machines mteducare icse 10th reference notes

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TEST/ASSIGNMENT FOR STD 7th /8th /9th /10th TOPIC: SUB: BOARD:CBSE/ICSE/SSC

Ans. In crowbar; F is in between L & E

Therefore, it is class -I lever (First order)

Load arm = 25cm; Effort arm = 150-25 = 125 cm.

So

What is the relationship between the mechanical advantage and the velocity ratio for: (1) An ideal machine; (2) A practical machine? Give two reasons why the efficiency of a single movable pulley system is not 100%. Explain why scissors for cutting cloth may have blades much longer than the handles; but shears for cutting metals have short blades and long handles. A woman draws water from a well using a fixed pulley. The mass of the bucket and water together is 6.0 kg. The force applied by the woman is 70 N. Calculate the mechanical advantage. PROB. 1. The diagram shows the use of a lever.

i) State the principle of moments as applied to the above lever.

ii) Give an example of this class of lever.

iii) If FA = 10. cm, AB = 500 cm, calculate the mechanical advantage and minimum effort

required to lift the load.

Ans.

i) M.A. = 1 ,

ii) Effort = 2550 N

PROB. 2.

× h h =

[3][2]

(ii) (Take g = 10 m/s2)

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(b) The radius of the driving wheel of a set of gears is 18 cm. It has 100 teeth and rotates at a speed of 30 rpm. The driven wheel rotates at a speed of 150 rpm. Calculate: (i) The gear ratio. (ii) The number of teeth on the driven wheel. (iii) The radius of the driven wheel.(b) (i)

(b) Name the type of single pulley that can act as a force multiplier. Draw a labelled diagram of the above named pulley. [3](c) A pulley system has a velocity ratio of 4 and an efficiency of 90%. Calculate:- (i) the mechanical advantage of the system.

(ii) the effort required to raise a load of 300 N by the system. (a) Which class of levers has a mechanical advantage always greater than one? What change can be brought about in this lever to increase its mechanical advantage? [2]

Q (b) Diagram 1 below shows a weightless lever in equilibrium. Neglect friction at the fulcrum F.

(i) State the principle of moments as applied to the above lever.(ii) Define mechanical advantage and calculate its value for the given lever.(iii) Name the type of lever which has mechanical advantage greater than 1. [4]

(ii) Mechanical advantage. The mechanical advantage or the force ratio of a machine is the number which expresses the ratio of the resistance overcome(LOAD) to the effort applied to the machine to produce equilibrium. Machines are usually so constructed that the mechanical advantage is greater than unity orW > P.(iii)

PROB. 2. The diagram shows the use of a lever.

i) State the principle of moments as applied to the above lever.

ii) Give an example of this class of lever.


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iii) If FA = 10 cm, AB = 500 cm, calculate the minimum effort required to lift the load.

Ans.

i) The vector sum of moments of all coplanar forces acting on a body in

the equilibrium state about any point is zero. In the above case

moment of effort and moment of load about F is zero.

ii) This is class III type of the lever, as the effort lies in between the

fulcrum and load. ex. Fire Tongs

iii) Taking moments about F,

moment of Effort about F = moment of load

510×10 = 500 about FE

Therefore, E = 2550 N

Effort ARM Load ARM

It consists of two sets of pulleys Each set contains two or more pulleys These pulleys are mounted on a common axle and are of same radius

(A) MACHINES, LEVERS AND INCLINED PLANE

3.1 MACHINESIt is our common experience that when we find difficult to open a nut, we use wrench to

open it. It is difficult to lift up a bucket of water directly from a well, but it becomes easier to lift it

up with the use of a pulley. We can find many such examples in our daily life where the use of a

machine (such as wrench, pulley, etc.) makes the job easier. Thus we define a machine as

below :


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Functions and uses of simple machines :Technical Terms Related to a Machine

Load : The resistive or opposing force to be overcome by a machine is called the load (L).Effort : The force applied on the machine to overcome the load is called the effort (E).

Mechanical advantage (M.A.) : The ratio of the load to the effort is called the mechanical advantage of the machine, i.e.,

..(3.1)

A machine, which needs a small effort to overcome a certain load, is said to have

mechanical advantage greater than one, while the machine which needs a large effort to

overcome a load is said to have mechanical advantage less than one. A machine having

mechanical advantage greater than one, works as a force multiplier, while the machine having

mechanical advantage less than one, is used to obtain gain in speed.

Since mechanical advantage is the ratio of two similar quantities, so it has no unit.

Velocity ratio (V.R.) : The ratio of the velocity of effort to the velocity of load is called the

velocity ratio of machine i.e.,

If d and d are the distances moved in time t by the load and the effort respectively, then

Velocity of load =


Load (L)Mechanical advantage (M.A.) = Effort (E)

Velocity of effortVelocity ration (V.R.) = Velocity of load

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Velocity of effort =

Velocity ratio (V.R.) =

Or ……..(3.2)

Thus the velocity ratio is also defined as the ratio of the displacement of effort to the

displacement of load.

