LPC Physics
Static Equilibrium
Static Equilibrium
Purpose:
To determine that, for a body in equilibrium, the following are true:
• The sum of the torques about any point is zero
• The sum of forces is zero
Equipment:
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Non-Concurrent Forces Apparatus
Hooked Mass Set
Long Rods (3)
Pulleys (2)
Right-Angle Clamps (4)
Table Clamps
Balance
String
Carpenter’s Level
Theory:
The conditions for the mechanical equilibrium of a rigid body are
∑F = 0
∑τ = 0
(a)
Eq. 1a and b
(b)
That is, the (vector) sums of the forces F and the torques τ acting on the body are zero.
The first condition, ∑ F = 0 , is concerned with translational equilibrium and
ensures that the object is at a particular location (not moving linearly) or that it is moving
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with a uniform velocity (Newton’s first law of motion). In this experiment, the rigid
body (the non-concurrent forces apparatus) is restricted from linear motion and ∑ F is
automatically satisfied.
To be in static equilibrium, a rigid body must also be in rotational static
equilibrium. Although the sum of the forces on the body may be zero and it is not
moving linearly, it is possible that it may be rotating about some fixed axis of rotation.
However, if the sum of the torques is zero, ∑τ = 0 , the object is in rotational
equilibrium, and either it dies not rotate (static case) or it rotates with a uniform angular
velocity. (Forces produce linear motion and torques produce rotational motion.)
A torque, or moment of force, results from the application of a force acting at a
distance from an axis of rotation. The magnitude of the torque is equal to the product of
the force’s magnitude and the perpendicular distance from the axis of rotation to the
force’s line of action, or τ = Fd (Figure 1). The distance d is called the lever arm or the
moment arm of the force.
Force
Line of action
Axis of rotation
(perpendicular
to plane of
paper)
d
F
Figure 1
The magnitude of the torque is equal to the product of F and the perpendicular
d from the axis of rotation to the force’s line of action τ = Fd.
Relative to an axis of rotation, a rigid body can rotate in only two directions:
clockwise and counterclockwise. It is therefore customary to refer to clockwise torques
and counterclockwise torques, that is, torques that may produce clockwise rotations and
torques that may produce counterclockwise rotations. For example, in Figure 2, F1 and
F2 produce counterclockwise torques and F3 and F4 produce clockwise torques, but no
rotation would take place if the system were in rotational equilibrium.
d4
d1
d2
d3
50 cm
m1
m2
F1 = m1g
F2 = m2g
m3
F3 = m3g
m4
F4 = m4g
Figure 2
An example of torque in different directions. F1 and F1 give rise to
counterclockwise torques and F3 and F4 to clockwise torques.
The condition for rotational equilibrium is
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∑τ = ∑τ
cc
+ ∑τ cw = 0
Eq. 2
where τcc and τcw are counterclockwise and clockwise torques, respectively. Designating
the directions arbitrarily by plus and minus signs, Eq. 2 can be written
∑τ
cc
∑τ
− ∑τ cw = 0
or
cc
= ∑τ cw
Eq. 3
sum of counterclockwise torques = sum of clockwise torques
For example, for the rod in Figure 2, we have
Clockwise
Counterclockwise
τ1 +τ 2 = τ 3 +τ 4
or
F1 d 1 + F2 d 2 = F3 d 3 + F4 d 4
The forces are due to weights suspended from the rod. Then, with F = mg,
m1 gd 1 + m2 gd 2 = m3 gd 3 + m4 gd 4
m1 d1 + m2 d 2 = m3 d 3 + m4 d 4
Eq. 4
Center of Gravity and Center of Mass
The gravitational torques due to “individual” mass particles of a rigid body define what is
known as the body’s center of gravity. The center of gravity is the point of the body
about which the sum of the gravitational torques around an axis through this point is
equal to zero. For example, consider the rod shown in Figure 3. If the rod is visualized
as being made up of individual mass particles and the point of support is selected such
that ∑τ = 0 , then
∑τ
or
cc
= ∑τ cw
∑ (m g )d = ∑ (m g )d
i
cc
i
i
i
cw
and
(m1 d1 + m2 d 2 + m3 d 3 + L) cc = (m1 d1 + m2 d 2 + m3 d 3 + L) cw
With the rod in rotational equilibrium, it may be supported by a force equal to its weight,
where the support force is directed through the center of gravity. Hence, it is as though
all of the object’s weight (Mg) is concentrated at the center of gravity. That is, if you
were blindfolded and supported an object at its center of gravity on your finger,
weightwise you would not be able to tell if it were perhaps a rod or a block of equal mass.
