Lifting Machines

UNIT 12 LIFTING MACHINES

Lifting Machines

Structure 12.1 Introduction Objectives

12.2 Simple Lifting Machines 12.2.1

Definitions

12.2.2

Law of Lifting Machine

12.2.3

Maximum Mechanical Advantage

12.2.4

Maximum Efficiency

12.2.5

Friction Force in a Lifting Machine

12.3 Various Lifting Machines 12.3.1

System of Pulleys or Pulley Blocks

12.3.2

Differential Wheel and Axle

12.3.3

Worm and Worm Wheel

12.4 Summary 12.5 Key Words 12.6 Answers to SAQs

12.1 INTRODUCTION Lifting machines are the devices which are used for materials handling in the industries. They are used for shifting heavy components, heavy machines, etc. from one place of shop to the other place. They are used in loading and unloading cargo for ships at the harbours. Even for loading and unloading heavy loads in goods trains lifting machines are used. In a simple lifting machine, the effort is applied at one point of the machine and the heavy load is lifted at the other point of the machine. The peculiarity of a lifting machine is that by applying smaller force, large loads can be lifted. Objectives After studying this unit, you should be able to understand •

basic theoretical background of simple lifting machines,

•

pulleys and system of pulleys,

•

differential wheels and axle, and

•

worm and worm wheel.

12.2 SIMPLE LIFTING MACHINES There are some technical terms which are used vis-à-vis, lifting machines. These terms are defined in the following section.

12.2.1 Definitions Let a load ‘W’ be lifted by applying an effort ‘P’. When load lifting operation is done, effort and load move unequal distances. Let the load move by a distance ‘x’, and effort move by a distance ‘y’. Mechanical Advantage (M. A.) It is defined as the ratio of load ‘W’ lifted to the effort ‘P’ applied.

M. A. =

Load or weight W = Effort P

87

Introduction to Structural Mechanics

Velocity Ratio (V. R.) It is defined as the ratio of the distance moved by effort ‘y’ to the distance moved by load or weight ‘x’

V. R. =

Distance moved by effort y = Distance moved by load x

Efficiency of a Lifting Machine Efficiency of a machine is defined as the ratio of the output of the machine to the input of the machine. Input of a machine

= work done on the machine = effort × distance moved by the effort =P×y

Output of a machine = work done by the machine = load × distance moved by load =W×x Therefore efficiency of machine W MA W×x η= = P = y P× y VR x

Ideal Machine If a machine has no friction, it will have no power loss. It will have 100% efficiency. Therefore, output of machine will be equal to the input into the machine. Input of machine = Output of machine P×y=W×x Or,

y W = P x

Or,

MA = VR

An actual machine can not be an ideal machine because it will have friction and there will be some power loss due to friction. Reversible and Irreversible Machine Effort is applied in the machine to lift the load. Suppose the load has been lifted by a distance ‘x’. If effort is removed and the load starts falling down, the machine is said to be reversible. If load stays there even if the effort is removed the machine is said to be irreversible. In case of irreversible machine, friction in the machine is sufficient to hold the load in case of effort being removed. Work done by the friction force = Input – Output = Py – Wx Force irreversible machine Work done by friction force > Output of machine

88

∴

Py − Wx > Wx

or

Py > 2 Wx

or

Wx 1 < Py 2

or

η<

Lifting Machines

1 or 50% 2

Therefore, if efficiency of the machine is less than 50%, it is an irreversible machine. Example 12.1 A lifting machine can lift 800 N by the application of 100 N. Distance moved by the effort is 100 cm. At the same time load moves by 10 cm. Determine the mechanical advantage, velocity ratio and efficiency of the machines. Solution Load lifted = 800 N Effort applied = 100 N Distance moved by effort = 100 cm Distance moved by load = 10 cm Mechanical Advantage (M. A.) = Velocity Ration (V. R.) = Efficiency (n) =

W 800 = =8 P 100

y 100 = = 10 x 10

MA 8 = = 0.8 or 80% VR 10

MA = 8, VR = 10 and η = 80%

Ans.