A machine in which displacement of load is more than the displacement of effort is said

to have the velocity ratio less than I and such a machine is used to obtain gain in speed. On the

other hand, a machine works as force multiplier if its velocity ratio is more than I, i.e.,

displacement of load is less than the displacement of effort.

Since the velocity ratio is also the ratio of two similar quantities, so it has no unit just

like M.A.

Work input : The work done on the machine by the effort, is called the work input (W

input), i.e.,

Work input = Work done by effort.

Work output : The work done by the machine on the load, is called the work output (W

output), i.e.,

Work output = Work done on load.

Efficiency ( ) : Efficiency of a machine is the ratio of the useful work done by the machine to

the work put into the machine by the effort.


V.R = d d

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In the words, efficiency is the ratio of the work output to the work input. It is denoted by

the symbol (eta).

Efficiency is usually expressed as a percentage, so we may write

.. (3.3) .. ….(3.3)

It has no unit since it is also the ratio of two similar quantities.

Principle of a machine

When energy is supplied to a machine by the effort, it does some useful work. The

energy supplied to a machine is called the input (W input) and the useful work done by the

machine is called the output (W output.)

The point at which energy is supplied to a machine by applying by applying the effort, is

called the effort point and the point where energy is obtained by overcoming the load is called

the load point.

Input energy = Work done at the effort

Point

= Effort x displacement of

point of application of the

effort.


Efficiency = Work output (W output) Work input (W input X 100%

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Output energy = Work done at the load

Point

= Load x displacement of

Point of application of the

Load.

From the law of conservation of energy, it is obvious that he work done by a machine

(i.e., output energy) can never be greater than the work done on a machine (i.e., input energy).

Thus, no machine can have efficiency greater than 1 or 100%.

For an ideal machine, in which there is no dissipation of energy in any manner, the work

output will be equal to the work input. i.e., the efficiency of an ideal machine is 100%.

In practice, however, no machine is ideal or 100% efficient for the following

reasons :

(i) the moving parts in it are not perfectly smooth (or frictionless).

(ii) the string in it (if any ) is not perfectly elastic.

(iii) its different parts are not perfectly rigid, and

(iv) its moving parts are not weightless.

Due to above factors there is always some loss of energy in the machine. Thus, the

output energy from a machine is always less than the input energy to it No machine is

100% efficient.


Note : The energy lost in overcoming the force of friction between the moving parts of a machine is the most common type of loss of energy in it.

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If a machine is 80% efficient, it implies that 80% of the total energy supplied to the

machine is obtained as useful energy. The remaining 20% of the energy supplied is lost in

overcoming friction etc.

3.2 RELATIONSHIP BETWEEN EFFICIENCY ( ), MEHANICAL ADVANTAGE (MA) AND

VELOCITY RATIO (VR)

et a machine overcomes a load L by the application of an effort E. In time t, let the

displacement of effort be d and the displacement of load be d and the displacement of load be

d.

Work input = Effort x displacement of

Effort

= E x d

Work output = Load x displacement of

Load

= L x d

Work outputEfficiency = Work input

or …..(3.4)

Thus, the mechanical advantage of a machine is equal to the product of its efficiency and

velocity ratio.


M.A. = V.R. x

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For an ideal machine (free from friction, etc.) the efficiency is equal to 1 (or 100%), so

the mechanical advantage is numerically equal to the velocity ratio.

3.3 LEVERS

Levers are the simplest kind of machines used in our daily life.

The axis, about which the lever turns, passes through a point of the lever which is called

the fulcrum. This point does not move, but remains fixed when the lever is in use.

Principle of a lever (M.A. of a lever)

Fig 3.1 shows a lever ( or a straight rod) AB with the fulcrum at F. An effort E, applied at

a point A of the lever, overcomes a load L at the point B. The distance AF of point A, at which

effort is applied, from the fulcrum F is called the effort arm and the distance BF of point B, at

which the load acts, from the fulcrum F is called the load arm.

Fig. 3.1 Levers and their kinds

An ideal lever works on the principle of moments. Neglecting the friction and assuming

that the lever is weightless, in the equilibrium position of the lever, by the principle of moments,


In actual practice, the mechanical advantage for all practical machines is always less than its velocity ratio (i.e., M.A. < V.R.) or efficiency is less than 1 (i.e., n < i).

A lever is a rigid, straight or bent bar which is capable of turning about a fixed axis.

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Clockwise moment of the load about the fulcrum = Anticlockwise moment of the effort about the

fulcrum.

or Load x Load arm = Effort x Effort arm

or L x BF = E x AF

or L = AFE BF

But L = M.AE

…. (3.5)

This relation is known as the law of levers. Thus

From eqn. (3.5), it is clear that if effort arm = load arm, M.A. = 1; if effort arm < load arm,

M.A. < 1 and if effort arm > load arm, M.A. > 1. The mechanical advantage of a lever can be

increased by increasing its effort arm.

Kinds of levers

Depending upon the relative positions of the effort, load and fulcrum, there are following

three types of levers : (a) Class I levers, (b) Class II levers and (c) Class III levers.

