PASCO ME-9502 User Manual
®
PASCO Mechanics
Statics System
ME-9502
Instruction Manual
012-12876B
*012-12876*
®
The cover page shows a a Friction Block on the Statics System Inclined Plane with a PASCO Mass Hanger from
the ME-8979 Mass and Hanger Set suspended by a thread over a Small Pulley. Most of the components of the
Statics System are held magnetically to the included workboard. This manual contains fifteen copy-ready experiments about the fundamentals of statics.
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Table of Contents
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
Equipment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1
Recommended Equipment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
Spares Package. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
Components Package . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
About the Manual. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
EXPERIMENTS
1. Hooke’s Law - Measuring Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2. Adding Forces - Resultants and Equilibriants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3. Resolving Forces - Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
4. Torque - Parallel Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
5A. Center of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
5B. Equilibrium of Physical Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
6. Torque - Non-Parallel Forces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
7. The Inclined Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
8. Sliding Friction and Static Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
9. Simple Harmonic Motion - Mass on a Spring. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
10. Simple Harmonic Motion - The Simple Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . 53
11A: Simple Harmonic Motion - Physical Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
11B: Minimum Period for a Physical Pendulum. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
11C: Simple Harmonic Motion - Beam on a Spring . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
12. Simple Machines - The Lever . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
13. Simple Machines - The Inclined Plane. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
14. Simple Machines - The Pulley . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
15. Forces on a Boom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
Technical Support, Warranty, and Copyright. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
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Statics System
iv 012-12876B
Statics System
®
If F 0 then a = 0=
a
F
net
m
----------
=
Statics Board
ME-9502
Introduction
The study of mechanics often begins with Newton’s Laws of Motion. The first law describes the conditions for an
object to maintain its state of motion. If the net force on an object is zero, the acceleration of the object is zero.
The second law describes what happens if the net force on an object is not zero. The acceleration is directly proportional to net force, in the same direction as the net force, and inversely proportional to mass, or
Much of what is studied in an introductory course deals with the ways that forces interact with physical bodies.
The PASCO Statics System is designed to help you investigate the nature of forces for the special case in which
there is no acceleration. In other words, the vector sum of all the forces acting on the body is zero.
One reason to study the case of no acceleration is because it is easier to measure non-accelerating systems than it
is to measure accelerating systems. A great deal can be learned about the vector nature of forces by studying the
many ways in which forces can be applied to an object without causing acceleration.The second reason is that in
our everyday experience, non-accelerated systems are the rule, not the exception.
Equipment
The ME-9502 Statics System consists of the Statics Board, Spares Package, Components Package, and Mass and
Hanger Set.
ME-9503 Statics Board
The Statics Board is a ferrous metal plate approximately
45 by 45 centimeters (18 by 18 inches) with a writable
white board finish on both sides.)
The board has rubber bumpers on its base. White board
pens (or “dry erase” pens) can be used to write and draw
directly on the board. The pens are available from stationery stores and the ink can be erased with a cloth or a
white board eraser.
The Mounted Scale Assembly, Large and Small Pulley
Assemblies, Balance Arm Assembly , Force Wheel Assembly, Torque Wheel Assembly, Inclined Plane Assembly,
and Utility Mount Assembly all attach magnetically to t he
board. Components can be stored on the rear side of the
board when not in use. Except for the Inclined Plane
Assembly, all the components that can mount on the board
have a rubber ring on the base that protects the board and
the component. The round magnetic base is 6.5 centimeters (cm) in diameter.
1
Statics System Equipment
®
No
No
Yes
Yes
Moving Components on the Board
Rear view of Statics Boards
Connector
screws
Use the Connector
screws to join two
Statics Boards together.
Important
When moving or removing any of
the magnetically mounted components from the board, handle the
component by the magnetic base
rather than by the component itself.
This will reduce the strain on the
component.
A second board (available separately) can be attached to the first
board with the three included connector screws. To join the boards,
remove the connector screws from
the threaded holes on the side of
the first board. Put the edge of the
second board next to the first
board. Align the holes and join the
two boards with the connector
screws
.
2
012-12876B
®
Model No. ME-9502 Recommended Equipment*
ME-8979 Mass and Hanger Set
PS-2201 5 N Load Cell
Amplifier Model
Load Cell Amplifier (PASPORT) PS-2198
Dual Load Cell Amplifier PS-2205
Load Cell Amplifier (ScienceWorkshop) CI-6464
ME-9504 Spares Package
Item Qty Item Qty
String Tie Assembly 3 Thumbscrew 6-32 by 5/8” (for Cord Clip) 2
Torque Indicator Assembly 6 Thumbscrew 4-40 by 1/2” (for balance arm protractor) 2
Nylon Thread, spool 1 Thumbscrew 6-32 by 1/4” (for balance arm pivot) 1
Cord Tensioning Clip 2 Washer 0.285” OD (for balance arm protractor) 2
Angle Indicator Arm 2 Plumb bob, brass (for inclined plane) 1
ME-9505 Components Package
Item Qty Item Qty
Large Pulley Assembly 1 Utility Mount Assembly 1
Small Pulley Assembly 2 Mass Cart Assembly 1
Mounted Scale Assembly 1 Torque Wheel Assembly 1
Balance Arm Assembly 1 Inclined Plane Assembly 1
Force Wheel Assembly 1 Friction Block Assembly 1
Double Pulley Assembly 1 Asymmetrical Plate 1
ME-8979 Mass and Hanger Set
Components from the ME-8979 P ASCO Mass and Hanger Set are used
in most of the statics experiments. The set includes four mass hangers,
a storage box, and twenty-seven masses ranging from 0.5 g to 100 g
made from brass, aluminum, or polycarbonate plastic.