Example 12.2 An effort of 400 N is applied in a lifting machine. The distance moved by effort is 4 m and that by load is 40 cm. If the efficiency of the lifting machine is 70%, determine (i)

load lifted by the machine

(ii)

mechanical advantage

(iii)

velocity ratio

Solution Effort applied (P) = 400 N Distance moved by effort (y) = 4 m Distance moved by load (x) = 40 cm = 0.4 m Efficiency of the lifting machine η = 70% or 0.7. (i)

Output = Load × distance moved by load = W × 0.4 Input = 400 × 4 = 1600 Nm

η= or, W =

Output 0.4 W , or = 0.7 = Input 1600 0.7 × 1600 = 2800 N 0.4

Ans. 89


(ii)

Mechanical Advantage (M. A.) =

(iii)

Velocity Ratio (V. R.) =

W 2800 = = 7.0 P 400

y 4 = = 10 x 0.4

Ans. Ans.

SAQ 1 How much mechanical advantage will be for an ideal machines?

12.2.2 Law of Lifting Machine The law of the lifting machine is obtained by drawing a graph between effort and corresponding load lifted by the machine. Figure 12.1 shows such a graph. The various values of effort and the corresponding loads have been plotted in this figure. The average line AB can be drawn through the points. Mathematically, the law of the machine can be represented by a straight line P=mW+C

. . . (12.1)

where, m is slope of the line AB and C is the intercept of the line on Y-axis, m and C are constants.

Figure 12.1

12.2.3 Maximum Mechanical Advantage Law of the machine is P=mW+C

W =

or

90

M. A. =

P−C P ⎛ C⎞ = ⎜1 − ⎟ m m⎝ P⎠ W 1 ⎛ C⎞ = ⎜1 − ⎟ P m⎝ P⎠

. . . (12.2)

⎛C ⎞ When an effort (P) increases, ⎜ ⎟ reduces and mechanical advantage increases as ⎝P⎠ C is very indicated by the expression (12.2). If the value of P is very large such that P very small, results in maximum mechanical advantage. Therefore, maximum mechanical advantage can be expressed as :

Maximum M. A. =

Lifting Machines

1 m

. . . (12.3)

12.2.4 Maximum Efficiency From Eq. (12.2), the expression of mechanical advantage is given by : M. A. = Efficiency =

W 1 ⎛ C⎞ = ⎜1 − ⎟ P m⎝ P⎠ MA 1 C⎞ ⎛ = ⎜1 − ⎟ VR m × VR ⎝ P⎠

Efficiency of machine will be maximum when

C is negligible or effort is very very P

large. Therefore, Maximum efficiency, (η) max =

1 m × VR

. . . (12.4)

12.2.5 Friction Force in a Lifting Machine The effort required to lift the load W is P. In an ideal lifting machine MA = VR.

Ideal load = VR P

or,

Ideal load lifted = P × VR

or,

Therefore, Friction force (F) = Ideal Load lifted – Actual load lifted F = P × VR – W The above expression determines friction force in terms of load lifted. The friction force in terms of effort applied will be expressed as :

FP =

F W =P− VR VR

Example 12.3

For a lifting machine 15 N effort is required to raise a load of 70 N with an efficiency equal to 60%, and 25 N effort is required to lift a load of 1300 N. Determine the law of machine. What will be the effort required to raise a load of 1000 N? Determine maximum mechanical advantage and the maximum efficiency. Solution

Given : Effort P1 = 15 N, and load W1 = 700 N Effort P2 = 25 N, and load W2 = 1300 N Let m be the gradient and C be the intercept on y-axis. The law of machine can be expressed as : P=mW+C

. . . (a)

Substituting for P and W in Eq. (a), we get :

and

15 = m 700 + C

. . . (b)

25 = m 1300 + C

. . . (c) 91


Subtracting Eq. (b) from Eq. (c), 10 = 600 m, or m =

10 1 = 600 60

Substituting m in Eq. (b)

1 × 700 + C 60

15 =

C = 15 −

or

70 = 3.33 6

Substituting values of m and C, we get the law of machine as follows :

P= When load

W + 3.33 60

Ans.

. . . (d)

W = 1000 N

P=

1000 + 3.33 = 19.99 N 60

Ans.

Maximum Mechanical Advantage Max. M. A. =

1 1 = = 60 1 m 60

Ans.