(a) Class I levers : In these types of levers, the fulcrum F is in between the effort E and

the load L as shown in Fig. 3.1 (a).

Some of these are shown in Fig. 3.2.


Effort arm AFM.A. = Load arm BF

The mechanical advantage of a lever is equal to the ratio of the length of its effort arm to the length of its load arm.

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Fig. 3.2. Class I levers

When the effort arm is longer than the load arm, the mechanical advantage and the

velocity ratio of the lever are greater than 1. Such a lever serves as a force multiplier, i.e., it

enables us to overcome a large resistive force (load) by a small effort e.t. shears, used for

cutting the thin metal sheets, have much longer handles as compared to its blades. Similarly a

crowbar, claw hammer, pliers and spoon, used to open the lid of a container, have long handles

(or long effort arms). If a lever has effort arm and load arm of equal lengths, then its mechanical

advantage and velocity ratio are both equal to 1 e.t. a physical balance whose both arms are

equal. When a lever has effort arm shorter than the load arm, its mechanical advantage and

velocity ratio both are less than 1 e.g., a pair of scissors whose blades are longer than its

handles. Such levers are used to obtain gain in speed because the velocity ratio less than 1

implies d > d i.e., the displacement of load is more as compared to the displacement of effort.

For example, a pair of scissors is used to cut a piece of cloth (or paper) so that the blades

move longer on the cloth (or paper) than its handles.

(b) Class II levers : In these types of levers, the load L is in between the effort I and the

fulcrum F as shown in Fig. 3.1 (b). The effort arm is thus always longer than the load arm. From

eqn. (3.5), M.A. > 1 and since M.A. is either equal to or less than V.R., so V.R. > 1. Thus


For class I levers, the mechanical advantage and velocity ratio can have any value greater than I, equal to 1 or less than 1.

The mechanical advantage and velocity ratio of class I levers are always more than 1.

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In other words, second class of lever always acts as a force multiplier i.e., a less effort is

needed to overcome a large load by using the second class of lever. For example, a hard nut

is broken by applying a little effort in a nut cracker.

Examples : A nut cracker, a bottle opener, wheel barrow, a lemon crusher, a paper cutter, a

mango cutter, an oar used for rowing a boat, a bar used to lift a load, raising the weight of the

human body on toes, are examples of Class II levers. Some of these are shown in Fig. 3.3.

(c) Class III levers : In these types of levers, the effort E is in between the fulcrum F and

the load L (Fig. 3.1 (c) and the effort arm is always smaller then the load arm. Therefore from

eqn. (3.5) M.A. < 1 and since M.A. is either equal to or less than VR i.e., V.R. < 1 for these

levers. Thus

With these types of levers we do not get gain in force, but we get gain in speed i.e., a

larger displacement of load is obtained by a smaller displacement of effort. For example, the

blade of a knife moves longer by a small displacement of its handle.


The mechanical advantage and velocity ratio of class III levers are always less than 1.

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Examples : Sugar tongs, the forearm used for lifting a load (or action of the bicep muscle), fire

tongs, foot treadle, knife, a spade used to lift coal (or soil), etc., Class III levers. Some of these

are shown in Fig. 3.4.

Fig. 3.4 Class III levers

Examples of each class of levers as found in the human body

In the human body, we can find the examples of all the three classes of levers. In a

human body, the muscles exert force (i.e., effort) by contraction.

(i) Class I lever in the action of nodding of the head : Fig. 3.5 shows the action of

nodding of the head. In this action, the spine acts as the fulcrum, load is at its

front part, while effort is at its rear part. Thus this is an example of Class I lever.

Fig. 3.5 Nodding a action of head (Class I lever)


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(ii) Class II lever in raising the weight of the body on toes : Fig. 3.6 shows how the weight

of the body is raised on the toes. The fulcrum is at toes at one end, the load (i.e., weight of

the body) is in the middle and effort by muscles at other end. Thus this is an example of

Class II lever.

Fig. 3.6 Raising weight of the body on toes

(Class II lever)

(iii) Class III lever in raising a load by forearm : Fig. 3.7 shows the action of bicep. The

elbow joint acts as fulcrum at one end, biceps exerts the effort in the middle and a load

on the palm is at other end. Thus this is an example of Class III lever.

Fig. 3.7 Raising a load kept on a hand

(Forearm as Class III lever)

3.4 INCLINED PLANE

It is a common experience n our daily life that a sloping plank (or an inclined plane) is

used to load a truck or to take the scooter from road into the house on a higher level. Similarly,

inclined planes or ramps (or stairs on an inclined plane) are used to reach the bridge over the

railway tracks at a railway station. Thus we can define an inclined plane as follows :


An inclined plane is a sloping surface that behaves like a simple machine whose mechanical advantage is always greater than 1.

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Thus, less effort is needed in lifting a load to a higher level by moving over an inclined

plane as compared to that in lifting the load directly (i.e., vertically up). An inclined plane thus

acts as a force multiplier. However we may note that the work done in lifting a load up while

using an inclined plane will be same as in lifting the load directly up to the same height.

