UNESCO-NIGERIA TECHNICAL & VOCATIONAL
EDUCATION REVITALISATION PROJECT-PHASE II
NATIONAL DIPLOMA IN
MECHANICAL ENGINEERING TECHNOLOGY
MECHANICAL ENGINEERING
SCIENCE[DYNAMICS]
COURSE CODE: MEC124
YEAR I- SEMESTER 2
PRACTICALS
Version 1: December 2008
1
COURSE: MECHANICAL ENGINEERING SCIENCE II (DYNAMICS)
COURSE CODE: MEC 124
CONTACT HOURS 2HRS/WK
Course Specification: PRACTICAL CONTENT
TABLE OF CONTENT
Week 1
PAGE
1. Experiment no 1 part 1: the simple wheel and axle……………………………..4
Week 2
2. Experiment no 1 part 2: the differential wheel and axle…………………………7
Week 3
3. Experiment no 3: the screw jack ............................................................................9
Week 4
4. Experiment no 4: the worm and wheel drive ……………………………………11
Week 5
5. Experiment no 5: simple pulleys ..........................................................................13
Week 6
6. Experiment no 6 : pulley blocks ...........................................................................15
Week 7
7. Experiment no 7: the Weston differential chain blocks..........................................17
Week 8
8. Experiment no 8: chain drive ……………………………………………………..19
Week 9
9. Experiment no. 9 simple pendulum..........................................................................21
2
Week 10
10.
Experiment no 10.compound pendulum...................................................22
Week 11
11.
Experiment no.11 axial oscillations..........................................................24
Week 12
12.
Experiment no. 12 torsional oscillations....................................................26
Week 13
13.
Experiment no. 13 wheel and axle.............................................................28
Week 14
14. Experiment no. 14 hydraulic jack ..........................................................................30
Week 15
15. Experiment No. 15 Moment of Inertia...................................................................35
3
EXPERIMENT NO 1 Part 1: THE SIMPLE WHEEL AND AXLE
Teaching Element:
The simple wheel and axle is used for hoisting or hauling a load through a considerable distance, and is usually
called a winch windlass. It consists of a wheel fixed to an axle or drum of smaller radius.
The load is attached to a rope or wire which is wound round the drum when the wheel is turned. The wheel may
be replaced by a handle, as in the windlass used for hoisting water from a well, or it may be turned by projecting
bars as in a ship's capstan. For heavy duty winches the wheel is usually power driven and may have gearing.
The wheel and axle is basically a lever. The applied force acts at the end of a lever arm equal to the radius of the
wheel and lifts a resisting load acting at the end of a short lever arm equal to the radius of the drum. Neglecting
friction, the Mechanical Advantage of the wheel and axle is equal to the Velocity Ratio.
Although the theoretical Mechanical Advantage is equal to the Velocity Ratio the actual Mechanical Advantage
is reduced because of friction in the spindle which increases with the load. The applied force, the load and the
friction losses gives rise to the relationship called the LAW OF THE MACHINE.
Student Objectives:
The object of the experiment is:
1.To determine the Law of the Machine for a simple wheel and axle.
2.To investigate the variation of Mechanical Advantage and Efficiency with load.
4
Apparatus
Wheel and axle, cord, weights, metre rule.
Belts, Chains and Gears
Theory
Calculated V.R.
Consider one turn of the axle.
The load is raised a distance equal to the circumference of the axle = πd.
The effort moves a distance equal to the circumference of the wheel = πD.
Calculated V.R. = distance moved by the effort/distance moved by the load
= πD / πd
.
=D/d
Method
1. Place a load and an effort on the machine and measure the distance moved by the load and the distance
moved by the effort in the same time.
2. Commencing with a small load, find the effort required to raise the load.
3. Repeat for a series of increasing loads.
5
Observations
Calculations
Show one complete set of specimen calculations in your laboratory book.
Graphs
Plot graphs of (a) load against effort;
(b) Load against effect of friction;
(c) Load against efficiency.
And find the law of the machine from graph (a).
Conclusion
The conclusion may be determined as stated in the introductory notes on graphs.
6
EXPERIMENT NO 1 Part 2: THE DIFFERENTIAL WHEEL AND AXLE
Teaching Element:
In Experiment 11 Part 1, it is shown that with a Simple Wheel and Axle the mechanical advantage could be
increased by making the axle smaller compared with the wheel. A more practical way is to use the
DIFFERENTIAL WHEEL AND AXLE. This has the axle made in two parts of different diameter. The axle
rope is wound in opposite directions round the two parts of the axle and passes under a movable pulley to which
the load is attached. When the wheel is turned the rope winds onto the larger axle but off the smaller axle. For
one turn of the wheel the length of the pulley rope taken up will be equal to the difference of the circumference
of the two parts of the axle.
