Name: ___________________________
Group Members: ___________________________
___________________________
___________________________
Torques & Rotational
Equilibrium
PURPOSE & OBJECTIVES
An important condition of rigid bodies in many practical applications is of static equilibrium.
Examples include beams in bridges and beam balances. When a rigid body, such as these examples is
“in balance”, it is at rest, or in static equilibrium. The criterion for rotational static equilibrium is that
the sum of the torques, or moments of force acting on a rigid body, be equal to zero. To
In this experiment, we will study torques and rotational equilibrium, using a beam balance in the
form of a meter-stick and suspended weights. The mass of an object will be determined
experimentally and the experimental value compared to the mass of the object as measured on a
laboratory balance. In addition, we will investigate the concept of center of gravity.


Explain mechanical equilibrium and how it is applied to rigid bodies.
Describe how a laboratory beam balance measures mass.
MATERIALS
Meter-stick
Four knife clamps with hangers
Unknown mass with hook
Support Stand
Hooked masses
Laboratory balance.
PRELIMINARY QUESTIONS
1. When you push an object that is free to pivot about a point, how does the magnitude of the
force affect its rotation? How does the distance from the pivot point that a force is applied affect
the rotation?
2. The conditions for mechanical equilibrium of a rigid body are that the object is a state of
translational equilibrium and rotational equilibrium. Express these two conditions
mathematically.
3. If the mass of a 1 m long bar of metal is uniformly distributed, how much mass would be in 1 cm
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1-7 Torques and Equilibrium
of the bar? How could you determine how much mass was in 30 cm of the bar?
PROCEDURE
Part I – Apparatus with Support Point at Center of Gravity
1.
Determine the mass of the meter-stick without any of the hangers attached and record.
2.
Mass the weight hangers and determine the average mass of a hanger.
3.
Place a knife edge clamp (without hanger) on the meter-stick near its center and place it on the
support stand. Adjust the position of the clamp on the meter-stick until it is balanced on the
stand. Tighten the clamp and record the distance of the balance point (x0) from the end of the
meter-stick.
Case 1 – Two Known Masses
4.
Suspend a mass (m1 = 100 g) on a hanger at the 15 cm position on the meter-stick.
5.
Suspend a mass (m2 = 200 g) on a hanger on the opposite side of the support clamp on the
meter-stick. Adjust the position of the hanger until the system is in static equilibrium. Record
the masses and the moment arms. Do not forget to include the masses of the clamps. Lever
arms are the distances from the pivot point to the masses.
6.
Record the data in the Data Table. Compute the torques and find the percent difference in the
computed values.
Case 2 – Three Known Masses
7.
With the meter-stick on the support stand at x0 ,suspend m1 = 100 g at the 30 cm position and
m2 = 200 g at the 70 cm position. Suspend m3 = 50 g and adjust the lever arm of this mass
until static equilibrium is achieved.
8.
Record the data in the Data Table. Compute the torques and find the percent difference in the
computed values.
Case 3 – Unknown Mass
9.
With the meter-stick on the support stand at x0 ,suspend an unknown mass (m1) near the end
of one end of the stick (ex. at the 10 cm position). Suspend from the other side of meter-stick
m2 = 200 g to act as a counterweight. Just like a laboratory beam balance scale, you will adjust
the position of the counterweight until the system is in static equilibrium.
10. Remove the unknown mass and determine its mass by using a laboratory balance.
11. Compute the value of the unknown mass by using the conditions for rotational equilibrium.
Use the sum of the torques and lever arms to calculate the unknown mass. Compare the
calculated value with the measured value by calculating percent error.
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Part II – Apparatus Supported at Different Pivot Points
Case 4 – Meter-stick with One Mass
12. Suspend a mass, m1 = 100 g, at or near the zero end of the meter-stick. Move the support
clamp on the meter-stick until the system is in equilibrium.
13. The torque opposite the suspended mass is due to the mass of the meter stick beyond the
support point. The lever arm r2 is located at half of the length (L2) of the meter-stick to the
right of the support point. Using the total mass of the meter-stick as m2 with a lever arm r2 ,
compute the counterclockwise and clockwise torques and compare by percent difference.