Mechanical advantage (M.A.) of an inclined plane

Let a load L be lifted through a height h by using an inclined plane. In Fig. 3.8, AC is an

inclined plane which makes an angle CAB = Q with the horizontal AB. The length of the inclined

plane is AC = and the vertical height of the inclined plane is BC = h.

Fig. 3.8 Inclined plane

Let the load L be raised from the bottom A to the top C of the plane by applying an effort E

along the inclined plane. At any position on the inclined plane, following three forces act on the

load :

(i) The weight of the load equal to L vertically downwards,

(ii) the reaction R of the plane on the load normal to the plane upwards and

(iii) the effort E along the plane upwards.

The load L can be resolved in two components :

(i) L cos Q. the component normal to AC downwards, and

(ii) L sin Q, the component along AC downwards.


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If the inclined plane is frictionless, then in equilibrium, the component L cos Q balances

the normal reaction R and the component L sin Q is balanced by the effort E, i.e., L cos Q

= R L sin Q = E.

Thus, the minimum effort needed to pull the load along the inclined plane is L sin Q and

distance moved by effort is d = AC = , while the useful or required distance moved by the load

is d = BC = h.

Now mechanical advantage (M.A.) = Load (L)

Effort (E)

= L 1

L sin Q sin Q

From right angled triangle ABC,

Sin Q perpendicular h

Hypotenuse

Hence

The velocity ratio of an inclined plane

Now efficiency = M.A.

V.R.


=

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In actual practice, due to some friction between the plane and the body which is pushed

on it, the effort needed is more than L sin Q, so the mechanical advantage is less than /h,

hence, efficiency is less than 1 (since velocity ratio remains /h).

It is obvious that for a given height h, longer the length of the inclined plane (i.e., smaller

the angle of inclination Q), greater is the mechanical advantage of the inclined plane.

If inclined plane is not used, the effort needed will be equal to the load to be lifted i.e., L.

Thus the use of inclined plane decreases the effort needed from L to L sin Q. The usefulness of

an inclined plane can be understood by the following example.

Example : Suppose a boy an exert a maximum force of 20 kgf, i.e., he cannot lift

vertically a load of mass more than 20 kg. Now if he wants to raise a load of mass 40 kg on to a

high wall, he can do it with the help of an inclined plane whose slope is given as

1 Effort 20 1Sin Q = M.A Load 40 2

Thus, by placing a wooden plank at an angle equal to 300 with the horizontal surface, he

can push the load of mass 40 kg to any height by exerting a force (or effort) of 20 kgf. Note that

if there is some friction between the plank and the body, the angle of inclination of the plank will

then be kept less than 300.

1. Calculate the ideal mechanical advantage of a lever in which the effort arm is

60 cm and the load arm is 4 cm.

Given : Effort arm = 60 cm, load arm = 4 cm. Ideal mechanical advantage

= effort arm 60

Load arm 4


Or Q = 300= ==

= = 15

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2. Draw a diagram to illustrate the action of the biceps muscle as a lever. Name

the class of lever.

The action of the bicep muscle as a lever is shown in Fig. 3.9. This belongs to Class

III type of levers in which effort E is between the load L and fulcrum F.

Fig. 3.9 Action of bicep muscle.

3. A machine is driven by a 100 kg mass that falls 8.0 m in 4.0 s. It lifts a load of mass

500 kg vertically upwards. Upwards. Taking g = 10 ms -2 , calculate :

(i) the force exerted by the falling mass.

(ii) the work done by the falling mass in its displacement by 8.0 m,

(iii) the power input to the machine,

(iv) the power output of the machine if its efficiency is 60%,

(v) the work done by the machine in 4.0 s.

Given : m = 100 kg, d = 8.0 m, t = 4.0 s, L = 500 kg

(i) Force exerted by the falling mass = mg

= 100 x 10 = 1000 N.

(ii) Work done by the falling mass

= Force exerted by the falling

Mass x displacement

= 1000 x 8.0 = 8000 J


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(iii) Power input = Work done on machine

Time

= 8000 J

4.0 s

(iv) Given, efficiency = 60% = 0.6, power output = ?

Since efficiency = Power output

Power input

Power out put = Power input x Efficiency

= 2000 W x 0.6 = 1200 W

(v) Work done by the machine in 4.0 s

= Power output x time

= 1200 W x 4.0 s = 4800 J

4. A crowbar 2 m long is pivoted about a point 10 cm from its tip. Calculate the

mechanical advantage of the crowbar. What is the least force which must be applied at

the other end to displace a load of 100 kgf ?

Given : L = 100 kgf, since load is at tip which is at distance 10 cm from the fulcrum, so load arm

= 10 cm,

Total length of crowbar = 2 m = 200 cm

Effort arm = 200 – 10 = 190 cm, E = ?

Mechanical advantage = effort arm

Taking moments about the pivot (fulcrum),

Alternative method :


= 2000 W

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5. A uniform seesaw, 5 m long is supported at its centre. A boy weighing 40 kgf sits at a

distance of 1 m from the centre of the seesaw. Find where a girl of weight 20 kgf must sit on the

other side of the seesaw so as to balance . the weight of the boy. To which class of lever does it

belong ?