Although a large theoretical Mechanical Advantage can be obtained with a Differential Wheel and Axle, the
practical Mechanical Advantage is smaller due to friction and also because the movable pulley has to be lifted
as well as the load.
Student Objectives:
The object of the experiment is:
1.To determine the Law of the Machine for a differential wheel and axle.
2.To verify that Mechanical Advantage and Efficiency increase with load up to a limiting maximum.
3. To find the effort required to raise each of a range of loads.
4.To find the effect of friction at each load.
Apparatus
Wheel and axle, cord, weights, metre rule.
7
Belts, Chains and Gears
Theory
Calculated V.R.
Consider one turn of axle.
The load is raised a distance equal to one-half the difference between the circumferences of the axles.
= ½(πa – πb)
=1/2π (a-b).
The effort moved a distance equal to the circumference of the effort wheel = πD.
Calculated V.R. =distance moved by the effort / distance moved by the load
=
π .D
1
2π (a − b )
2D
=
a −b
Observations
Calculations:
All complete as in Experiment No.1.
Conclusion:
The conclusion may be determined as stated in the introductory notes on graphs.
8
EXPERIMENT NO 3: THE SCREW JACK
Teaching Element:
The Screw Jack is a simple device for raising a heavy load with comparatively little effort and when only a
short lift is required. The motor car jack is a typical example. It consists essentially of a nut and screw in which
the nut is held stationary while the screw is turned and lifts the weight. The same principle is used in the bench
vice and screw press.
A long lever such as a crowbar will also lift a heavy load through a short distance with only a small effort. With
the screw jack the friction of the screw is sufficient to hold the load in the raised position when the effort is
removed so that it will not run back or "overhaul". This is an important feature of the screw jack.
Student Objectives:
The object of this experiment is:
1.To measure the effort required to raise various loads using a simple form of screw jack and to determine how
the Mechanical Advantage and Efficiency varies with load.
2.To test whether the screw jack "overhauls".
Apparatus :
Screw jack, rope, weights, metre rule.
9
Theory
Calculated V.R.
Consider one turn of wheel.
Distance moved by effort is equal to circumference of wheel =2πR.
Distance moved by the load = pitch of thread = p.
V.R. = distance moved by the effort / distance moved by the load = 2πR / p.
Method :
Observations :
Graphs :
All complete as in Experiment No.1.
(In the apparatus illustrated below, the effort is applied in two parts.)
The sum of these parts should be entered as E in the table of
Conclusion :
10
EXPERIMENT NO 4: THE WORM and WHEEL DRIVE
Teaching Element:
In many rotating machines the input shaft is driven by a high speed electric motor or internal combustion
engine. The output shaft, however, is often required to run at a much lower speed. A method of speed reduction
must therefore be provided and this is usually some form of gearing.
In Simple Gear Trains as in Experiment No 1, it is shown that a pair of gear wheels could be used to provide a
speed reduction in order to gain a Mechanical Advantage. With a single pair of gear wheels the reduction in
speed is produced by making the small wheel drive the large wheel, the speed reduction being determined by
the number of teeth in the gear wheels. When a very large reduction of speed is required, a single pair of gear
wheels would not be practical and additional gear wheels are needed. An alternative method is to use a WORM
AND WHEEL.
The worm is a screw with "threads" which engage in correspondingly shaped (helical) teeth on the wheel. As
the worm revolves, its threads slide along the teeth on the wheel and push them in the direction of the worm
axis, making the wheel turn.
A worm gear provides means of obtaining a large speed ratio and a large Mechanical Advantage. It also enables
the driver and driven shafts to be at 90 degrees, unlike the ordinary gear drive where the shafts are parallel. In
the ordinary gear drive, the gear wheels have "straight teeth" parallel to the shaft. These engaging teeth have a
rolling action, whereas with the worm and wheel, the worm thread has a sliding action along the wheel teeth
and the friction loss is greater. The efficiency of the drive is therefore correspondingly reduced.
A modified form of worm and wheel is used in the steering mechanism of some motor cars to enable the
movement of the steering wheel to turn the front wheels of the car. The worm and wheel is also widely used in
industry for high speed reduction gears.
Student Objectives:
The object of the experiment is:
1.To verify the Speed Ratio of a simple worm and wheel.
2.To measure the Efficiency of the drive under various loads.
Object:
As in Experiment No.2.
Apparatus:
Worm and wheel, cord, weights, metre rule.
11
Belts, Chains and Gears
Theory :
Calculated V.R.
Let T = number of teeth in the wheel, and the worm be single start. One revolution of the effort wheel will
cause the worm to move the wheel to which the load is attached one tooth, i.e. I/T of a revolution.