Case 5 – Meter-stick with One Mass and Meter-stick Masses Taken Into Account
14. Assuming that the mass of the meter-stick is uniformly distributed, you can determine the mass
per unit length (µ) by dividing the total mass of the meter-stick by 100 cm. Using this value,
you can compute the mass of the lengths of meter-stick on both sides of the support point.
15. Using the information from Case 4, calculate the mass of the meter-stick on the right side of
the support point by multiplying L2 by µ from above and record as m2. Repeat the process for
the mass of the meter-stick on the left side of the support point and record that as m3.
16. You will need to determine the lever arm of the mass of the meter-stick on the left side of the
support. The lever arm r3 is located at half of the length (L3) of the meter-stick to the left of the
support point. Using the masses of each side of the meter-stick as m3 with a lever arm r3 and
m2 with a lever arm r2, compute the counterclockwise and clockwise torques and compare the
differences with Case 4.
Case 6 – Center of Gravity
17. Suspend a mass, m1 = 100 g, at or near the zero end of the meter-stick and another mass, m2 =
100 g, at the 60 cm position. Move the support point until the system is in equilibrium. This is
the center of gravity (x’0). Find the lever arms r1 and r2.
18. Repeat this procedure for m2 at the 70 cm position.
19. Notice that the position of the center of gravity moves as the mass is distributed. Where do
you predict it would be located if m2 were moved to the 90 cm position? Using this prediction,
calculate the clockwise and counterclockwise torques.
20. Experimentally determine the center of gravity for this case and compute the percent
difference in the experimental and predicted values.
DATA
Part I – Apparatus with Support Point at Center of Gravity
Mass of Meter-stick ______________
Total mass of clamps
Balancing position (Center of Gravity)
x0 ______________
Average mass of one clamp ______________
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______________
Case 1 Diagram
Masses
Positions
Lever arms
Torques
m1 = _________
x1 = 15 cm
r1 = _________
τcc = _________
m2 = _________
x2 = _________
r2 = _________
τcw = _________
Percent
Difference
Case 2 Diagram
Masses
Positions
Lever arms
Torques
m1 = _________
x1 = 30 cm
r1 = _________
τcc = _________
m2 = _________
x2 = 70 cm
r2 = _________
τcw = _________
m3 = _________
x3 = _________
r3 = _________
Percent
Difference
Case 3 Diagram
Masses
Positions
Lever arms
m1 =Unknown
x1 = _________
r1 = _________
m2 = _________
x2 = _________
r2 = _________
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1-7 Torques and Equilibrium
Mass 1
Measured
= _________
Calculated
= _________
Percent Error
Case 4 Diagram
Masses
Positions
Lever arms
Torques
m1 = _________
x1 = _________
r1 = _________
τcc = _________
m2 = _________
x2 = _________
r2 = _________
τcw = _________
Percent
Difference
x’0 = _________
Case 5 Diagram
Masses
Positions
Lever arms
m1 = _________
x1 = _________
r1 = _________
Torques
τcc = _________
m2 = _________
x2 = _________
r2 = _________
m3 = _________
x3 = _________
r3 = _________
τcw = _________
x 0 = _________
’
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1-7 Torques and Equilibrium
Percent
Difference
Masses
Case 6
Positions
m1 = _________
x1 = 0 cm
r1 = _________
m2 = _________
x2 = 60 cm
r2 = _________
Masses
Positions
m1 = _________
x1 = 0 cm
r1 = _________
m2 = _________
x2 = 70 cm
r2 = _________
Masses
Positions
x’0
Torques
m1 = _________
x1 = 0 cm
Predicted
= _________
τcc = _________
m2 = _________
x2 = 90 cm
Measured
= _________
τcw = _________
Diagram
Diagram
Diagram
x’0
x’0
Lever arms
Lever arms
Percent Difference
ANALYSIS
1. Each case in Part I satisfied the condition of ΣF = 0. Explain how this is true.
2. In many instance, the balancing position x0 of the meter-stick by itself is not at the 50 cm
position. Explain why this is the case.
3. Explain how a triple beam laboratory balance works.
4. Suppose in a situation as in Case 2 in the experiment, m1 = 200 g, were at the 20 cm position
and m2 = 100 g at the 65 cm position. Would there be a problem in experimentally balancing
the system with m3 = 50 g ? Explain. If so, how could you resolve the problem?
5. Explain the effects of taking the mass of the meter-stick into account when the balancing
position is not at 50 cm.
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