The between the load and the effort. It is a Class I lever.

Let the girl be sitting at a distance x metre from the centre, then

Effort arm = x metre, load arm = 1 metre,

L = 40 kgf, E = 20 kgf

Taking moments about F,

40 x 1 = 20 X x

x = 40 20Hence the girl must sit at a distance 2 m from the centre on the opposite side of boy.

6. A cook uses a fire tong of length 28 cm to lift a piece of burning coal of mass 250 g. If

he applies the effort at a distance of 7 cm from the fulcrum, find the effort.

Take g = 10 m s-2

A fire tong is a leer of Class III which has the fulcrum and load at the ends, with effort in

between.

Given :

Load L = 250 gf = 0.25 x 10 N = 2.5 N,

Load arm = 28 cm, Effort arm = 7 cm.

By the principle of moments,

Load x load arm = Effort x effort arm

Effort = Load x

7. This diagram below shows a lever in use.


= 2 m.

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(i) To which class of lever does it belong ?

Give one example of this class.

(ii) State the principle of moments as applied to the above lever and hence calculate its

mechanical advantage if AB = 2.0 m and AF = 20 cm.

(iii) Calculate the effort needed to lift the load.

(i) The fulcrum F is in between the load and the effort, so it is a Class I lever. A pair of scissors

is an example of this class.

(ii) By the principle of moments,

Load x AF = Effort x BF

Given : AB = 2.0 m,

AF = 20 cm = 0.2 m,

BF = AB – AF = 2.0 – 0.2 = 1.8 m.

Mechanical advantage = Load BFEffort AF

= 1.8 m 0.2 m

(iii) Given : Load = 18 kgf

Effort = Load 18 M.A. 9


= 9

= 2 kgf

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8. A truck driver can load oil drums into the back of the truck by pushing them up a

sloping plank, or by lifting them directly. Each drum has a mass of 80 kg. the plank is 3

m long and the back of the truck is 0.8 m above the ground.

(a) How much force world be needed to lift a drum into the truck directly, without using

the plank ?

(Take g = 10 N kg -1)

(b) Assuming that there is no friction between the plank and the drum, calculate : (i) the

mechanical advantage of the inclined plank, (ii) the force (or effort) needed to push a

drum up the plank.

(c) If the actual force needed to push a drum up the plank is 300 N, why is this less than

the answer to part (a), but more than the answer to part (b) (ii) ?

Given : m = 80 kg,g = 10 N kg -1, length of the plank = 3 m, height of the back of the truck h =

0.8 m.

(a) Force needed to lift a drum directly = mg

= 80 x 10 = 800 N

(b) (i) Mechanical advantage = length of the plank l height of the plank h

= 3 m 0.8 m

(ii) The effort needed to push a drum up the plank

E = Load 800 NM.A. 3.75


= 3.75

= 213.33 N=

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(c) The actual force needed to push a drum up the plank (i.e., 300 N ) is less than the weight (or

load = 800 N) because the sloping plank acts like a machine of which the mechanical

advantage is greater than 1.

The actual force needed to push a drum up the plank (i.e., 300 N) is more than the effort

needed to push a drum up the plank ( = 213.33 N ) as calculated in part (b) (ii) because there is

some friction between the plank and the drum. So a part of the force ( = 300 – 213.33 = 86.67

N )overcomes the force of friction.

EXERCISE 3(a)

(B) PULLEY

3.5 PULLEY

Fig. 3.17 (a) and (b) shows a simple pulley. It is a metallic (or wooden) disc with a

grooved rim. A string is passed around the groove. The disc rotates about an axle passing

through its centre. The axle is fixed to a frame or a block.

Generally a single pulley or a combination of two or more pulleys fixed in a frame, is

called a block, while a string (rope or chain) that winds around the pulleys in different blocks is

known as tackle.

A pulley can be used in two ways : (i) by keeping its axix of rotation fixed, and (ii) without

keeping its axis of rotation fixed. When the axis of rotation of a pulley is kept fixed, it is called a

fixed pulley and if the axis of rotation is not fixed, it is called a movable pulley.

3.6 SINGLE FIXED PULLEY


A pulley which has its axis of rotation fixed, is called a fixed pulley.

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Fig. 3.18 shows a single fixed pulley in which the axle is fixed to a rigid support and an inextensible string of negligible mass passes around the grooved rim of the pulley. One end of the string is connected to the load L and the effort E is applied at the other end of the string. This type of pulley is used for lifting a small load (such as a water bucket or a basket).

M.A. V.R and of a single fixed pulleyThe load L to be lifted is tied to one end of the string and an effort E is applied at the

other end. Tension T acts upwards on the string on both sides of the pulley and it is the same

throughout the string. Neglecting (i) the mass of the string, (ii) friction between the string and

surce of the rim of the pulley” and (iii) the friction in the pulley bearings, in the balanced

position of the load, we have

L = T ….(i)

And E = T (when pulley is not rotating ) … (ii)

…. (308)

Thus, in this arrangement there is no gain in the mechanical advantage. In other words,

using a single fixed pulley, the effort needed to lift a load is equal to the load itself (in the ideal

situation). In actual practice, there is always some friction between the string and surface of the

rim of the pulley, so the effort needed is grater than the load to be lifted.