Recommended Equipment*
Load Cell and Load Cell Amplifier
A PS-2201 5 N Load Cell can be used to measure force. When it is connected through an amplifier to a PASCO interface or to a PASCO
hand-held data logger, the force data from the Load Cell can be
recorded, displayed, and analyzed.
.
*See the PASCO catalog or Web site at www.pasco.com for information about PASCO load cells, load cell
amplifiers, interfaces, hand-held data loggers, and data acquisition software.
012-12876B 3
Statics System Spares Package (ME-9504)
®
Mass Balances
Stop Watch
Inner part
(force disk)
Outer part
Tab
Put
threads
through
this hole.
String Tie Assembly
T orque Indicator Assembly
Tie a
thread to
this hole.
Snap connects
to Torque Wheel
Stopwatch (ME-1234)
The PASCO Stopwatch has a liquid crystal display (LCD) with two display modes. Its precision
is 0.01 seconds up to 3599.99 seconds, and 1 second up to 359999 seconds. Its memory holds up to
nine event times.
Mass Balance
Use a mass balance such as the OHAUS Scout
Pro Balance 2000 g (SE-8757A) or the OHAUS
Cent-O-Gram Balance (SE-8725).
Mounted Scale Assembly (ME-9824A)
The Mounted Scale Assembly (included in the ME-9505 Components Package) is also available separately as a
replacement or as an extra spring scale for experiments where measuring more than one force is required.
Spares Package (ME-9504)
The Spares Package contains some items which are used with assemblies from the Components Package. For
example, the String Tie Assembly snaps into the center of the Force Wheel Assembly and the Torque Indicator
Assembly snaps into the Torque Wheel Assembly. The Cord Tensioning Clip is used on the Utility Mount Assembly. Other items such as the Angle Indicator Arms, thumbscrews, washers, and brass weight are replacement parts.
String Tie Assembly.
When the String Tie Assembly is snapped into the center of the Force
Wheel Assembly, you can use the apparatus to study force vectors. The
String Tie Assembly has two parts: a clear plastic inner part (the “force
disk”) that is free to move around in the center of the outer part, which
has a center hole and two tabs that snap into the Force Wheel Assembly.
(The two parts are not meant to be separated.)
Initial Setup
Get three threads that are 38 cm (15”) long. Put the threads through the
hole in the inner part of the String Tie Assembly. Tie the ends of the
threads together (for example, with an overhand knot) so that they cannot be pulled back through the inner part.
Torque Indicator Assembly
Note that the Torque Indicator Assembly has a
snap that can be connected to the Torque Wheel
Assembly . The other three assemblies are
spares.
Initial Setup
Tie a 30 cm (12”) thread to each of three Torque Indicator Assemblies.
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012-12876B
®
Model No. ME-9502 Components Package (ME-9505)
Spring
Indicator
disk
Scale
Base
Bottom
hook
Thumbnut
Top hook
Tab
Tube
Mounted Scale Assembly
Zero
mark
Small Pulley (2) Large Pulley Double Pulley Block
Mass Cart Friction Block
Felt
Hook
Rod for
stacking
masses
Wood
Components Package (ME-9505)
Mounted Scale Assembly
The mounted scale has four strong magnets in its base. The tube is marked in
newtons (N), ounces (oz.) and millimeters (mm). The thumbnut allows the top
hook to be raised or lowered in order to align the red indicator disk with the
zero mark on the tube.
T o zero the scale, mount the scale on the statics board. Leave the bottom hook
empty. Unscrew the thumbnut a few turns. Rotate the top hook clockwise
(left-to-right) to lower the indicator disk and counter-clockwise (right-to-left)
to raise the indicator disk. When the disk is aligned with the zero mark on the
tube label, tighten the thumbnut to hold the top hook in place.
To move the scale on the statics board, hold the scale by the tabs on the base
and push or pull the scale to the desired location. To remove the scale from the
statics board, lift one or both of the tabs on either side of the base away from
the board.
Large and Small Pulleys and Double Pulley Block
The Statics System comes with two Small Pulleys and one Large Pulley
mounted on magnetic bases. It also includes a Double Pulley Block (that does
not have a magnetic base). The Double Pulley Block is designed for
block-and-tackle experiments. A support thread can be tied to either end of the
Double Pulley Block.
All the pulleys have low friction ball bearings. The smaller pulleys are 2.41
cm (0.850”) outside diameter (OD) and 1.65 cm (0.650”) inside diameter (ID).
The larger pulleys are 2.79 cm (1.10”) OD and 2.28 cm (0.90”) ID.
Please use the tabs on the base of the components when moving them around or removing them from the statics
board.
Mass Cart and Friction Block
The Mass Cart and Friction Block are
designed to be used with the Inclined Plane
Assembly. The Mass Cart has low friction ball
bearings in each wheel and a metal rod in the
center for stacking extra masses. A thread can
be attached at either end.