Maximum Mechanical Efficiency Max. η =

1 m × VR

For 15 N effort to raise a load equal to 700 N, M. A. = Therefore,

0.6 = V. R. =

or, Therefore,

Max. η =

700 = 46.66 15 46.66 VR 46.66 = 77.77 0.6 1 60 = = 0.77 or 77% Ans. m × VR 77.77

Example 12.4

The efficiency of a machine is 75% when an effort of 20 N is required to lift a load of 160 N. Calculate the velocity ratio and frictional force of the machine in terms of effort and load. What is the law of the machine if frictional resistance of the machine is constant? Determine the effort required to run this machine at a load of 200 N. Solution

Given : Efficiency (η) = 75% Effort (P) = 20 N 92

Load lifted (W) = 160 N

Mechanical Advantage (M A) = Velocity Ratio (V R) =

Lifting Machines

160 =8 20

MA 8.0 = η 0.75

= 10.67

Ans.

Frictional force in terms of load (FW) = P × V R – W FW = 20 × 10.67 – 160

or

= 53.4 N Frictional force in terms of effort ( FP ) = P −

FP = 20 −

or

W VR 160 =5N 10.67

The effort required to overcome friction is 5 N, which is constant. In the law of machine, the factor C is constant which is equal to effort required when load lifted is zero. C = FP = 5 N

Therefore,

Therefore, law of the machine can be written as : P=mW+5 For

P = 20 N and W = 160 N 20 = 160 m + 5

m=

or

20 − 5 15 3 = = 160 160 32

Substituting for m, the law of machine is given by :

P= For load

3 W +5 32

Ans.

W = 200 N

P=

3 × 200 + 5 = 23.75 N 32

Ans.

SAQ 2 (a)

The effort applied to raise a load is 50 N. The effort lost in overcoming friction is 10% of it, and the velocity ratio of lifting machine is 10. Determine the load lifted.

(b)

What load will be lifted by an effort of 12 N if the velocity ratio is 20 and efficiency of the machine at this load is 60%. If the machine has constant frictional force, determine the law of the machine, and find the effort required to lift a load of 100 N and efficiency of the machine.

(c)

Under which condition the mechanical advantage will be equal to the velocity ratio?

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12.3 VARIOUS LIFTING MACHNIES Lifting machines are used to lift large loads. Their use becomes necessary in handling of heavy machines, loading and unloading of ships, railway wagons, etc. There are many types of lifting machines; a few of them are being explained in the following sections.

12.3.1 System of Pulleys or Pulley Blocks Pulley is a wheel which is made of metal or wood. It has a groove around its circumference for accommodating rope or chain. Several pulleys can be arranged to make a pulley system. The following assumptions are made while deriving formulae for different pulley systems : (i)

The weight of pulley is very small as compared to the weight lifted and therefore, it is neglected.

(ii)

The friction between the pulley and rope is negligible. Therefore, the tensions on the two sides of the rope passing round the pulley are taken equal in magnitude.

The pulleys are used for lifting loads employing their suitable combinations. These combinations are used to obtain higher mechanical advantage. There are four system of pulleys which are extensively used : (i)

First system of pulleys

(ii)

Second system of pulleys

(iii)

Third system of pulleys

(iv)

Weston’s differential pulley block.

First System of Pulleys

Figure 12.2 shows the arrangement of pulleys in the first system of pulleys. In this system one pulley remains in the same position with the help of fixed block, whereas other pulleys move with their blocks. One end of the string or rope is tied to the fixed support and the other end is fastened to the block of the next higher pulley after it passes round the periphery of bottom pulley. The load is attached to the lower most pulley. The number of pulleys that are used depends on the velocity ratio required.

Figure 12.2

94

Let y be the distance covered by effort. This length of the string has been contributed by the string around pulley 3. For movement of the string by a distance y y y, pulley centre moves by . The upward movement of the pulley 3 is . The 2 2 1 ⎛ y⎞ upward movement of the centre of pulley 2 will be ⎜ ⎟ by virtue the same 2 ⎝2⎠ argument. Applying the same reasoning, the centre of pulley 1 will move by 1⎛ y ⎞ y . ⎜ ⎟= 2 ⎝ 2 2 ⎠ 23

Lifting Machines

Since weight is attached to the block of pulley 1, therefore, movement of the y load = 3 . 2 Therefore, velocity ratio (V. R.) =

y = 23 = 8 ⎛ y ⎞ ⎜ 3⎟ ⎝2 ⎠

If there are ‘n’ moving pulleys V. R. = 2n Mechanical advantage (M. A.) =

W . P

Second System of Pulleys

Figure 12.3 shows the pulley arrangements for second system of pulleys. It consists of two pulley blocks. Each block consists of several pulleys. Upper block frame is connected to the fixed support whereas lower block frame can move up and down. Both the blocks may have equal number of pulleys, or upper block may have one pulley more than the lower block. If the upper block has one pulley more than the lower block, one end of rope or string is tied to the frame of lower block. The string then goes around different pulleys and effort is applied at the other end. The load is suspended to the frame of lower block.