Mechanical advantage = load L T Effort E T = 1

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If the effort E moves a distance d downwards, the load L also moves the same distance

d upwards, i.e., if d = d then d = d.

……(3.9)

Hence efficiency = M.A = 1 or 100% …..(3.10) V.R(when mass of the string and friction in the pulley bearings is ignored)

Now the question arises that when there is no gain in mechanical advantage or the velocity

ratio, then why is it used > The reason is that a single fixed pulley is used only to change the

direction of the force applied, i.e. with its use the effort can be applied in a more convenient

direction. To raise a load directly upwards is difficult. But with the help of a fixed pulley, the

effort can be applied in the downward direction to move the load upwards. One can

conveniently make use of his own weight also for the effort.

3.7 A SINGLE MOVABLE PULLEY

Fig. 3.19 shows a single movable pulley in which the load L to be raised, is attached to its axle. An inextensible string of negligible mass passes around the grooved rim of the pulley. One end of the string is tied to a hook H at a rigid support and the effort E is applied at its free end. The tension T acts on the string on both sides of the pulley as shown in Fig. 3.19. The effort balances the tension at the free end.


Velocity ration = d d d d

A pulley whose axis of rotation is not fixed in position, is called a movable pulley.

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Fig. 3.19 Single movable pulley

M.A., V.R. and of a single movable pulley

In Fig. 3.19, the load L is balanced by the tension in two segments of the string, so

L = T + T + 2T and E + T (Here we have assumed that the weight of pulley is negligible)

Mechanical advantage = Load L Effort E

or

Thus, using a single movable pulley, the load can be lifted by applying an effort equal to half the load (in ideal situation), i.e., the single movable pulley acts as a force multiplier.

When the load is pulled up through a distance d, the free end of the string is pulled up by the effort through a distance 2d, i.e.,

If d = d then d = 2d.

Velocity ration = distance moved by the effort Distance moved by the load

or ….(3.12)

… (3. 13)

Here we have not taken into consideration the friction in the pulley bearings and also the

weight of the pulley and string, otherwise M.A. will be less than 2 and efficiency will be less than

100% since V.R. always remains 2.


M.A. = 2T = 2 T

V.R = 2d = 2 d

Hence efficiency = M.A. = 2 V.R. 2 = 1 or 100%

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It may be noted here that with a single movable pulley, the effort has to be applied in the

upward direction. However, it is inconvenient to apply the effort in an upward direction,

therefore the movable pulley is used along with a single fixed pulley to change the direction of

effort.

Fig. 3.20 Single movable pulley with a fixed pulley

Fig. 3.20 shows the arrangement where B is the fixed pulley and A is the movable pulley to

which the load is attached. Now effort can be applied in the downward direction which is quite

convenient. The mechanical advantage and the velocity ratio of this arrangement remain the

same as that of a single movable pulley.

3.8 COMBINATION OF PULLEYS (BLOCK AND TACKLE SYSTEM)

When large loads are to be lifted or shifted, a single pulley is not enough, but a system of

pulleys is then used. This is called the block and tackele system.

Here two blocks of pulleys are used. One block (upper) is fixed to a rigid support and the

other block (lower) is movable.

The number of pulleys used in the movable lower block is either equal to or one less than

the number of pulleys in the fixed upper block. An inextensible single string of negligible mass

passes around all the pulleys. One end of the string is attached to hook of the lower block (if the

number of pulleys in the upper block is more than that in the lower block) or it is attached to

hook of the upper block (if the number of pulleys is equal in both the blocks) so as to apply the

effort in the downward direction. Fig. 3.21 shows a block and tackle arrangement of 5 pulleys in


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which one end of sting is attached to the hook of the lower block, while Fig. 3.22 shows a block

and tackle arrangement of

L (= 10 kgf) L (= 10 kgf)

Fig. 3.21 Block and Fig. 3.22 Block and

tackle for 5 pulleys tackle for 4 pulleys

4 pulleys in which one end of the string is attached to the hook of the upper block.

The load L is attached to the movable lower block and the effort E is applied at the free end

of the string. The tension along the entire length of the string is the same and is denoted as T.

M.V., V.R. and for a block and tackle system

In Fig. 3.21, the tension in the five segments of string supports the load L, therefore,

L = 5 T and E = T.

Hence M.A. = L 5 T E T

Similarly in Fig. 3.22, the tension in the four segments of string supports the load L, therefore,


= = 5

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L = 4 T and E = T

Hence M.A. = L 4 T E T

In general, if the total number of pulleys used in both the blocks is n and the effort is being applied in the downward direction, then the tension in n segments of string supports the load, therefore,

L = n T and E = T

…..(3.14)


= = 4

M.A. = Load L n T Effort E T= Total no. of pulleys in both the blocks

= = n

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Thus, the effort required to balance the load

In a block and tackle system if the load moves up through a distance d, the effort moves through a distance n d because each section of the string supporting the load is loosened by a length d, i.e., if d = d, then d = n d.

