The Equilibrium of Forces: The Second Law©98
Experiment 8
Objective: To study the methods of vector addition as applied to be equilibrium of
torques.
DISCUSSION:
The second law of the equilibrium of forces states that the vector sum of the
torques, acting on a rigid unaccelerated object, is zero. The torque due to a force depends
upon three things:
1.
The force acting on the body.
2.
The point at which this force acts.
3.
The point relative to which the torque is measured.
Since this latter point is arbitrary, that is, since it is not determined by the physical
situation, the value of the torque is arbitrary. Nevertheless, the vector sum of the torques
acting on a body in equilibrium is zero, provided all the torques have been measured
relative to the same reference point.
The magnitude of the torque  due to a force is defined to be the product of the
magnitude of the force F, the magnitude of the displacement r of the point of application
of the force from the reference point, and the sine of the angle  between the vectors F
and r. That is,
(1)
  Fr sin 
The direction of the torque is always
perpendicular to the plane formed by the vectors
r and F. For example, if r and F lie in the plane
of this page, the torque is out of or into the page.
All the forces and displacements we shall be
considering in this laboratory lie in a common
plane, their torques are either parallel or
antiparallel to one another, lying perpendicular
to the plane. In such a situation, it is more
convenient to specify whether a torque is
clockwise or counterclockwise. These ideas are
diagramed in Fig. 1. This figure shows a body
being acted upon by a force F. The point
relative to which torques are measured is the
point O, which is chosen arbitrarily, and for
simplicity it was chosen to be the center of mass
of the object. The torque is clockwise if the angle , when measured from r to F is
designated by an arrow pointing clockwise. The torque is counterclockwise if the arrow
points in the counterclockwise direction. The torque in Fig. 1 is counterclockwise.
8-1
Returning now to the second law of the equilibrium of forces, we translate its
"vector" language into one using "clockwise" and "counterclockwise": the sum of the
clockwise torques acting upon a rigid body must equal the sum of the counterclockwise
torques if body is to be in equilibrium. Written thusly, this law makes sense, of course,
only when all the forces, Fn, and the displacements, rn, lie in the same plane.
The apparatus in this experiment
consists of a meter stick supported by two
strings. Weights hang from these strings as
well as from the meter stick. A diagram of the
apparatus is shown in Fig. 2. The meter stick is
the body on which the forces act, and the forces
themselves are due to the weights hanging
from the strings and meter stick. Since the
forces do not all act at the same point on the
meter stick, the torque due to these forces must
balance if equilibrium is to be a maintained. In
addition, the forces themselves must balance.
F2
F1
F5
F3=W F4
r1
r2
r3
r4
r5
Figure 2: The forces on a meterstick.
The supporting strings are not vertical, and so they
Fv
pull horizontally as well as vertically on the stick. In
establishing whether or not the system is theoretically in
F
equilibrium, we need to decompose the forces due to the
supporting strings into their horizontal and vertical
components. This decomposition is effected by constructing
Fh
a right triangle whose hypotenuse lies along the string and
whose arms are horizontal and vertical, as seen in Fig. 3.
Figure 3: Vertical and horizontal
Letting F be the force on the string, Fh its horizontal
vector components.
component and Fv its vertical component, we have
F
Fh 2  Fv 2
(2)
Having decomposed the forces due to the supporting strings, we may test the first
law of equilibrium, that the vector sum of the forces must be zero. Using the forces as
shown in Fig. 2, we have
(3)
 F1  F2v  W  F4  F5v  0
and
F5h  F2h  0
(4)
where W is the weight of the meter stick acting at the center of the meter stick.
To test the second law of equilibrium, we measure r from some arbitrarily chosen
point. In this example that point is zero-point of the meter stick. Hence all the r
measurements were from the end of the meterstick in Fig. 2. Thus the torque equation is:
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 r1 F1  r2 F2v sin  2  r3W3  r4 F4  r5 F5v sin  5  0
(5)
Here, the counterclockwise torques are positive and the clockwise torques are negative.
Notice that there is no reference to sin1, sin3 or sin4; all these angles are 90° and the
sine of 90° is 1. Since the point from which the torques are measured is immaterial, we
could just well have measured them from say, the middle of the meter stick as long as we
are careful about the signs.
EXERCISES:
1. Determine the forces on the meter stick.
a. Weigh the meter stick and determine the position of its center of gravity.
b. Add weights to the supporting strings and meter stick so that the meter stick is in
equilibrium when it is horizontal.
c. Measure the forces F1, F2, F4 and F5, by noting the masses of the weights hanging
from the supporting string and meter stick.
2. Determine the h- and v-components of the forces.
a. Measure the angle of elevation each string makes with the horizontal.
b. On paper, construct right triangles whose base angles equal the angles of
elevation of the strings.
c. Letting the hypotenuses of the right triangles be proportional to the tensions in the
strings, measure the arms of the triangles to find the horizontal and vertical
components of the tension.
3. Find the percent deviation of the sum of the upward forces from the sum of the
downward forces by inserting the measured values of the vertical components of the
forces into Eq. (3).
4. Find the percent deviation of the sum of the forces in the +x-direction from the sum
of the forces in the -x-direction by inserting the measured values of the horizontal
components into Eq. (4).
5. Analyze the torques.
a. Calculate the torques exerted by each of the forces (measuring r from the zeropoint of the meter stick).
b. Label whether the torques are clockwise or counterclockwise.
c. Find the percent deviation of the sum of the clockwise torques from the sum of
the counterclockwise torques using Eq. (5).
6. Repeat Exercise 5, measuring r from the center of gravity of the meter stick.
8-3