If an object’s weight were concentrated at its center of gravity, so would be its
mass, and we often refer to an object’s center of mass instead of center of gravity. These
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Static Equilibrium
points are the same as long as the acceleration due to gravity, g, is constant (uniform
gravitational field). Notice how g can be factored and divided out of the previous
“weight” equations, leaving “mass” equations.
Also, it should be evident that for a symmetric object with a uniform mass
distribution, the center of gravity is located at the center of symmetry. For example, if a
rod has a uniform mass distribution, its center of gravity is located at the center of the
rod’s length. Why?
Center of gravity
m1g
m4g
m4g
m1g
Mg
Figure 3
A rod can be considered as being made up of individual masses in rotational
equilibrium when the vertical support force is directed through the center of gravity
Experiment:
x1 = 0
x2
θ1
xcg
mg
m4
m1
c.g
θ3
m2
adj.
m3
x3
x4
Figure 4
1. Determine the mass of the non-concurrent force apparatus.
2. Suspend the apparatus horizontally by strings attached to weights, as shown in the
sketch. Weights should not be attached symmetrically.
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3. Adjust weights and the center of gravity adjustment to obtain equilibrium.
4. Measure the distances and angles necessary to determine the torques. Record
uncertainties for all measurements.
5. Remove m2 and m4 and adjust m1 and m3until the apparatus is again horizontal.
6. DO NOT change the center of gravity adjustment. Measure θ3 and calculate the center of
gravity.
Analysis:
1. Carefully make a large sketch of the apparatus indicating the forces acting on the
apparatus and their corresponding angles.
2. Assuming the pulleys are frictionless, write Newton's 2nd Law for torques and forces
as applied to the equilibrium of the apparatus.
3. Which of the torques are positive? Calculate each positive torque and its uncertainty.
Add the positive torques to find the total positive torque.
4. Calculate the error in the total positive torque.
5. Repeat Step 3 in the analysis for each negative torque. Calculate the total negative
torque and its uncertainty.
6. Does the absolute value of the total negative torque equal the total positive torque
within the uncertainty? Why or why not?
Note: How would we calculate the uncertainty in sinθ when the uncertainty in θ ,
δθ, is known? One way is to calculate:
sin(θ + δθ) and sin(θ - δθ)
i.e., δ(Sinθ) =
7.
Sin(θ + δθ) - Sin(θ - δθ)
2
Calculate % difference between positive and negative torques as well as the % uncertainty in your
measurements.
Results:
Write at least one paragraph describing the following:
• what you expected to learn about the lab (i.e. what was the reason for conducting
the experiment?)
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•
•
•
Static Equilibrium
your results, and what you learned from them
Think of at least one other experiment might you perform to verify these results
Think of at least one new question or problem that could be answered with the
physics you have learned in this laboratory, or be extrapolated from the ideas in
this laboratory.
1
1
The theory section of this lab was shamelessly borrowed from:
Jerry D. Wilson. Physics Laboratory Experiments, 2nd Edition. Lexington MA: D.C. Heath and
Company, 1986. Experiment 13.
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Clean-Up:
Before you can leave the classroom, you must clean up your equipment, and have your
instructor sign below. How you divide clean-up duties between lab members is up to you.
Clean-up involves:
• Completely dismantling the experimental setup
• Removing tape from anything you put tape on
• Drying-off any wet equipment
• Putting away equipment in proper boxes (if applicable)
• Returning equipment to proper cabinets, or to the cart at the front of the room
• Throwing away pieces of string, paper, and other detritus (i.e. your water bottles)
• Shutting down the computer
• Anything else that needs to be done to return the room to its pristine, pre lab form.
I certify that the equipment used by ________________________ has been cleaned up.
(student’s name)
______________________________ , _______________.
(instructor’s name)
(date)
STATIC EQUILIBRIUM
DATA TABLES AND ANALYSIS
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m1
m2
Static Equilibrium
m3
Experimental Equilibrium Data
x1
θ1
θ2
θ3
θ4
m4
x2
x3
x4
xcg
Net Torque
Positive Torque (τ+ )
Negative Torque (τ− )
% Difference
τ + −τ −
(τ + + τ − )
2
Error Analysis
% Uncertainty
δθ1
δθ2
δθ3
δθ4
δx1
δx2
δx3
δx4
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δxcg
δτ+
δτ− ⎛⎜ δτ + + δτ − ⎞⎟ × 100
⎜τ
τ − ⎟⎠
⎝ +