Figure 12.3

95


If both the blocks have equal number of pulleys, one end of the string is tied to the frame of upper block. The string then goes round the pulleys and effort is applied at the other end. A friction is assumed negligible, the tension in the string will be the same throughout and equal to the effort (P). Let n be the number of string segments supporting the lower block. For an ideal system, ∴

W=nP

or

W =n P

or

Velocity ratio = n.

If weight of lower block is appreciable in comparison to the load, it should be taken into account along with the load W. Let w be the weight of lower block. Therefore,

nP=W+w

W w =n− P P Mechanical advantage (MA) =

W . P

It may be noted that the number of string segments will be equal to the total number of pulleys in the two blocks. Third System of Pulleys

Figure 12.4 shows the third system of pulleys. The frame of the top most pulley is connected to the fixed support, and other pulleys are arranged in such a manner that they can move up and downs shown in the figure.

(a)

(b) Figure 12.4

Considering free body diagram T1 + T2 + T3 + T4 = W

96

If there is no friction, and treating the system to be ideal, we have :

Lifting Machines

T1 = P Considering the free body diagram of the lower most pulley, T2 = 2T1 = 2 P

Similarly,

T3 = 2T2 = 2 2 P

and

T4 = 2T3 = 23 P

Therefore,

( 23 + 2 2 + 2 + 1) P = W

or

W = 1 + 2 + 2 2 + 23 P

Since the system is ideal, VR =

W = 1 + 2 + 22 + 23 P

which can be written as = 24 − 1 . If there are ‘n’ pulleys, we can write : V R = 2n − 1 Weston’s Differential Pulley Block

Figure 12.5 shows a Weston’s Differential Pulley Block. It consists of upper block A comprising two pulleys 1 and 2 that rotate together and their frame is suspended from fixed supports. The lower block has only one pulley which moves up and down and its frame supports the load to be lifted. A single rope or string or chain passes around the pulleys as shown. If chain is used, it is better the upper block pulleys are provided with the required teeth. The effort P is applied to the chain passing over the bigger pulley of the upper block.

Figure 12.5

97


Let D be the diameter of pulley 1, and d be the diameter of pulley 2. As the effort is applied, the string unwinds from pulley 1 and it winds on pulley 2. this results in shortening of the string on the load side. For one rotation of pulleys 1 and 2, net shortening of string = π D − π d = π (D − d ) . This shortening of the string will result in the lifting of the load by a distance

= VR =

π (D − d ) 2 πD 2D = π (D − d ) D − d 2

If T1 and T2 are the member of teeth on pulley 1 and 2, respectively, VR =

2T1 T1 − T2

Example 12.5

A Weston differential pulley consists of a lower block and an upper block. The upper block has two cogged grooves – one of which has a radius of 12.5 cm and the other has a radius of 11.5 cm. If the efficiency (η) of the machine is 40%, calculate the effort required to raise a load of 150 kg. Solution

Diameter of larger pulley, D = 2 × 12.5 Diameter of smaller pulley, d = 2 × 11.5 = 23 cm Let the required effort be = P VR =

2D 2 × 25 = = 25 D − d 25 − 23

MA =

W 150 = P P

150 MA 6 η= ⇒ ∴ 0.4 = P = 25 VR P ∴

P=

6 = 15 kg 0.4

(Ans.)

Example 12.6

The efficiency of the Weston differential pulley block is 60%. The upper block has two pulleys of diameters 22 cm and 20 cm, respectively. Find the load which can be lifted by a human being who can apply effort of 300 N. Solution

Given

Efficiency (η) = 60% = 0.6 Diameter of larger pulley (D) = 22 cm Diameter of smaller pulley (d) = 20 cm Effort applied (P) = 300 N V. R. =

98

2D 2 × 22 44 = = = 22 D − d 22 − 20 2

η=

Lifting Machines

MA MA , or 0.6 = VR 22

MA = 0.6 × 22 = 13.2

or

MA =

W W , or 13.2 = P 300

W = 13.2 × 300 = 3960 N

or

Ans.

