a situation is much more likely to appear on
the C than the B exam—see the Physics C section below.)
Step 3: Equations.
Torquenel = counterclockwise - clockwise = 0.
(FB)(80 m) - [(100 kg-10 N/kg)(20 m) + (10,000 N)(40 m)] = 0.
Step 4: Solve. ^=5300 N
This is reasonable because pillar B is supporting less than half of the 11,000 N weight of the
bridge and Bob. Because Bob is closer to pillar A, and orherwise the bridge is symmetric,
A should bear the majority of the weight.
Of Special Interest to Physics C Students
It must be emphasized that even Physics C students must go through the four-step
problem-solving process described in this chapter. Frequently, Physics C students try to take
shortcuts, thinking that equilibrium problems are easy, only to miss something important.
If free-body diagrams are good enough for professional physicists to use, they are good
enough for you.
The Physics C exam will often expect you to find the torque provided by a force that
acts at an angle. For example, consider a force F acting on a bar at an angle 6, applied a
distance x from a pivot. How much torque does this force provide? See Figure 10.7.
,— Pivot
/
F
Figure 10.7