Distance moved by effort = πd
’’
”
” load = I/T × πD
”
”
” V.R. = Distance moved by effort/Distance moved by load
= πd × T / πD
= Td/D
Method:
As in Experiment No2
Observations:
Calculations:
All as in Experiment No.2
Conclusion:
12
EXPERIMENT NO 5: SIMPLE PULLEYS
Teaching Element:
A pulley is a wheel used to guide ropes or belts in selected directions so that a force transmitted in one direction
can be changed to another direction which is more convenient. The pulley spindle is made small compared with
the diameter of the wheel so that the pulley is easily turned. There is then little loss of effort in turning and the
pull or tension transmitted by the rope is practically the same on both sides of the pulley.
Pulleys can be "fixed" (they do not move bodily relative to the earth) or they can be "movable". Movable
pulleys are supported by the rope passing under the pulley, thus forming a cradle. The pulley block then moves
as the rope moves. By combining fixed and movable pulleys it is possible to lift heavy loads with reduced effort
and so gain a Mechanical Advantage.
Student Objectives:
The object of this experiment is:
1.To test whether the tension in a pulley cord is affected by a change in direction of the cord as it passes over a
pulley.
2.To determine the Mechanical Advantage of a simple combination of fixed and movable pulleys.
Apparatus:
Pulleys, cord, weights, metre rule. (A suggested arrangement of the apparatus is as shown below.)
13
Pulleys and Chain Block
Theory:
Calculated V.R.
If load is lifted a vertical distance of 1 foot each of slack, therefore the effort rope will have to be pulled 3
feet to tighten the ropes. Hence the V.R.is 3 (i.e. the same as the number of ropes supporting the load pulley).
Observations:
Calculations:
Conclusion:
14
EXPERIMENT NO 6: PULLEY BLOCKS
Teaching Element:
Pulley blocks are widely used to lift heavy loads when only "manpower" is available and are sometimes used in
conjunction with motorized lifting tackle. Experiment No 9 shows that by using "fixed" and "movable" pulleys
it is possible to gain a Mechanical Advantage. To develop this principle further, this experiment shows that by
using multiple pulley blocks the Mechanical Advantage can be increased.
Student Objectives:
The object of this experiment is to investigate the mechanical characteristics of a set of pulley blocks which has
three sheaves in the upper block and two sheaves in the lower block.
Apparatus:
Pulleys, cord, weights, metre rule. (A suggested arrangement of the apparatus is as shown below.)
Pulleys and Chain Block
Theory:
Calculated V.R.
If load is lifted a vertical distance of 1 foot each of slack, therefore the effort rope will have to be pulled 3
feet to tighten the ropes. Hence the V.R.is 3 (i.e. the same as the number of ropes supporting the load pulley).
Observations:
15
Calculations:
Conclusion:
16
EXPERIMENT NO 7: THE WESTON DIFFERENTIAL CHAIN BLOCKS
Teaching Element:
The Weston Differential Chain Blocks are used for hoisting heavy loads. This type of hoist is frequently used
on gantry for loading and off-loading heavy equipment because of its simple and effective design. The Weston
Chain Blocks are a variation on the Differential Wheel and Axle, dealt with in Experiment No 1.
Whereas other types of hoists are usually made up of ropes and pulleys, the Weston Blocks consist of an endless
chain and chain wheels. These are very strong and the engagement of the chains and notched chainwheels
prevent slipping.
The top block consists of two wheels which are fixed to rotate together, one wheel being slightly larger than the
other. The bottom block contains one wheel only and acts as a snatch block. The chain is endless and passes
round the larger top wheel, then round the snatch block wheel, then round the smaller top wheel and finally the
two ends are connected and allowed to hang loose.
By making a small difference between the size of wheels in the top block a large Velocity Ratio and a
theoretical Mechanical Advantage can be obtained. In practice the Mechanical Advantage is reduced mainly
due to friction in the system and as a result the Efficiency is low. This, however, is not a disadvantage in that if
Efficiency is less than 50% the machine will not "overhaul". That is to say the system will support a load on the
snatch block without applying a force on the effort chain. This feature makes it a "safe" hoisting machine.
Student Objectives:
The object is to carry out tests on Weston Differential Chain Blocks and show:
1.Load is directly proportional to effort.
2.Mechanical Advantage increases with load up to a limiting value.
3. The method of finding the Velocity Ratio.
4.The Efficiency is such that "overhauling" will not occur.
Apparatus:
Pulleys, cord, weights, metre rule. (A suggested arrangement of the apparatus is as shown below.)
17
Pulleys and Chain Block
Theory:
Observations:
Calculations:
Conclusion:
18
EXPERIMENT NO 8: CHAIN DRIVE
Teaching Element:
When power is transmitted between two shafts by a belt drive, the belt is liable to slip on the pulleys under
heavy loads. In a Chain Drive, the chain consists of a number of pivoted links which fit over suitably shaped
teeth or "sprockets" on the chain wheels. This ensures that no slip takes place, so that the chain drive gives a
"positive' drive". A common example of the chain drive is that used on bicycles.