…….(3.16)

Thus, the velocity ratio is always equal to the number of strands of tackle (or sections of the string) supporting the load.

..(3.17)

Note that the efficiency is 100% only in the ideal situation, i.e., when there is no friction in

bearings of pulleys and the weight of the string and the weight of pulleys in the lower block, are

negligible.

It is interesting to note that the pulley system is only a force multiplier and there is no gain in

energy.

The work done by the effort (or the input energy)

= Effort x distance moved by the effort


E = L (load) N (number of pulleys)

Thus in a block and tackle system, the effort gets multiplied n times, where n is the total number of pulleys in the system. It therefore acts as a force multiplier.

= n Velocity ration = n d D

Efficiency n = M.A. n V.R. n = 1 or 100% =

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= E x nd = nEd

The work done on the load (or the output energy)

= Load x distance moved by the load

= L x d

= nE x d = nEd (since M.A. = L = n) EThus, there is no gain in energy.

Effect of weight of pulleys on M.A., V.R. and

Consider a system of n pulleys. Let w be the total weight of pulleys and frame in the lower block. In the balanced position,

L + w = nT and E = T

Thus the mechanical advantage is less than the ideal value n.

Now….. (3.19)

Hence, efficiency

or

Thus, efficiency is reduced due to the weight of the pulleys of the lower block. More is the weight of the pulleys, less is the efficiency.


V.R. = n

For greater efficiency, the pulleys in the lower block should be as light as possible and the friction in bearings of pulleys should be minimized by the use of lubricants.

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EXAMPLES

1. The diagram below shows a fixed pulley used by a boy to lift a load of 400 N through a

vertical height of 5 m in 10 s. The effort applied by the boy on the other end of the rope is 480

N.

(i) What is the velocity ratio of the pulley ?

(ii) What is the mechanical advantage ?

(iii) Calculate the efficiency of the pulley.

(iv) Why is the efficiency of the pulley not 100% ?

(v) What is the energy gained by the load in 1o s ?

(vi) How much power was developed by the boy in raising the load ?

(vii) The boy has to apply an effort which is greater than the load he is lifting. What is the

justification for using the pulley ?

(i) When the effort moves a distance d downwards, the load moves a distance d upwards .

Velocity ration = displacement of effort


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Displacement of load

(ii) Mechanical advantage = Load Effort

(iii) Efficiency = M.A.

(iv) The efficiency is less than 100% because some energy is wasted in overcoming the friction of the pulley bearings.

(v) Energy gained by the load = load x displacement of load = 400 x 5 = 2000 J

(vi) Power developed by the boy

= Effort x displacement Time

(vii) Use of pulley helps in changing the direction of applied force to a convenient direction. One may also use his own weight as effort.

2. The adjacent diagram shows the combination of a movable pulley P1 with a fixed pulley P2 used for lifting up a load W.

(i) State the function of the fixed pulley P2.(ii) If the free end of the string moves through a distance x, by what distance is the load W raised ?


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(iii) What effort has to be applied at C to just raise the load W = 20 kgf ? Neglect the weight of the pulley P1 and the friction.

(i) The fixed pulley P2 is used to change the direction of effort to be applied from upwards to downwards.

(ii) If the free end of the string moves through a distance x, the load will rise by a distance x/2

(iii) Given W = 20 kgf, E = ? In equilibrium, W = 2T and at C, effort E = T

Effort needed E = W = 20 kgf = 10 kgf. 2 2

3. A block and tackle has two pulleys in each block, with the tackle tied to the hook of the lower

block and the effort being applied upwards. Draw a neat diagram to show this arrangement and

calculate its mechanical advantage. If the load moves up a distance x, by what distance will the

free end of the string move up ?

The arrangement is shown in Fig. 3.25 Here the load is being supported by five segments of

the string, therefore

L = 5T and E = T

M.A. = L = 5T = 5 E T

If the load moves up a distance x, the free end of the string will move up by a distance 5 x.

4. A pulley system with a velocity ratio of 4 is used to lift a load of 150 kgf through a vertical height of 20 m. The effort required is 50 kgf. Calcuate : (a) the distance moved by the effort, (b) the work done by the effort, (c) the mechanical advantage, (d) the efficiency of the pulley system, and (e) the total number of pulleys and the number of pulleys in each block (g = 10 N kg -1).


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Given : V.R. = 4, L = 150 kgf, dL = 20 M,

E = 50 kgf = 50 x 10 N = 500 N, dE = ?

(a) V.R. = Distance moved by effort dE

Distance moved by load dL

Or 4 = dE

20 Distance moved by the effort dE = 20 x 4 = 80 m

(b) Work done by the effort = effort x distance

= 500 N x 80 m

= 40000 J

(c) Mechanical advantage =

(d) Efficiency =

(e) The total number of pulleys = 4

The number of pulleys in each block = 2

(Note that tackle will be tied to the hook of the upper block so as to apply the effort

Downwards).