Example 12.7

A load of 1000 N is to be lifted by using 5 pulleys. If the efficiency of the system is 70%, determine the effort required in the I, II and III systems of pulleys. Solution

Given

Efficiency (n) = 70% Load lifted (w) = 1000 N

I System of Pulleys V. R. = 25 = 32

η= or

MA MA , or 0.7 = VR 32

MA = 0.7 × 32 = 22.4

MA =

W = 22.4 P

or

1000 = 22.4 P

or

P=

1000 500 = = 44.64 N 22.4 11.2

Ans.

II System of Pulleys V. R. = 5 MA = η VR = 0.7 × 5 = 3.5

MA = P=

or

W = 3.5 P 1000 = 285.7 N 3.5

Ans.

III System of Pulleys V. R. = 2 5 − 1 = 32 − 1 = 31 MA = η VR = 0.7 × 31 = 21.7

W 1000 = 21.7 or P = = 46.08 N P 21.7

Ans.

SAQ 3 For the same number of pulleys which pulley system out of I, II and III system is more efficient and why?

99


12.3.2 Differential Wheel and Axle Figure 12.6 shows the block diagram of differential wheel and axle. The shaft is supported by the bearings at its ends. Over this is mounted a formation consisting of three cylinders of difference diameters. It can either be forged with the shaft or these three cylinders can be keyed to the shaft. This is one of the most efficient lifting machines and very large amount of velocity ratio can be achieved in this machine.

Figure 12.6

There are two strings or ropes. One of them is wound around the cylinder of the largest diameter as shown in Figure 12.6. The ends of other string are tied on the cylinders B and C, and the string is wound on both of them in the opposite sense. The winding of strings on cylinders A and C is in same sense. As the effort is applied on one end of the string wound on A, it winds on cylinder B and unwinds from C at the same time. The load is mounted on the pulley which rests on the string as shown in Figure 12.6. The load is lifted up by application of effort due to the difference in diameters of B and C. Let P be the effort, W be load lifted, D be the diameter of largest cylinder A, d1 be the diameter of B, and d2 be the diameter of C. Let the axle be rotated by one revolution, the net length of string wound = π d1 − π d 2 = π ( d1 − d 2 )

∴

The height by which the load is lifted =

π (d1 − d 2 ) . 2

The length of string unwound from cylinder A is πD. Therefore, velocity ratio, VR =

or

VR =

Mechanical advantage M A =

100

πD π (d1 − d 2 ) 2

2D ( d1 − d 2 ) W P

Therefore, efficiency,

η=

MA W (d1 − d 2 ) = VR 2 DP

Hence,

η=

W (d1 − d 2 ) 2 DP

Lifting Machines

Example 12.8

An effort of 180 N is required to lift a load of 7200 N in differential wheel and axle. The diameter of wheel is 25 cm. The larger and smaller diameters of the differential axle are 11 cm and 10 cm, respectively. Determine the velocity ratio, mechanical advantage and efficiency. Solution

Given data : Effort (P) = 180 N Load (W) = 7200 Diameter of largest cylinder or wheel (D) = 25 cm Diameter of larger cylinder (d1) = 11 cm Diameter of smaller cylinder (d2) = 10 cm V. R = M. A. =

η=

2D 2 × 25 = = 50 d1 − d 2 11 − 10 W 7200 = = 40 180 P MA 40 = = 0.8 or 80% VR 50

SAQ 4 In which respect is a differential wheel and axle a better system than a pulley systems?

12.3.3 Worm and Worm Wheel Figure 12.7 shows a worm and worm wheel. It consists of a worm which may be single start, double start or of higher starts. It gears with a toothed wheel which is called worm wheel. A wheel ‘A’ is attached to the shaft of worm over which a string or rope passes. The effort is applied at the free end of this string. The load is suspended through rope on the drum attached to the worm wheel as shown in the Figure 12.7.