Since the chain has to fit on both sprocket wheels, the "pitch" of the teeth must be the same on each wheel, and
the number of teeth on each wheel must be directly proportional to their diameters. The speed ratio of a chain
drive can therefore be found by counting the number of teeth on each wheel.
The most common type of chain used in a chain drive is the "roller" chain. The links of this are held together
with pins, and a hardened steel sleeve is fitted over each pin to make a roller which forms the surface in contact
with the teeth. There is, however, some friction set up in the chain and between the chain and the sprocket
wheel. The ratio of the work output to the work input can be used to measure the efficiency of the chain drive.
Student Objectives:
The object of this experiment is:
1.To check the Speed Ratio of a chain drive.
2.To measure the Efficiency of drive transmission.
19
Belts, Chains and Gears
Theory:
Observations:
Calculations:
Conclusion:
20
EXPERIMENT NO. 9 SIMPLE PENDULUM
Object:
(a) To find how the periodic time varies with the length of a simple pendulum.
(b) To determine the value of the acceleration due to gravity.
Apparatus:
Theory:
A small bob suspended on a string, a rule and a stop clock.
Let the bob be displaced slightly from the equilibrium position.
Restoring torque T = WL sin θ
( for small oscillations sin θ = θ ).
Therefore Restoring torque = Wlθ
Now T = Iά where I = mass moment of inertia
=
W
g
l
2
Wlθ =
therefore T =
W
g
2
lα
W
g
ά=
2
lα
g
θ
l
Hence motion is simple harmonic and
l
θ
=
α
g
and tp= 2π√
l
g
Hence tp α√l
Method:
1.
2.
3.
4.
5.
6.
7.
Suspend the pendulum from a rigid support.
Displace the bob slightly from its equilibrium position (ensuring that it oscillates in one plane ).
Take the time for 50 oscillations and hence calculate the time for 1 oscillation, i.e. the periodic time tp
Repeat with 6 different lengths of pendulum.
Plot a graph of t against l.
Plot a graph of t against √l.
Determine the slope of te graph of t against √l and hence determine the acceleration due to gravity (slope
of the graph =2π/√g).
Observations:
Length, l (m)
Time for 50 oscillations,
Sec.
Periodic time, tp
Conclusions:
1. Comment on the shape of the graphs and state the relationship between the periodic time of vibration
and the length of the pendulum as found from the graphs.
2. State the value found for g and compare with the generally accepted figure. Give reason for any
difference in the values.
21
EXPERIMENT NO 10.COMPOUND PENDULUM
Object:
To determine the radius gyration of a pulley by oscillating it as compound pendulum, and hence determine its
moment of inertia.
Apparatus:
Pulley, knife edge support, rule, stop clock and balance.
Theory:
O
θ
h
G
Fig.10
W
Let k be the radius of gyration of the pendulum about its center of gravity G and let h be the distance of the
point of suspension from the center of gravity.
Restoring torque T = W hsinθ (if θ is small sinθ = θ).
Therefore T=Whθ.
Now T= Ioά
Where Io= moment of inertia about O
=IG +Mh2
=W/gk2+W/gh2
=W/g (k2+h2)
Therefore Whθ= W/g(k2+h2)ά
α = gh / (k2+h2) θ
ie ά is proportional to θ
Hence motion is simple harmonic and θ/ά = (k2 + h2)/gh
Therefore tp = 2π√(k2+h2)/gh.
Method:
1.
2.
3.
4.
5.
6.
7.
Weigh the pulley.
Suspend the pulley on the knife edge as shown in Fig.10
Measure the distance from the point of support to the center of gravity of the pulley.
Displace the pulley slightly from its equilibrium position (ensuring that it oscillates in one plane).
Take the time for 50 oscillations and hence calculate the periodic time.
Calculate the value of its radius of gyration about the center of gravity.
Calculate its moment of inertia.
Observations:
weigh of pulley =W= (N)
Distance from knife edge to center of gravity = h = mm.
22
Calculations:
tp = 2π√(k2+h2)/gh.
Hence calculate k.
I=W/gk2
Conclusions:
State the values for k and I.
23
EXPERIMENT NO.11 AXIAL OSCILLATONS
Object:
(a) To find how the periodic time of vibration of a weight suspended on a helical spring varies with the load.
(b) To determine the effective weight of the spring.
Apparatus:
Helical spring, rigid support, weight carrier, weights stop clock and rule.
Theory:
Let W = weight on spring
S = spring stiffness
X = displacement of the load on the spring from its equilibrium position.
When the load is displaced (x) and released,
Restoring force = Sx
But force = mass x acceleration
Therefore Sx=W/gf
f=Sg/W x
i.e. f α x.
Hence motion is simple harmonic and tp = 2π√ displacement/acceleration
=2π√W/Sg
hence tp2α W
Method
1. Hang a weight carrier on the end of the spring.
2. Place a weight on the carrier and note the deflection.
3. Increase the load in small increments and note the deflection for each load.
4. Plot a graph of load against deflection and hence determine the stiffness of the spring (slope of the
graph).