The Friction Block is made of beech wood. It
has an “eye” hook at one end and is covered
with felt on the bottom and one side. Its dimensions are 2.54 cm by 5.08 cm by 6.35 cm (1.0” by 2.0” by 2.5”).
012-12876B 5
Statics System Components Package (ME-9505)
®
Inclined Plane Assembly
End stops
Plumb bob
Brass Weight
Put thread
through
the hole.
Tie a
knot.
Thread
Inclined Plane Assembly
The Inclined Plane Assembly has four strong magnets on its back side that hold it in position on the Statics Board.
CAUTION: The magnets on the Inclined Plane Assembly are not covered with protective material.
Be careful to firmly hold the inclined plane when placing it on the board so that the magnets are
not damaged by “snapping” against the board.
The inclined plane has end stops at each end of the plane and a degree scale with a brass plumb bob for determining the angle of the incline.
Replacing the Plumb Bob
The ME-9504 Spares package includes a spare plumb bob (brass weight). To
replace the plumb bob, use a Phillips head (“crosshead”) screwdriver to loosen
the screw on the back of the degree scale. Remove the piece of thread and get a
new piece approximately 10 cm (4”) long. Put one end of the thread through the
hole in the brass weight and tie a double knot in that end of the thread so it will
not slip back through the hole. Put the other end of the thread through the small
hole near the top of the degree scale. Adjust the length of the thread so the plumb
bob can swing freely and then wind the end of the thread around the screw on the
back. Tighten the screw to hold the thread in place.
Magnets
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012-12876B
®
Model No. ME-9502 Components Package (ME-9505)
Torque Wheel Assembly
Torque Indicator Assembly
Tie a thread
to the hole.
Hole
Base
Torque
Indicator
Force Wheel Assembly
String Tie Assembly
Notch
String
Tie
Bubble
level
Base
Degree
scale
Notch
Tab
Tab
Torque Wheel Assembly and Torque Indicator Assembly
The Torque Wheel Assembly has a 5 cm radius disk with a
label that has concentric circles 2 mm apart. The disk rotates
freely on a ball bearing and has five holes for attaching a
T orque Indicator Assembly. The Torque Indicator snaps into
the hole and rotates freely. Tie a 30 cm (12”) thread to the hole
in the end of each indicator.
To remove a Torque Indicator, firmly grasp the indicator at the
pivot in the middle and pull outwards while pushing on the end
of the snap from the other side of the disk.
Force Wheel Assembly and String Tie Assembly
The Force Wheel Assembly has a 5 cm radius disk with a 360°
label and a bubble level for leveling the Force Wheel. Unlike
the Torque Wheel, the Force Wheel does not rotate freely but
the disk can be turned in order to level the wheel. Once leveled,
the wheel will stay in position until moved again.
REMINDER: Be sure to put tie three threads through the hole
in the center of the String Tie before putting the String Tie into
the Force Wheel.
The String Tie Assembly snaps into the hole in the center of the
Force Wheel Assembly. Align the tabs on the String Tie with
the notches inside the hole at the center of the Force Wheel.
Push the String Tie firmly into the Force Wheel until both tabs
“snap” into place. Note that the surface of the outer part of the
String Tie will be very slightly below the surface of the Force
Wheel, but the inner part of the String Tie will be flush with the
Force Wheel.
Removing the String Tie Assembly
If you need to remove the String Tie (in order to replace the threads, for example), remove the Force Wheel from
the Statics Board. On the underside of the disk are instructions about removing the String Tie.
Remove String Tie
012-12876B 7
Statics System Components Package (ME-9505)
®
String Tie
Base
Force Wheel
Remove String Tie
Beam
Pivot
base
Bubble
level
Protractor
Tie a
thread to
the hole.
Angle Indicator
Thumbscrew
Thumbscrew
Washer
Balance Arm Assembly
Indicator
line
This tick
mark is at
50 mm.
This tick
mark is at
60 mm.
Note: The Angle Indicator is
not shown in this example.
Reading the Scale
(Imagine that the Force Wheel disk is transparent so that you can see the String Tie.) Hold the Force Wheel as
shown and push the clear plastic tabs inward with your forefingers until the String Tie pops out of the Force
Wheel.
Balance Arm Assembly
The Balance Arm Assembly consists of a Beam, a Pivot, and three Protractors that can be mounted on the beam.
Each protractor has a transparent Angle Indicator. The pivot has a bubble level for leveling the beam.
Adjust the beam by loosening the thumbscrew on the top of the pivot and sliding the beam one way or the other.
Move the protractors by loosening the thumbscrew at the center of the protractor and sliding the protractor along
the beam. Note that not all experiments need three protractors.
Reading the Scale
Read the position of the protractor relative to the midpoint through the two rectangular windows on the protractor.
The bottom window shows the position to the nearest centimeter, and the top window shows the position to the
millimeter. The top window has a small indicator line on its bottom edge. The bottom window in the example
shows the position as between 50 mm and 60 mm and the top window shows it at 55 mm.
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012-12876B
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Model No. ME-9502 Components Package (ME-9505)
Utility Mount
Base
Rod for
Asymmetrical
Plate
Thumbscrew and
post for Load Cell
Attach a Cord
Clip here.
