Equilibrium of Rigid Body
When a rigid body is acted upon by a system of forces that 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 force’s tendency to
produce rotation. It is equal to the product of the force and the lever arm (Eqn. (1)). The
lever arm (r) is the perpendicular distance from the axis of rotation to the line of action of
the force. The line of action is a line extended through the force in both directions (Fig.
1).
F
τ=Fr
r
(1)
Line of Action
Rigid Body
Torques that produce clockwise motion are considered negative and torques that
produces 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.
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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 torque
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 lever arms of the various forces and to the algebraic signs of
the torques.
APPARATUS:
1.
2.
3.
4.
5.
6.
7.
8.
Meter stick
4 - meter stick knife-edge clamps
4 - weight hangers
2 - pulleys
3 - support rods
4 - clamps
2 - set of slotted weights
Balance
MD
MB
MC
MA
Fig. 2
FA
rA
rB
FB
Conditions of Equilibrium
C.G.
∑F
y
rC
FC
rD
FD
FR
∑ τ = −F r
A A
C. M .
Fig. 3
54
= FA + FB − FC − FD − FR = 0
+ FB rB + FC rC − FD rD = 0
PROCEDURE
1.
Mass the meter stick and calculate its weight. Record these value in the data
table.
2.
Label the meter stick clamps A thru D and weigh each clamp individually and
record their values in the data table.
3.
Set up the apparatus as shown in Figure 2 so that it is equilibrium. No two
masses can be the same magnitude. (This is done by trail and error)
4.
Record the position and the magnitude of three of the forces (FA, FB, and FC as
shown on Fig. 3) 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 using the labels shown in Fig. 3
5.
Apply the first condition of equilibrium to find the magnitude of the FD.
Remember this force includes the weight of the clamp. Show work on the data
sheet.
6.
Apply the second condition of equilibrium to find the position of the FD. Use a
direction of counter clockwise as positive and clockwise as negative. Show
work on the data sheet.
DATA
Mass of the meter stick = __________________ FR = __________________
Center of gravity of the meter stick ______________________________
Hanging Mass
Clamp Mass
F
r
Force A
g
g
N
m
Force B
g
g
N
m
Force C
g
g
N
m
DIAGRAM:
Measure x
from here
FA
FC
FR
55
FB
FD
Question 1: How much weight (FD) do we need to balance the ruler? Apply the first
condition of equilibrium to find the magnitude of FD. Record this below as the
Calculated Weight of FD.
Question 2: How far from the center do we put this weight? Apply the second condition
of equilibrium and find the position rD. Use the center of gravity as the axis of rotation.
Calculated
Weight of FD
Force
D
N
Calculated
rD
Experimental
mass at FD
Experimental
Weight of FD
g
N
m
Percent Error
of FD
* At this point you should have a calculated Force for FD and a place to put it (rD). Even though we already have an
experimental value for FD and rD, we will not use those numbers. Instead of finding errors for both, we need to confine
it to one. Move the clamp for FD to the location of rD. Then using trial and error, find the Experimental mass at F4 that
brings the ruler into balance. Record the value of mass plus clamp above. Calculate the Experimental Weight of FD
next. Now you can find the percent error.
56
Question
The choice of the axis of rotation does not make any difference in the calculation of
procedure 5. Choose a different axis of rotation and calculate the position of the fourth
force over again, relative to this new axis of rotation. State where the new axis of
rotation is located and show that it is actually the same position on the meter stick as
before.
Use a
different axis
point
FA
FC
FR
57
FB
FD