5. Place a small weight on the hanger.
6. Give the weight a small axial displacement.
7. Take the time for 50 oscillations and calculate the periodic time.
8. Repeat (7) for 6 different weights.
9. plot a graph of W against tp2
Observations from items 2 & 3
W, Ibf
In
24
Observations from items 5-8
W,Ibf
Time for
50
oscillation
Periodic
time, tp
tp2
Conclusions
1. Comment on the shape of the graph of W vs. tp2.
2. Is tp2 directly proportional to w? (i.e.is the graph a straight line passing through the origin?)
3. If tp2 is not directly proportional to W what weight would have to be added to w to make tp2 α (W+w)?
(w is called the effective weight of the spring.)
4. What fraction of the weight of the spring is w?
5. State the corrected formula for the periodic time of vibration.
25
EXPERIMENT NO. 12 TORSIONAL OSCILLATIONS
Object To determine the moment of inertia of a pulley by torsional oscillation and to compare it with the value
found in Experiment No. 2.
Apparatus
Coil spring, pulley, disk, rule and stop clock.
Theory
Let t1=periodic time of oscillation for the disk
t2 =periodic time of oscillation for the pulley
I1=moment of inertia of the disk
I2= moment of inertia of the pulley
S= torsional stiffness of the spring
W1=weight of the disk
x1=radius of the disk
I=W1x12/ 2g
Disk:
t1=2π√I1/S
t12=4π2I1/S
pulley:
t2=2π√I2/S
t22 =4π2I2/ S
Therefore t12/t22=I1/I2
I2=(t22/t12 )I1
=(t22 /t12 )W1x12 /2g
26
Method
1. weigh the disk and measure its radius.
2. Attach the disk to the spring.
3. Give the disk a small torsional displacement from its equilibrium position.
4. Note the time for 50 oscillations and calculate the periodic time of oscillation.
5. Repeat (4) several times and find the mean periodic time of oscillation t1.
6. Attach the pulley to the spring and repeat (3), (4)& (5),t2.
7. Calculate the value of I2.
Conclusions
1. State the value of I2 .
2. How does it compare with the value found in Experiment No.2?
3. If there is a difference in the value, comment on the accuracy of each experiment and the reasons for the
difference.
27
EXPERIMENT NO. 13 WHEEL AND AXLE
Object To measure the radius of gyration of a small wheel and axle.
Apparatus Wheel and axle mounted in bearings with axis horizontal, cord, selection of small weights, calipers,
rule and stop clock.
Belts, Chains and Gears
Theory
Let W1=falling weight
W2=weight of wheel and axle
h=height through which W1falls
t=time of fall
N1=number of revolutions made by the wheel while W1 falls through h ft.
N2=number of revolutions made by the wheel in coming to rest after W1 strikes
ground
N=total revolutions made by wheel =N1+N2.
After weight strikes ground, k.e. of ½ Iw2 is used to overcome friction in N2 revolutions; therefore friction work
overcome in one revolution=1/2(Iw2/N2).
Work done against friction when weight is falling in N1 revolution=1/2Iw2(N1/N2).
By conservation of energy
p.e. of weight =k.e. of falling weight on striking ground + k.e. of rotation of wheel +work done against friction
W1h=W1v2/2g+1/2Iw2+1/2Iw2N1/N2…………………(1)
28
w=v/r and v=(mean velocity)×2
=2h/t
I=W2k2/g where W2 is weight of wheel and axle.
Substitution in Equation (1) gives a value of k, the radius of gyration of wheel and axle.
Method
1. Attach a small weight to a piece of cord which is wound round the axle.
2. Adjust the length of the cord so that when the wheel is allowed to rotate the cord leaves the axle as the
weight strikes the ground.
3. Re-wind the cord evenly on the axle until the weight is a known height from the ground.
4. Allow the wheel to rotate under the action of the falling weight and note the time of fall.
5. Determine the number of revolutions N1made when the weight is falling and also the number of
revolutions N2 made after the weight strikes the ground until the wheel comes to rest.