Attaching a Load Ce ll
Thumbscrew
Post
Cord Clip
Thread
Load Cell
Figure A: Hold the
cord clip so the two
parts separate
Figure B: Loop the
thread back through
the cord clip
Figure C: The thread
goes around the
screw hole
Figure D: The cord clip is ready to
be attached to a Utility Mount or a
Load Cell
Thumbscrew
hole
Peg
Threaded
post
Peg
Attaching a Cord Clip
Small
diameter
post
Base
Align the peg of the Cord
Clip with the hole in the
small diameter post.
Use the 5/8” thumbscrew
to attach the Cord Clip to
the threaded post.
Utility Mount Assembly
The Utility Mount has several functions. It can support a PS-2201 5 N
Load Cell, it has a rod from which you can hang the Asymmetrical
Plate, and the included Cord Tensioning Clip allows you to connect a
thread to the mount. The thumbscrew for the Load Cell is “captured”
by a rubber O-ring so that it will not fall out of the mount.
Attaching a Load Cell
You can use a 5N Load Cell (available separately) with
the Statics System to measure a force in much the same
way that the Mounted Scale can be used to measure a
force. (In fact, after the Load Cell is calibrated, it could be
used to calibrate the Mounted Scale.) First, loop a thread
through a Cord Clip (see below) and then attach the clip
to the “LOAD” position that is imprinted on the side of
the Load Cell. Next, attach the Load Cell’s “FIXED” end
to the thumbscrew and post on the Utility Mount.
Using a Cord Clip
The Cord Clip connects to the Utility Mount so that to a
thread and be attached to the Mount itself. Use the Clip
to attach a thread to a Load Cell (see above) that is mounted on the Utility Mount.
It is best to attach the thread through the clip before the clip is installed on the mount or on a Load Cell. The Cord
Clip has two parts but it does not come apart. Start by holding the top part of the clip so the two parts are separate
as shown in Figure A, leaving an opening through which the thread can be threaded. Insert the thread into the flat
end (not the pointed end) of the clip. Loop the thread back through the clip as shown in Figure B and Figure C.
Attach the Cord Clip to the Load Cell or the Utility Mount using the thumbscrew to tighten the clip shut.
012-12876B 9
Statics System About the Manual
®
Rod
Asymmetrical Plate
Base
Suspend the Asymmetrical Plate
Utility
Mount
Asymmetrical Plate
The Asymmetrical Plate is a five-sided polygon with a small hole at
each corner. The Utility Mount has a slender metal rod designed to support the Asymmetrical Plate so that the plate can swing on the rod until
it reaches equilibrium.
About the Manual
The ME-9502 Statics System provides an introduction to static
mechanics. The experiments first introduce force as a vector quantity
and then build on this concept so that students will understand the equilibrium of a physical body under the application of a variety of forces
and torques. Experiments to demonstrate simple harmonic motion with
a pendulum and a spring/mass system are also included.
The experiments are presented in three groups: Basic Experiments,
Advanced Experiments, and Simple Machines. Each experiment is
copy-ready.
Basic Experiments provide the essentials for a solid introduction to
static mechanics. The concepts of vector forces, torques, and center of
mass are explored.
Advanced Experiments allow the student to combine the principles already studied to understand such phenomena as static equilibrium. Friction and simple harmonic motion are also investigated.
Simple Machines provide an opportunity for the student to investigate applications of the principles already studied and also to introduce important new concepts. Levers, inclined planes, and pulley systems are studied using the
principles of static equilibrium, work, and conservation of energy.
In addition to the equipment in the Statics System, a few common items such as pencils, rulers, protractors, paper,
a stop watch, and a mass balance will be needed for some experiments. Check the “Equipment Needed” section at
the beginning of each experiment.
NOTE: Vector quantities are designated with boldface font, such as F, W, or F
. When the same letters are used in
1
normal style font, they refer to the magnitude (size) of the vector quantity. Since the vector nature of torques is not
introduced in these experiments, boldface font is not used to designate torques.
10
012-12876B
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Model No. ME-9502 Exp. 1: Hooke’s Law—Measuring Force
Top
hook
Thumbnut
Indicator
Zero
mark
Bottom
hook
Figure 1.1: Setup and Procedure
Mass
hanger
10
mm
Thread
Exp. 1: Hooke’s Law—Measuring Force
Equipment Needed
Item Item
Statics Board Mounted Spring Scale
Mass and Hanger Set Thread
Introduction
At a practical level, a force is simply a push or a pull. A force is also a
vector quantity that has magnitude (size) and direction. There are different ways to apply and measure a force. One way to apply a force is to
hang a known mass, and determine the force based on the assumption
that gravity pulls the mass downward toward the center of the earth
with a magnitude F = mg, where m is the known mass and g is the
acceleration due to gravity (9.8 m/s
be expressed as 9.8 N/kg. Another way to apply a force is to pull with a
spring. A spring stretches when it is pulled, and if he amount it stretches
is directly proportional to the applied force, then the spring can be calibrated to measure unknown forces. In this experiment you will use the
known force associated with gravity pulling on calibrated masses to
investigate the properties of the Mounted Spring Scale.
2
). Note that the value of g can also
Hooke’s Law
Hooke’s Law describes the relationship between the amount of force
and the amount of stretch for an “ideal” spring. The law states that the
force and the stretch are directly proportional. In other words, the ratio
of the force divided by the stretch is a constant, k. The constant is called
the “spring constant”.