Figure 12.7

101


Let W be the load lifted, P be the effort applied, R be the radius of wheel, r be the radius of load drum, T be the number of threads on work wheel, and n be the start of the worm. For one rotation of the effort wheel, Distance moved by the effort ‘P’ = 2 πR. Distance moved by the load ‘W’ = 2 π r × VR =

Velocity ratio,

Mechanical advantage, M A =

η=

∴ Efficiency

2πR n 2πr × T

n T =

RT rn

W P MA VR

Example 12.9

In a single-threaded worm and work wheel, the number of teeth on the worm wheel is 50. The diameter of the effort wheel is 20 cm and that of the load drum is 10 cm. Calculate the velocity ratio; and if the η of the machine is 20%, determine the effort required to lift a load of 30 kg. Solution

T = 50 Diameter of effort wheel = 20 cm ∴ Radius of effort wheel =

20 = 10 cm 2

Diameter of load drum = 10 cm ∴ Radius of the load drum = VR =

10 = 5 cm 2

Radius of effort wheel × No. of teeth on worm wheel Radius of load drum

W = 30 kg =

10 × 50 = 100 (Ans.) 5

If P is the effort required to lift the load, MA =

W 30 = P P

30 MA 3 η= ⇒ 0.2 P = VR 100 10 P ∴

P=

3 1 × = 1.5 kg 10 0.2

(Ans.)

Example 12.10

102

The number of teeth on the worm wheel are 80. The radius of the effort wheel is 30 cm and that of load cylinder is 10 cm. Determine the load which can be lifted by applying 200 N effort. Take efficiency as 20% when worm is (i) single start and (ii) double start.

Lifting Machines

Solution

Given data : Radius of effort wheel (R) = 30 cm Radius of load wheel (r) = 10 cm Efficiency of machine = 20% or 0.2 Number of teeth on worm wheel = 80. Load (W) = ? Single Start Worm VR =

η= ∴

TR 80 × 30 = = 240 r 10 MA MA , or 0.2 = VR 240

M A = 0.2 × 240 = 48 MA =

W = 48 P

W = 48 × P = 48 × 200 = 9600 N

∴ Double Start Worm

VR =

η=

80 × 30 TR = = 120 m×r 2 × 10 MA MA , or 0.2 = VR 120

∴

M A = 24

or

W = 24 P W = 24 × 200 = 4800 N

Ans.

SAQ 5 (a)

Out of all the lifting machines discussed herewith which one is an irreversible machine?

(b)

Can any ideal lifting machine be made? If not why?

(c)

A system of pulleys (Figure 12.8) is given. Obtain the accelerations of the weight if P = 300 N, and Q = 200 N. Assume pulleys to be weightless and frictionless; and the string to be inextensible.

String P

Q

Figure 12.8

103


12.4 SUMMARY In this unit the basic definitions of the lifting machines, and some of the lifting machines have been explained. For each of these machines, derivation of the formulae for mechanical advantage, and velocity ratio was explained. The use of formulae was also explained through some examples. Therefore, after going through this unit you must have understood the following : •

The basic theoretical background of simple lifting machines.

•

Pulleys and various systems of pulleys.

•

Differential wheel and axle.

•

Worm and worm wheel.

The lifting machines are mostly used in materials handling which is very common in our daily life.

12.5 KEY WORDS Mechanical Advantage

: It is the ratio of load lifted to the effort applied.

Velocity Ratio

: It is the ratio of the distance travelled by effort to the distance travelled by load.

Efficiency of Machine

: It is the ratio of mechanical advantage to the velocity ratio.

Ideal Machine

: This machine does not result in any power loss; therefore, input to the machines is equal to the output of the machine.

Reversible and Irreversible Machine

: If efficiency of the machine is more than 50%, it is a reversible machine. If efficiency is less than 50%, it is an irreversible machine.

Law of Lifting Machine

: It is the expression through the depiction of a straight line by which its effort-load relationship can be expressed.

System of Pulleys

: It is the arrangement of several pulleys in order to make a lifting machine.

Differential Wheel and Axle

: It is a lifting machine in which three cylinders/discs are used to generate differential movement of string/rope.

Worm and Worm Wheel

: This is also a lifting machine in which worm and worm wheel are used. They are the spiral gears for which axes make 90o and lie in different parallel planes.