6. Repeat (4)& (5) several times to obtain mean values.
7. Repeat (4)-(6) using a series of in creasing weights.
8. Calculate k for the wheel and axle for each different weight.
9. Calculate the theoretical value of k from k2=R2/2 for solid disk.
Observations
Weight of wheel and axle W2=1bf
Diameter of wheel
d1=in
Diameter of axle
d2=in
Falling weight W1,
1bf
Height of fall h, ft
Time of fall t, sec
Total revolutions of
wheel N
Revolutions
beforeW1strikes
ground N1
Revolutions afterW1
strikes ground N2
Radius of gyration
k, in
Conclusions
1. State the experimental and theoretical values obtained for radius of gyration.
2. Comment on any differences in these values.
29
EXPERIMENT NO. 14 Hydraulic Jack Laboratory Experiment
Mechanical Advantage and Efficiency
Introduction
Hydraulic jacks provide a means of lifting loads that otherwise could not be lifted by conventional mechanical
(screw and scissors) jacks. A common hydraulic jack with a lever arm and pump has two Mechanical
Advantages (MA) built into it. One MA is due to the lever arm and the other is due to the ratio of the ram piston
diameter squared (D2) to the pump piston diameter squared (d2), D2 / d2. The two MA’s combine in a manner
that results in a multiplying effect rather than an adding effect, which yields a much larger overall MA. The
efficiency of a jack can be determined from the basic definition of efficiency, which is output / input. This
laboratory exercise is appropriate for mechanical engineering technology or mechanical engineering students
for a course in applied fluid mechanics, fluid power or machine design. The needed equipment is readily
available off-the-shelf, at a very low cost.
Background
A hydraulic jack has a relatively large overall MA due to the lever arm and the ratio of the ram piston diameter
squared to the pump piston diameter squared. The MA due to the lever arm can be demonstrated and
determined by analyzing the simple lever arm shown in Figure 1. Point A is the lever arm pivot point, Point B is
where the lever arm applies an output force (FOLA = force out, lever arm) to the jack’s pump and Point C is
where the input force (FILA = force in, lever arm) to the system is applied, usually by hand.
Figure 1. Hydraulic jack lever arm.
Summing moments about Point A yields;
Σ MA = 0 = (FOLA * l1) – (FILA * l2)
or
(FOLA * l1) = (FILA * l2)
30
or
FOLA / FILA = l2 / l1.
The mechanical advantage due to the lever arm is either of the above ratios, or
Equation (1)
MAlever arm = FOLA / FILA = l2 / l1.
It is usually difficult to determine FOLA, but easy to measure l1 and l2, and therefore possible to calculate
MAlever arm.
The mechanical advantage due to the ratio of the ram diameter squared to the pump diameter squared can be
developed by analyzing the basic fluid principles of the jack. A simplified hydraulic jack is shown in Figure 2.
Point A is where the force into the system is applied (FILA) and Point B is where the force out of the system
(FOHS = force out, hydraulic system) is used to lift a load. In the case of a jack the force out of the lever arm
(FOLA) is equal to the force into the hydraulic system (FIHS) at the pump.
The definition of fluid pressure is a force per unit area, or in equation form,
Equation (2)
P=F/A
where P = pressure (N/m2, psi),
F = force (N, lbf), and
A = area (m2, in2).
Figure 2. Simple hydraulic jack system.
The pressure throughout the hydraulic fluid is assumed to be constant, and for Figure 2 the following applies,
P = FOHS / AD = FIHS / Ad
31
where P = the pressure of the hydraulic fluid in the jack (N/m2, psi)
AD = the area of the ram piston (m2, in2) and
Ad = the area of the pump piston (m2, in2).
Rearranging the above yields
FOHS / FIHS = AD / Ad = (π D2/4) / (π d2/4) = D2 / d2.
The mechanical advantage of the hydraulic system is any of the above ratios, or
MAhydraulic system = FOHS / FIHS = AD / Ad = D2 / d2.
Equation (3)
Measuring the piston diameters is usually easier than measuring the forces into and out of the hydraulic system.
When students are posed the question "How do you determine the total mechanical advantage in a system with
multiple MAs?", most will reply with "You add them." This, however, is not true. One way to determine how to
deal with multiple MAs is to think of them as being in series inside of a machine. For example, the following
block diagram, Figure 3, demonstrates how two MAs, MA1 and MA2, can be manipulated to determine the
overall MAtotal.
Figure 3. One system with two mechanical advantages, MAtotal = Out / In = 30 / 5 = 2 * 3 = 6.
Figure 3 shows that the total system MA is determined by taking the ratio of the output to the input, or by
multiplying the individual MAs. For the topic at hand, the total mechanical advantage of the hydraulic jack
would be determined by multiplying the MA of the lever arm by the MA of the hydraulic system, or
MAtotal = MAlever arm * MAhydraulic system .
The efficiency of the hydraulic jack can now be determined from the following,
η = (Output / Input) * 100 = (Output Force / (Input Force * MAtotal)) * 100
Equation (4)
= (FOHS / (FILA * MAtotal)) * 100
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where η = the jack efficiency (%).
Equipment
The following is a list of equipment required to perform the experiment.
1. Small 1 or 2 ton hydraulic jack
2. Assortment of small weights
3. Assortment of large weights
4. Weight hanger
5. Load Cell
6. Scale
7. Testing apparatus
The testing apparatus can be constructed out of readily available materials. A testing apparatus that could be
used is shown in Figure 4.
Figure 4. Testing apparatus example.
Experimental Procedure
1. Weigh the weight hanger.
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2. Weigh the assorted small weights.