Setup
Place the Spring Scale on the Statics Board so that the spring hangs vertically in the tube. Do not hang anything on the bottom hook. The indicator must be aligned with the zero mark on the label. To zero the
Spring Scale, loosen the thumbnut at the top of the scale. Turn the top
hook clockwise to lower the indicator, and counter-clockwise to raise
the indicator. Once the indicator is aligned with the zero mark on the
label, tighten the thumbnut.
Procedure
1. Attach a thread to the bottom hook and hang a mass hanger from the thread.
2. Add mass to the mass hanger until the indicator is aligned with the 10 mm mark on the label. Adjust the mass
3. Record the total amount of mass (including the mass hanger) in the data table. Record the uncertainty.
so that the indicator is as close as possible to the mark. Estimate the uncertainty in your measurement. (If you
add or remove 0.5 g, can you see a change? What happens if you add or remove 1 g or 2 g?)
4. Add more mass to the mass hanger until the indicator is aligned with the 20 mm mark on the label and record
the total amount of mass and the uncertainty.
5. Repeat the process to move the indicator down by 10 mm each time until the indicator is aligned with the 80
mm mark. Record the total mass and the uncertainty each time.
012-12876B 11
Statics System Exp. 1: Hooke’s Law—Measuring Force
®
Weight - Spring Scale
Weight
---------------------------------------------------- -
X100
Data Table
Spring Displacement (m) Mass (kg) Uncertainty Weight (N)
0.010 m (10 mm)
0.020 m (20 mm)
0.030 m (30 mm)
0.040 m (40 mm)
0.050 m (50 mm)
0.060 m (60 mm)
0.070 m (70 mm)
0.080 m (80 mm)
Calculations
1. Using the formula F = mg, where m is the mass and g is the a cceleration due to gravity, calculate the weight in
newtons for each trial. Record the weight in the data table. (To get the correct force in newtons, you must convert the mass value to kilograms.)
2. On a sheet of graph paper, construct a graph of Weight (N) versus Spring Displacement (m) with Spring Dis-
placement on the x-axis.
3. Draw the line that best fits your data points on the graph. The slope of this line on the graph is the ratio of the
force that stretched the spring divided by the amount of stretch. In other words, the slope is the spring constant, k, for the spring in the Spring Scale.
4. Determine the spring constant, k, from your graph and record the result. Remember to include the units (new-
tons per meter).
Spring constant = _____________________________
Using a Spring Scale to Measure Force
• Hang 160 g (0.160 kg) on the Spring Scale. Calculate the weight based on F = mg. Read the force in newtons
from the Spring Scale.
Weight = _______________ Spring Scale reading = ______________
• How does the measurement from the Spring Scale compare to the actual weight?
• Calculate the percent difference:
Percent Difference = _______________
Questions
1. Hooke’s Law states that the relationship between force and displacement in springs is a linear relationship. If
2. In what way is Hooke’s Law useful when calibrating a spring for measuring forces?
3. On your graph of Weight versus Spring Displacement, did the best fit line go through the origin (zero)? If it
12
Hooke’s Law was not valid, could a spring still be used successfully to measure forces? If so, how?
didn’t, what does that mean?
012-12876B
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Model No. ME-9502 Exp. 2: Adding Forces—Re sultants and Equilibriants
Figure 2.2: Setup
Force
Wheel
Bubble
level
Force
disk
Spring
Scale
Thread
Force
disk
Fg = mg
F
E
= mg
Figure 2.1: Resultant and Equilibriant
F
A
F
B
F
R
F
E
Car
Exp. 2: Adding Forces—Resultants and Equilibriants
Equipment Needed
Item Item
Statics Board Mounted Spring Scale
Force Wheel Large and Small Pulleys
Mass and Hanger Set Thread
Theory
In figure 2.1, Person A and Person B are pulling with forces represented by FA and
F
on a car stuck in the mud. Since these forces are acting on the same point of the
B
car, they are called concurrent forces. Each force is defined both by is direction
(the direction of the vector arrow), and by its magnitude, which is proportional to
the length of the vector arrow. (The magnitude of the force is independent of the
length of the tow rope.)
The total force applied by the two people can be determined by adding vectors F
and F
. The parallelogram method is used in the example. The diagonal of the par-
B
allelogram is called the resultant, F
combination of F
and FB.
A
. It shows the magnitude and direction of the
R
Because the car is not moving, the net force on the car must be zero. The friction
between the car and the mud is equivalent to the resultant force, F
opposing force is called the equilibriant, F
the resultant, F
Setup
, and it has the opposite direction of the resultant.
R
. This force has the same magnitude as
E
. This equivalent
R
Set up the Spring Scale and Force Wheel on the Statics Board as shown. Twist the
Force Wheel until the bubble level shows that the Force Wheel is level. Attach one
of the threads from the force disk (inner part of the String Tie) in the center of the
Force Wheel to the bottom hook of the Spring Scale. Connect a second thread to a
mass hanger (let the third thread dangle). Add 80 g (0.080 kg) to the mass hanger.
Adjust the Spring Scale up or down so that the force disk is centered in the Force
Wheel. The mass hanger applies a force downward, F
ity, where m is the total mass of the mass hanger). When the force disk is centered,
the system is in equilibrium, so the downward force F
equal and opposite force, the equilibriant, F
F
, is applied by the Spring Scale.
