Torques on a Seesaw
As you read, pay attention to Example 9.1, which shows how to do a statics problem. Here, children are balanced on a seesaw. We are given information about the masses of both children, and how far from the pivot point one child is sitting. We are asked to determine where the second child is sitting to balance.
In Figure 9.8 we see that the children are balanced and therefore are at equilibrium. The free-body diagram shows that there is no net force, and no net rotational acceleration. To determine the distance of the second child from the pivot point, we use the torque equation, and set torque equal to zero. To determine the upward balancing force from the pivot point, we use the fact that net force equals zero to solve for the individual force at the pivot point.
When two children balance a seesaw as shown in Figure 9.8, they satisfy the two conditions for equilibrium. Most people have perfect intuition about seesaws, knowing that the lighter child must sit farther from the pivot and that a heavier child can keep
a lighter one off the ground indefinitely.
Figure 9.8 Two children balancing a seesaw satisfy both conditions for equilibrium. The lighter child sits farther from the pivot to create a torque equal in magnitude to that of the heavier child.
Example 9.1 She Saw Torques On A Seesaw
The two children shown in Figure 9.8 are balanced on a seesaw of negligible mass. (This assumption is made to keep the example simple-more involved examples will follow). The first child has a mass of
and sits
from the pivot.(a) If the second child has a mass of
, how far is she from the pivot? (b) What is
, the supporting force exerted by the pivot?
Strategy
Both conditions for equilibrium must be satisfied. In part (a), we are asked for a distance; thus, the second condition (regarding torques) must be used, since the first (regarding only forces) has no distances in it. To apply the second condition for equilibrium, we first identify the system of interest to be the seesaw plus the two children. We take the supporting pivot to be the point about which the torques are calculated. We then identify all external forces acting on the system.
Solution
The three external forces acting on the system are the weights of the two children and the supporting force of the pivot. Let us examine the torque produced by each. Torque is defined to be
Here , so that
for all three forces. That means
for all three. The torques exerted by the three forces are first,
second,
and third,
Note that a minus sign has been inserted into the second equation because this torque is clockwise and is therefore negative by convention. Since acts directly on the pivot point, the distance
is zero. A force
acting on the pivot cannot cause a rotation, just as pushing directly on the hinges of a door will not cause it to rotate. Now, the second condition for equilibrium is that the sum of the torques on both children is zero. Therefore
or
Weight is mass times the acceleration due to gravity. Entering for
, we get
The quantities on the right side of the equation are known; thus, is
As expected, the heavier child must sit closer to the pivot (1.30 m versus ) to balance the seesaw.
Solution (b)
This part asks for a force . The easiest way to find it is to use the first condition for equilibrium, which is
The forces are all vertical, so that we are dealing with a one-dimensional problem along the vertical axis; hence, the condition can be written as
where we again call the vertical axis the -axis. Choosing upward to be the positive direction, and using plus and minus signs to indicate the directions of the forces, we see that
This equation yields what might have been guessed at the beginning:
So, the pivot supplies a supporting force equal to the total weight of the system:
Entering known values gives
Discussion
The two results make intuitive sense. The heavier child sits closer to the pivot. The pivot supports the weight of the two children. Part (b) can also be solved using the second condition for equilibrium, since both distances are known, but only if the pivot point is chosen to be somewhere other than the location of the seesaw's actual pivot!
Several aspects of the preceding example have broad implications. First, the choice of the pivot as the point around which torques are calculated simplified the problem. Since is exerted on the pivot point, its lever arm is zero. Hence,
the torque exerted by the supporting force
is zero relative to that pivot point. The second condition for equilibrium holds for any choice of pivot point, and so we choose the pivot point to simplify the solution of the problem.
Second, the acceleration due to gravity canceled in this problem, and we were left with a ratio of masses. This will not always be the case. Always enter the correct forces -do not jump ahead to enter some ratio of masses.
Third, the weight of each child is distributed over an area of the seesaw, yet we treated the weights as if each force were exerted at a single point. This is not an approximation -the distances and
are the distances to points directly
below the center of gravity of each child. As we shall see in the next section, the mass and weight of a system can act as if they are located at a single point.
Finally, note that the concept of torque has an importance beyond static equilibrium. Torque plays the same role in rotational motion that force plays in linear motion.
Take-Home Experiment
Take a piece of modeling clay and put it on a table, then mash a cylinder down into it so that a ruler can balance on the round side of the cylinder while everything remains still. Put a penny 8 cm away from the pivot. Where would you need to put two pennies to balance? Three pennies?
Source: Rice University, https://openstax.org/books/college-physics/pages/9-2-the-second-condition-for-equilibrium
This work is licensed under a Creative Commons Attribution 4.0 License.