Equilibrium of a Rigid Body
When a rigid body is acted upon by a system of forces which do not all pass through the same point,
a change may be produced in the linear velocity or in the rotation of the body. Under certain conditions the body will be in equilibrium. This experiment presents a study of the conditions for the
equilibrium of a rigid body under the action of several forces.
The torque (τ) or the moment of a force, is a measure of the forceʼs tendency to produce rotation. It
is equal to the product of the force and the lever arm (Equation 1). The lever arm, L, is the perpendicular distance from the axis of rotation to the line of action of the force. The line of action is a line
(1)
Figure 1
extended through the force in other direction (Figure 1).
Torques that produce clockwise motion are considered negative and torques that produce counterclockwise motion are considered positive.
A rigid body is a body whose particles do not change their distances from one another. If a system
of forces is acting on a rigid body and the forces considered collectively have no tendency to produce any motion of translation or rotation, then the body is in equilibrium. That is, a rigid body is in
equilibrium when it has no linear or angular acceleration. The two conditions for the equilibrium of
a rigid body are: (l) the vector sum of all the forces acting on the body must be zero; (2) the algebraic
sum of all the torques (or moments of force) about any axis must be zero. The first condition means
that the sum of the forces in any direction must be equal to the sum of the forces in the opposite
direction. The second condition means that the sum of the clockwise torques around any point must
be equal to the sum of the counterclockwise torques around the same point.
The most common force acting upon a body is its weight. For any ordinary object, no matter how
irregular its shape, there exists a point such that the entire weight may be considered as being concentrated at that point. This point is called the center of gravity of the body. The center of gravity
may be either within or outside the body, depending on the shape of the body. If a single force equal
to the weight of the body and acting vertically upward could be applied at the center of gravity, it
would support the body in equilibrium no matter how the body might be tipped about the center of
gravity.
Winona State University Physics Lab
Revised 3/04/04
Equilibrium of a Rigid Body
A knowledge of the position of the center of gravity is very useful in problems of equilibrium, for
that place is the point of application of the force vector representing the weight. When the axis of
rotation does not pass through the center of gravity the moment of force produced by the weight of
the body must be taken into consideration.
The solution of problems involving static equilibrium is simplified by drawing a suitable diagram
and the careful choice of the axis of rotation. The best choice is usually a point through which one
or more unknown forces act. This point may or may not be the actual pivot, or the center of gravity
of the body. Particular attention must be given to the correct evaluation of the moment arms of the
various forces and to the algebraic signs of the torques.
Apparatus:
Figure 2
Winona State University Physics Lab
Revised 3/04/04
Equilibrium of a Rigid Body
meter stick
4 - weight hangers
3 - support rods
2 - set of slotted weights
4 - meter stick knife-edge clamps
2 - pulleys
4 - clamps
equal arm balance
Procedure:
1.
Weigh the meter stick and the meter stick clamps and record the weight of the meter stick and
the average weight of a meter stick clamp.
2.
Find the center of gravity to the nearest millimeter by balancing it in one of the clamps.
3.
Choose the position and the magnitude of three of the forces acting on the meter stick. The
net force acting on the meter is the weight suspended from meter stick plus the vector sum of
the clamp. For example if the suspended weight is up on the meter stick and the weight of
the clamp is down, the net force is the suspended weight minus the clamp weight. Record the
values of the forces in the table and sketch a diagram of the forces on the meter stick.
4.
Apply the first condition of equilibrium to find the magnitude of the fourth force. Remember
this force includes the weight of the clamp. Show work on the back of the data sheet.
5.
Apply the second condition of equilibrium to find the position of the fourth force. Show work
on the back of the data sheet.
6.
Set up the apparatus as shown in Figure 2 and check to see if equilibrium occurs.
DATA TABLE
On the next page, sketch the meter stick and all the forces acting on it.
Weight of the meter stick _____________________________________
Average weight of a meter stick clamp ___________________________
Center of gravity of the meter stick _______________________________
Position and weight of force 1 ___________________ ,_______________
Position and weight of force 2 ___________________ ,_______________
Position and weight of force 3 ___________________ ,_______________
Calculated position and weight of 4 _______________ ,_______________
Winona State University Physics Lab
Revised 3/04/04
Equilibrium of a Rigid Body
Make a sketch of the meter stick and the forces acting on it.
Question:
The choice of the axis of rotation does not make any difference in the calculation of procedure 5.
Show that the answer is the same by choosing a different axis of rotation. State where the new axis
of rotation is located.
Winona State University Physics Lab
Revised 3/04/04