E
. In this case, the equilibriant force,
E
= mg (the force due to grav-
g
must be balanced by an
g
A
012-12876B 13
Statics System Exp. 2: Adding Forces—Resultants and Equilibriants
®
Figure 2.3: Find the Equilibriant
Small
Pulley
Small
Pulley
Large
Pulley
Force
Wheel
Force
disk
Spring
Scale
F
1
F
2
F
E
Vector diagram
1
2
E
M
2
M
1
Zero
line on
the
Force
Wheel
Procedure: Two Forces
1. Add or remove 0.5 g to the mass hanger. Did the force disk move away from the center position? How much
can you change the mass on the mass hanger without changing where the force disk is centered?
2. What are the magnitude and direction of F
• F
: Magnitude ________________ Direction ________________
g
, the gravitational force applied by the hanger, where Fg = mg?
g
3. Use the Spring Scale and Force Wheel to determine the magnitude and direction of F
• F
: Magnitude ________________ Direction ________________
E
Procedure: Three Forces
1. Attach the Large Pulley and the two Small
Pulleys to the Statics Board and move the
Spring Scale as shown in the figure.
2. Attach threads to the bottom hook of the
Spring Scale and to mass hangers over the
Small Pulleys.
3. Add 30 g (0.030 kg) to the upper mass
hanger and add 50 g (0.050 kg) to the lower
mass hanger.
4. Adjust the Large Pulley and the Spring
Scale so that the force disk is centered in the
Force Wheel.
, the equilibriant.
E
5. How much can you change the mass on the
hangers and still leave the force disk centered in the Force Wheel?
Data
Record the values of the hanging masses M1 and
M
(including the mass of the mass hangers), the
2
magnitude in newtons of the forces F
F
, and the angles 1, 2, and E with respect to
E
the zero-degree line on the Force Wheel..
Mass (kg) Force (N) Angle (°)
M
1
M
2
F
1
F
2
F
E
1
2
E
, F2, and
1
14
012-12876B
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Model No. ME-9502 Exp. 2: Adding Forces—Re sultants and Equilibriants
Analysis
1. On a separate piece of graph paper, use the values you recorded in the table to construct a vector diagram for
F
, F2, and FE. Choose an appropriate scale, such as 2.0 centimeters per newton, and make t he l engt h of eac h
1
vector proportional to the magnitude of the force. Label each vector and indicate the magnitude of the force it
represents.
2. On your vector diagram, use the parallelogram method to draw the resultant of F
F
. Measure the length of FR to determine the magnitude of the resultant force and record this magnitude on
R
and F2. Label the resultant
1
your vector diagram.
3. On your diagram, measure the direction of F
angle
.
R
relative to the horizontal axis of your diagram and label the
R
Questions
1. Does the magnitude of the equilibriant force vector, FE, exactly balance the magnitude of the resultant force
vector, F
2. Ho w does the dire ction of the equilibriant force vector, F
tor, F
. If not, what are some possible reasons for the difference?
R
?
R
E
, compare to the direction of the resultant force vec-
Extension
Vary the magnitudes and directions of F1 and F2 and repeat the experiments.
012-12876B 15
Statics System Exp. 2: Adding Forces—Resultants and Equilibriants
®
16
012-12876B
®
Model No. ME-9502 Exp. 3: Resolving Forces—Components
M
1
M
2
M
1
F
-F
x
-F
y
Spring
Scale
Pulley
Pulley
Force
Wheel
Force
disk
Figure 3.1: Setup
Exp. 3: Resolving Forces—Components
Equipment Needed
Item Item
Statics Board Mounted Spring Scale
Force Wheel Pulleys (2)
Mass and Hanger Set Thread
Theory
In experiment 2, you added concurrent forces vectorially to determine the magnitude and direction of the combined forces. In this experiment, you will do the reverse: you will find two forces which, when added together,
have the same magnitude and direction as the original force. As you will see, any force vector in the X-Y plane can
be expressed as a vector in the x-direction and a vector in the y-direction.
Setup
Set up the equipment as shown in the figure. As
shown, create a force vector F by hanging a
mass, M
center of the Force Wheel over a pulley.
, on a thread from the force disk in the
1
Set up the Spring Scale and a pulley so that the
thread from the Spring Scale is horizontal from
the bottom of the pulley to the force disk. Hang a
second mass hanger directly from the force disk
in the center of the Force Wheel.
Now pull the Spring Scale toward or away from
the pulley to adjust the horizontal or x-component of the force, F
on the vertical mass hanger , M
tical or y-component of the force, F
. Adjust the amount of mass
x
, to adjust the ver-
2
. Adjust the
y
x- and y-components in this way until the force
disk is centered in the Force Wheel.
Note: These x- and y-components are actually
the x- and y-components of the equilibriant, F
,
E
of the force F, rather than the components of F
itself.
Procedure 1
1. Calculate and record the magnitude of F (based on F = mg). Use the Force Wheel to measure the angle of F
and then record the angle.
• Magnitude, F = _________________ Angle, = ____________________
2. Record the magnitude of the x-component of the equilibriant of F and calculate and record the y-component
of the equilibriant of F.