3. Determine the mechanical advantage due to the lever arm on the jack by measuring the lengths l1 and l2,
and calculating the ratio l2 / l1.
4. Determine the mechanical advantage due to the ratio of the piston diameters squared by measuring the
pump piston and ram diameters, and calculating the ratio D2 / d2. The instructor may choose to
disassemble the jack and measure the diameters before the experiment actually takes place to save time.
5. Determine the total mechanical advantage of the jack by multiplying the individual MAs together.
6. Place the jack in the testing apparatus.
7. Place the weight hanger on the end of the hydraulic jack’s lever arm.
8. Add weights to the weight hanger until the jack just starts to lift the beam. Record the amount of weight
on the weight hanger and the load cell force.
9. Add large weight(s) to the beam.
10. Adjust the position of the lever arm so that it remains horizontal as the hanger weights are applied to
simplify the calculations.
11. Repeat steps 8-10 eight to ten times throughout the load range of the jack.
Student’s Laboratory Report
As a minimum the students should include the following in their report of the results.
1.
2.
3.
4.
5.
6.
Tabulated jack efficiency data for each weight lifted.
A graph of the force applied to the lever arm versus the jack’s output force.
A graph of the jack efficiency versus the jack’s output force.
An explanation of why the jack efficiency does not remain constant.
An explanation of what the system losses are, and how they effect the shape of the efficiency curve.
Determine what the internal pressure inside the jack is when the jack is used to lift a load equal to its
rated capacity.
References
1. Esposito, Anthony, Fluid Power with Applications, 3rd ed., Prentice Hall, Englewood Cliffs, NJ, 1994.
2. Mott, Robert L., Applied Fluid Mechanics, 4th ed., Prentice Hall, Englewood Cliffs, NJ, 1994.
Biography
Paul Ricketts is an Associate Professor of Engineering Technology at New Mexico State University (NMSU).
He has 5 years of industrial experience with an electric utility, and has taught in engineering technology for 11
years. He has a BS in Engineering Technology from NMSU and a MS in Mechanical Engineering from the
University of Texas at El Paso. He is active in the American Society of Mechanical Engineers (ASME) and the
American Society for Engineering Education (ASEE).
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Experiment 15 Moment of Inertia
The moment of inertia I of a body is a measure of how hard it is to get it rotating about some axis. The moment
I is to rotation as mass m is to translation. The larger the I, the more work required to get the object spinning,
just as the larger the mass m, the more work required to get it moving in a straight line. Alloy rim wheels on
bicycles have a lower I than steel rim wheels and so are easier to get spinning, making fast bicycle acceleration
easier.
The moment of inertia is always defined with reference to a particular axis of rotation — often a symmetry axis,
but it can be any axis, even one that is outside the body. The moment of inertia of a body about a particular axis
is defined as:
where the sum is over all parts of the body (labeled with an index i), mi is the mass of part i, and ri is the
distance from part i to the axis of rotation. Performing this sum is easy if the body consists of discrete point
masses. But if the body is a continuous object of some arbitrary shape, then performing the sum requires the
techniques of integral calculus. In this course, we simply tell you the answer for various shapes. For a disk with
an axis through the center of symmetry, the moment of inertia is
.
(1)
Notice that the thickness of the disk does not enter into the expression for Idisk,
which depends only on the radius and the total mass.
In this experiment you will measure I for a disk mounted on an axle. The axle
can be thought of as a very thick disk and you can use the same expression to
compute Idisk and Iaxle. The total I of the disk + axle is the sum of these two.
(2)
In this experiment, you will determine I in two ways. First, you will measure the masses and radii of disk and
axle and then compute I from the formula above. Then you will compute I by timing the wheel as it rolls down
inclined rails and using the principle of conservation of energy.
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Consider the wheel, consisting of disk and axle, rolling down an inclined set of rails after starting from rest at
the top, like so:
The total energy at any time is the sum of the translational kinetic energy, the rotational kinetic energy, and the
gravitational potential energy.
.
(3)
Here, M is the total mass of disk+axle, v is its translational speed , ω is its angular velocity, and h is the height
of the center of mass. Initially, the wheel is at rest at height ho, so its initial kinetic energy (both translational
and rotational) is zero and its total energy is all potential.
(4)
When the wheel reaches the bottom of the rails, h=0, and the energy is all kinetic:
(5)
where vf is the final translational speed and ωf is the final angular speed.
Because the rolling friction is very small, we can assume that the total energy is constant as the disk rolls down
the rails, and so the initial energy is equal to the final energy.
.
(6)
For an axle or wheel that rolls without slipping, the angular velocity ω and the translational speed v are related
by
.
(7)
Note that here and Eqn (8) below, r is the radius of the axle, NOT the radius of the big disk!
Using equations (6) and (7), one can find I in terms of M, r, g, vf, and ho.