12.6 ANSWERS TO SAQs SAQ 1

Available in the text. SAQ 2

(d)

Given data : Effort applied (P) = 50 N

104

Effort lost in overcoming friction = 10% of effort

=

Lifting Machines

10 × 50 = 5 N 100

Velocity Ratio (V R) = 10 Load lifted (W) = ? Friction force in terms of effort,

FP = P −

W VR

Substituting the values, we have

5 = 50 −

50 = 500 − W or W = 550 N

or (e)

W 10

Given data : Effort applied (P) = 12 N Velocity ratio (V R) = 20 Efficiency (η) = 60% or 0.6

η= or

0.6 = MA =

MA VR MA , or M . A. = 0.6 × 20 = 12 20 W P

or

W = 12 12

or

W = 12 × 12 = 144 N

Friction force in terms of effort,

FP = P −

W 144 = 12 − VR 20

= 12 − 7.2 = 4.8 N Law of the lifting machine is stated as : P = m W + 4 .8 For

P = 12 N

and

W = 144

or

12 = m × 144 + 4.8

or

m=

12 − 4.8 7.2 1 = = 144 144 20

Therefore,

P=

W + 4.8 20

For

W = 100 N

P=

100 + 4.8 = 5 + 4.8 = 9.8 N 9.8

105


MA =

100 9.8

100 MA 9.8 5 η= = = = 0.51 = 51% VR 20 9.8 (f)

Available in the text.

SAQ 3



(a)


(b)


(c)

Let T1 be the tension in the rope.

T1 T1 T1

T1 T1

T1

2T1 Q

2T1

P

Figure 12.9

x as the distance moved by Q will be 2 shared equally by the two strings of the pulley loop. Hence, the distance to be covered by Q will be twice that of P in the same time. If Q moves by a distance x, P will move by

∴ The acceleration of Q will be twice that of P. Let the acceleration of Q be 2a and that of P will be a. Let

MP = Mass of P = 300′ g

and

MQ = Mass of Q = 200′ g

Applying D’Alembert’s principle (≡ Newton’s second law of motion) to the motion of P and Q : M P × a = P − 2 T1

106

300 a = 300 g

or

2 T1 +

Similarly,

M Q × (2a) = T1 − Q

or

200 × 2a + 200 = T1 g

. . . (i)

. . . (ii)

Eqs. (i) and (ii) yield,

Lifting Machines

a = − 0.8918 m/sec 2

− ve sign meaning that the acceleration for P is in the downward direction. Hence, acceleration of Q will be = 1.7836 m/sec2 in upward direction.

FURTHER READING Dr. R. K. Bansal, “Engineering Mechanics”, Laxmi Publications.

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F. L. Shingar, “Engineering Mechanics”. Timoshanko, Young, “Engineering Mechanics”, Mc-Graw Hill Company. K. L. Kumar, “Engineering Mechanics”.

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INTRODUCTION TO STRUCTURAL MECHANICS

Lifting Machines

You have already studied about Statics and Dynamics in previous two blocks. Now you will be introduced to Structural Mechanics and Lifting Machines which involve analysis of structures and basic knowledge of lifting machines. Since a structure should be stable, therefore this block also deals with stability aspects of load carrying members or structures. This subject is inter-disciplinary. Mechanical engineers need structures for designing beams, wall brackets, etc. and lifting machines for loading and unloading whereas civil engineers require it in designing buildings and bridges. It is required by designers of offshore structures and by electrical and chemical engineers also for designing portions of their equipment. At the same time lifting machines are used for materials handling by almost every body. The subject of Structural Mechanics is very vast but we shall be dealing only with basic fundamentals. Depending on requirement of vast utility of these topics, it has been thought fit to explain this block in 4 units. Unit 9 introduces the reader to a very important tool of analysis. This tool can be used for analysis of problems in statics and dynamics. One of the advantages of this is that it provides very simple equations and they are also very few in number. Unit 10 is concerned with determination of axial forces in various members of different types of plane trusses by method of joints, method of sections and by graphical method. Unit 11 is devoted to the analysis of frames. The method of members has been explained for analysis of frames. This method can be used to determine reactions at supports and forces at the joints. Unit 12 is concerned with lifting machines. The basic fundamentals of lifting machines have been explained. The different type of pulley blocks which are very commonly used for loading and unloading also have been explained. Other machines like differential ‘wheel and axle’ and ‘worm and worm wheel’ also have been explained in this unit. At the end of the block, a list of books has been enclosed. These books may be helpful to understand more about the above mentioned topics.

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Lifting Machines

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