• X-component = _________________ Y-component = ____________________
012-12876B 17
Statics System Exp. 3: Resolving Forces—Components
®
F
Fx = F cos
F
y
= F sin
O
X
Figure 3.2: Vector components
F
R
Rx2Ry
2
+=
R
R
y
R
x
----- -
atan=
3. What are the magnitudes of F
• F
= _________________ Fy = ____________________
x
and Fy, the x- and y-components of F?
x
4. Change the magnitude and direction of the force vector F and repeat the experiment.
5. Record the magnitude and angle of the new force vector, F, and the magnitudes of F
• Magnitude, F = _________________ Angle, = ____________________
• F
= _________________ Fy = ____________________
x
Background
Why use components to specify vectors? One
reason is that using components makes it easier
to add vectors mathematically. The figure shows
the x- and y-components of a vector of length F,
and an angle with respect to the x-axis. Since
the components are at right angles to each other,
the parallelogram used to determine their resultant is a rectangle. Using right triangle AOX, the
components of F can be calculated:
and Fy.
x
• F
, the x-component of F is F cos
x
• F
, the y-component of F is F sin
y
If you want to add several vectors, find the x- and
y-components for each vector. Add all the
x-components together and all the y-components
together. The resulting values are the x- and
y-components for the resultant. The magnitude of the resultant, F
the resultant’s x-component (R
) and the y-component (Ry).
x
, is the square root of the sum of the squares of
R
The angle of the resultant is the arctangent of the y-component divided by the x-component.
Procedure 2
Make sure that the Force Wheel is level. Set up a new force, F. Put one thread from the force disk over a pulley and
tie a mass hanger to the end of the thread. Add mass to the mass hanger.
1. Calculate and record the magnitude of the force, F, that you set up with the pulley and mass hanger. Use the
Force Wheel to determine the angle.
• Magnitude, F = _________________ Angle, = ____________________
2. Calculate the magnitudes of F
and F
• F
= _________________ Fy = ____________________
x
= F sin .)
y
3. Now, set up the Spring Scale with a pulley and a thread from the force disk so it can apply the x-component,
F
, of the new force F. Adjust the Spring Scale so that it pulls the force disk horizontally by the amount, Fx.
x
4. Next, attach a mass hanger to the third thread from the force disk so the thread hangs vertically. Add mass to
the hanger so it pulls the force disk vertically by the amount, F
18
and Fy, the x- and y-components of the new force, F. (Remember, Fx = F cos
x
.
y
012-12876B
®
Model No. ME-9502 Exp. 3: Resolving Forces—Components
F
1
F
2
F
3
NOTE: The diagram is
not to scale.
Figure 3.3: Extension vectors
Question
1. Is the force disk at equilibrium in the center of the Force Wheel?
2. Why or why not?
Extensions
1. Generally it is most useful to find the components of a vector along perpendicular axes, as you did above.
However, it is not necessary that the axes be perpendicular. If time permits, try setting up the equipment to
find the components of a vector along non-perpendicular axes. Use pulley to redirect the component forces to
non-perpendicular directions.
2. The figure shows a classic vector combination. For the force disk to be in
equilibrium, the x-components of force
F
and force F2 must be equal and
1
opposite, and the y-components of F
and F
must add to equal the magni-
2
tude of force F
, the vertical force. Set
3
up the equipment so that the Spring
Scale applies force F
. Once the sys-
1
tem is in equilibrium, determine the xand y-components of the vectors and
compare them.
1
3. For the setup in the figure, change the
angles of F
are closer to the x-axis. Calculate the
x-components and notice the change.
What happens to the x-components as
the two forces become closer and
closer to parallel? What amount of
force would be needed if F
were both horizontal?
and F22 so that the vectors
1
and F2
1
012-12876B 19
Statics System Exp. 3: Resolving Forces—Components
®
20
012-12876B
®
Model No. ME-9502 Exp. 4: Torque—Parallel Forces
A
F
R
Figure 4.1: Non-concurrent forces
Object
F
1
F
1
Figure 4.2: Balance Arm
Base
Thumbscrew
Bubble level
Beam
Pivot
Scale
Figure 4.3: Setup
Angle
Indicator
Protractor 2
M
1
M
2
Protractor 1
Thread
Mass
hanger
Exp. 4: Torque—Parallel Forces
Equipment Needed
Item Item
Statics Board Balance Arm and Protractors
Mass and Hanger Set Thread
Theory
In experiment 2, you found resultants and equilibriants
for concurrent forces—forces that act upon the same
point. In the real world, forces are often not concurrent.
They act upon different points on an object. In the figure, for example, two forces are pulling on different
points of an object. Two questions can be asked:
1. In which direction will the object be accelerated?
2. Will the object rotate?
If the two forces were both applied at point A, the resultant would be the force vector shown, F
. In fact, FR points
R
in the direction in which the object will be accelerated. (This idea will be investigated further in later experiments,)
What about question 2? Will the object rotate? In this experiment you will begin to investigate the types of forces
that cause rotation in physical bodies. In doing so, you will encounter a new concept—torque.
Setup
Mount the Balance
Arm near the center of
the Statics Board.
Level the Beam
Loosen the thumbscrew and adjust the beam so that the zero mark on the beam is aligned with the indicator marks
on the pivot. When the beam is level, the bubble in the bubble level will be midway between the two lines on the
level.