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(8)
(You are asked to derive Eqn (8) in the questions.) You will use this expression to determine I using two
different choices of the initial height ho, thus yielding two new values of I that you can compare with I
calculated from Eqn (2).
Part 1. Measurement of I from dimensions and masses of disk and axle
In this experiment, it is a good idea to use centimeters and grams, rather than meters and kilograms for all your
measurements. This is because the moment of inertia of our disks turns out to be a very small number in MKS
units (roughly 10-3 kg⋅ m2) and it is a little awkward to work with small numbers. If you use cgs (centimetergram-second) units, you must be consistent and always use cgs units, so use g = 979.6 cm/s2 (not 9.796m/s2)
Gently slide the axle out of the disk and weigh both separately to find their masses. Measure their diameters to
find their radii, r for the axle and R for the disk. At this stage you do not need to know r very precisely, but you
will in part 2, so measure the diameter of the axle very carefully three or four times with the calipers. Use the
average of your measurements and estimate the uncertainty in r. (If you don't know how to use the calipers, ask
your instructor.)
Using Eqn (1), find Idisk and Iaxle separately and then compute
. (Is Iaxle significant, compared to
Idisk, or can it be ignored?) In using Eqn (1) to compute Idisk, we are making a small error by ignoring the hole in
the center of the disk. Compare the "missing mass" of the hole to the mass of the axle to determine whether this
omission is significant.
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Part 2. Measurement of I using energy conservation.
One end of the rails can be raised and lowered to one of three positions. Place the rails in the lowest position, at
which they are approximately level, and then using the adjustable screws in the base, make the rails exactly
level. Use the bubble level to get a rough level and then place the wheel on the rails to get a precise level. (If the
rails are exactly level, the wheel will not start rolling.)
Raise the movable end of the rails to one of the two upper positions and then fix the two starting blocks, one on
each rail, at some convenient position near the top of the track. Make sure that the starting blocks are level with
each other, so that the axle can be started resting against both blocks and will roll straight down the track when
released. Using the meter stick attached to one rail, record the positions of the sharp tip of the axle in the
starting and stopping positions and compute the distance d through which the wheel rolls. Leave the starting
blocks fixed from now on, so that the value of d is the same for all timings.
To determine the heights h1 and h2 through which the wheel descends, begin by measuring the height changes
H1 and H2 of the end of the rail when it is raised from the level position to the two upper positions. H1 and H2
can be measured quite accurately by measuring the separations of the "notches" that hold up the end of the rail.
Unfortunately, H1 and H2 are not the actual heights through which the wheel descends, since it does not roll the
whole length of the rail. Instead, the situation is as shown below, where d is the distance traveled by the wheel,
while D is the total length of the track (D is measured from the center of the pivot at the bottom to the center of
the support at the top.) The two triangles shown are similar triangles, therefore
calculate h1 and h2 from measured values of d, D, H1, and H2.
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. Use this relation to
Now use the stopwatch to measure the time t1 for the wheel to roll down the rail when it is in position 1 (height
H1). This is best done with the same person operating the stopwatch and releasing the wheel. Make a few trial
runs to determine the best procedure. Have each member of your team measure t1 a few times and record all
values. From your measurements, determine an average value of t1 and estimate its uncertainty. Repeat this
whole procedure for t2, measured when the rail is in position 2 (height H2).
Now calculate the wheel's final speed v at the bottom of its travel for each of the two positions. Be careful! The
quantity d/t is the average speed of the wheel. In the case of constant acceleration, the average speed is related
to the final speed by
.
Label the two final speeds v1 and v2.
Finally, with all your measurements, using Eqn (8), compute I for each of the two positions of the rail. Be
careful to display the correct number of significant figures in your final answers. Display your three final values
for I: your value from part 1 and your two values from part 2. If there is a large discrepancy among the values,
comment on possible sources of experimental error.
Question:
Equation (8) involves g, the acceleration of gravity. This seems to suggest that you would get a different value
for I if you conducted the same experiment on the moon, where g is different. But the definition of I (Eqn (8))
does not depend on location. Like the mass m, the moment I of an object is the same on the moon as on the
Earth or anywhere else. So how do you explain the presence of g in Eqn (8)?
Physics 2010 Pre-Lab Questions Experiment 4
1. Starting with equations (6) and (7), derive equation (8). Be sure to show all your steps.
2. What are the units of moment of inertia I? [Give the answer both in cgs (centimeter-gram-second)
units and in MKS (meter-kilogram-second) units.]
3. For a wheel that rolls without slipping, if you are given its angular speed ω , and its radius r, what is
its translational speed v?
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4. In part 2 of this lab (Eq'n 8) you will need to know vf , the final speed of the wheel. Why is it incorrect
to find vf by simply computing d/t? What is the correct expression to find vf ?
5. In part 2 of this lab (Eq'n 8 again) you will need to know r. Draw a quick sketch of the wheel, and
clearly indicate what "r" is.
6. Dry lab.
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