Add the Protractors
First, find the mass of two
of the protractors and
record the masses. (Note
that you can use the Spring
Scale to measure the mass.)
• Protractor 1 =
__________
• Protractor 2 =
__________
Loosen the thumbscrews on the two protractors and slide one onto each end of the beam. Tie a mass hanger to the
thread on the angle indicator of each protractor.
012-12876B 21
Statics System Exp. 4: Torque—Parallel Forces
®
d
1
d
2
M
1
M
2
Figure 4.4: Measuring Torques
center of
protractor 1
center of
protractor 2
pivot
M
1
M
2
d
1
d
2
Figure 4.5: More Torques
center of
protractor 1
pivot
center of
protractor 2
Procedure: Equal Distance, Equal Mass
Position one of the protractors near one end the beam and tighten its thumbscrew to hold it in place. Adjust the
position of the other protractor until the beam is perfectly balanced, and then tighten its thumbscrew to hold it in
place.
1. Measure d
and d2, the distances
1
from the pivot to the center of each
protractor.
•d
= _______________
1
•d
= _______________
2
2. Add a 50-gram mass to each mass
hanger.
• Is the beam still balanced?
3. Add an ad ditional 20-gram to one of the mass hangers.
• Can you restore the balance of the beam by repositioning the other protractor and mass hanger?
Procedure: Unequal Distance, Unequal Mass
Position one of the protractors approximately halfway between the pivot point
and the end of the beam and tighten its
thumbscrew to hold it in place. Add 75
grams of mass to the mass hanger, M
.
1
Place various masses on the other mass
hanger (M
) and slide it along the beam
2
as needed to rebalance the beam.
1. At the first balanced position, measure the total mass, M
and M2, on each side of the pivot (protractor, mass
1
hanger, added masses) and record the masses in the data table.
2. Measure the distances, d
and d2, between the centers of the protractors and the pivot and record the values in
1
the data table.
3. Take measurements for five more different valu es of M
and record your results in the data table. Be sure to
2
include the units of your measurements.
4. If there is time, vary M
and repeat the procedure.
1
• Reminder: For accurate results, include the mass of the protractor, mass hanger, and added masses when measuring M
and M2.
1
Calculations
Calculate the gravitational force (weight = mg) produced by the total mass on each side of the beam for each case.
Calculate the torques,
of the net force and the lever arm. Since the distance and the direction of the force are at right angles in this experiment, the torque, , is F d (where F
balanced position of the beam.
22
and 2, on each side of the beam for each case. Remember, torque, , is the cross product
1
= mg). Record your calculated values of weight, Fg, and torque, , for each
g
012-12876B
®
Model No. ME-9502 Exp. 4: Torque—Parallel Forces
Data Table
Case Total Mass
M
(kg)
1
1
2
3
4
5
6
Weight
F1 (N)
Distance
d1 (m)
Torque
1 = F1 d
1
Total Mass
M
(kg)
2
Weight
F2 (N)
Distance
d2 (m)
2 = F2 d
Questions
1. From your results, what mathematical relationship must there be between 1 and 2 in order for the beam to be
balanced?
2. What torque is exerted on the balance beam by the upward pull of the pivot point?
Extension
If you have time, try adding a third Protractor and Mass Hanger to the beam on the same side of the beam as the
second Protractor.
Torque
2
Question
What relationship must there be between
protractors and mass hangers?
, 2, and 3 in order for the beam to be balanced when there are three
1
012-12876B 23
Statics System Exp. 4: Torque—Parallel Forces
®
24
012-12876B
®
Model No. ME-9502 Exp. 5A: Center of Mass
Support
Center of mass
Object
Figure 5.1: Equipment Setup
M
1
M
2
Pivot
base
Protractor
Angle
Indicator
Exp. 5A: Center of Mass
Equipment Needed
Item Item
Statics Board Balance Arm and Protractors
Asymmetrical Plate Thread
Mass and Hanger Set
Theory
Gravity is a universal force; every bit of matter in the universe is attracted to every other bit of matter. Therefore,
when the Balance Arm is supported by the pivot, every bit of matter in the Balance Arm is attracted to every bit of
matter in the Earth.
Fortunately , the sum of all these individual gravitational forces produces a single resultant. This resultant acts as if
it were pulling between the center of the Earth and the center of mass of the Balance Arm. The magnitude of the
force is the same as if all the matter of the Earth were located at the center of the Earth, and all the matter of the
Balance Arm were located at the center of mass of the Balance Arm.
An object thrown so that it rotates tends to rotate about its center of mass,
and the center of mass follows a parabolic path. An object whose center of
mass is above a support tends to remain in rotational equilibrium (balanced on the support). In this experiment you will use your knowledge of
torque to understand and locate the center of mass of an object.
Setup
Measure and record the
mass of the beam.
Next, find the mass of
two of the protractors
and record the masses.
(Note that you can use
the Spring Scale to
measure the mass.)
• Beam = ________
• Protractor 1 = __________
• Protractor 2 = __________
Loosen the thumbscrews on the two protractors and slide one onto each end of the beam so each protractor is at the
165 mm mark. Tie a mass hanger to the thread on the angle indicator of each protractor
Mount the Balance Arm near the center of the Statics Board.
Record the total mass of the beam plus protractors plus mass hangers (5 grams each).
• Mass of system = _____________
012-12876B 25
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