LEVERS(b)Diagram given below shows a weightless lever in equilibrium. Neglect friction at the fulcrum F.


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(i) State the principle of moments as applied to the above lever.(ii) Define mechanical advantage and calculate its value for the given lever.(iii) Name the type of lever which has mechanical advantage greater than 1.(b)(i) The principle of moments states that the vector sum of the moments of all coplanar forces acting on a body in the equilibrium state is always zero. In the given case of the leverAnticlockwise moment of effort E about fulcrum F = Clockwise moment of load L about fulcrum Fi.e E x AF = L x BF.(ii)Mechanical advantage of a machine is defined as the ratio of the load lifted L to the effort applied E.i.e M.A = L / E = AF / BF(iii)We know that M.A of a lever M.A = L /E = AF /BFIf M.A > 1 Therefore AF (Effort Arm )/BF (Load Arm) > 1Hence, Effort Arm should be greater than Load Arm for a lever whose M.A is greater than 1Q1.(b)The diagram given below shows the use of a lever.

(i)State the principle of moments as applied to the above lever.(ii)Give an example of this class of lever.(iii)If FA=10 cm,AB=500 cm, calculate the minimum effort required to lift the load.

Ans:(i)The vector sum of moments of all coplanar forces acting on a body in the equilibrium state about any point is zero.In the above case moment of effort and moment of load about F is zero.(ii)This is class III type of the lever, as the effort lies in between the fulcrum and load.ex. Fire Tongs


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(iii)Taking moments about Fmoment of Effort about F = moment of load about FE x 10 = 50 x 510Therefore, E = 2550 N

PULLEYS(b) Q2.(b) (c) A crowbar 2 m long is pivoted about a point 10 cm from its tip. Calculate the mechanical advantage of the crowbar. What is the least force which must be applied at the other end to displace a load of 100 kgf? (2) Examples : A seesaw, a pair of scissors, crowbar, handle of water pump, claw hammer, a pair of pliers, the beam of a common balance, a spade used for turning the soil, a spoon used to open the lid of a tin can and the nodding of the human head are examples of Class I levers.

LOAD arm

In practice, however, no machine is ideal or 100% efficient for the following reasons :

(i) the moving parts in it are not perfectly smooth (or frictionless).

(ii) the string in it (if any ) is not perfectly elastic.

(iii) its different parts are not perfectly rigid, and

(iv) its moving parts are not weightless.

Due to above factors there is always some loss of energy in the machine. Thus, the output energy from a machine is always less than the input energy to it No machine is 100% efficient. In the action of nodding of the head, the spine acts as the fulcrum, load is at its front

part, while effort is at its rear part. Thus this is an example of Class I lever. The fulcrum is at toes at one end, the load (i.e., weight of the body) is in the middle and effort by muscles at other end. Thus this is an example of Class II lever. The elbow joint acts as fulcrum at one end, biceps exerts the effort in the middle and a load on the palm is at other end. Thus this is an example of Class III lever. (b)(i) The principle of moments states that the vector sum of the moments of all coplanar forces acting on a body in the equilibrium state is always zero. In the given case of the lever


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Anticlockwise moment of effort E about fulcrum F = Clockwise moment of load L about fulcrum Fi.e E x AF = L x BF.(ii)Mechanical advantage of a machine is defined as the ratio of the load lifted L to the effort applied E.i.e M.A = L / E = AF / BF(iii)We know that M.A of a lever M.A = L /E = AF /BFIf M.A > 1 Therefore AF (Effort Arm )/BF (Load Arm) > 1Hence, Effort Arm should be greater than Load Arm for a lever whose M.A is greater than 1(b) Diagram 6 below gives an arrangement of single moving pulleys.Copy the diagram

If the effort applied at the free end of the string is E,(i) show the direction and magnitude of the forces exerted by the four strings marked (1) to (4).(ii) what is the load which can be lifted by the effort ?(iii) calculate the mechanical advantage of the system of pulleys.(c) A block of wood is floating in water. The portion of the block inside water measures 50 cm x 50 cm x 50 cm. What is the magnitude of the buoyancy force acting on the block ?(b)

(ii) Suppose the load lifted = LBut effort applied = E (given)Let the effort Arm = xThen the load Arm = x/4Now, by the principle of machineLoad X load Arm = Effort x Effort Armi.e L (x/4) = E(x)Therefore , L = 4E(iii) M.A is defined as the ratio of the load lifted by the machine to the effort applied on the machinei.e M.A = L/E = 4E/ E = 4Hence, M.A = 4Lever


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A lever is a rigid rod capable of rotating about a fixed point called fulcrum. A lever has three parts - effort (E), fulcrum (F) and load (L). The fulcrum is the point on which the lever turns. Pulley A pulley is a flat circular disk, having a grooved rim and capable of revolving around a fixed point passing through its centre. It is used to change the direction of force. Inclined Plane

An inclined plane is a smooth, flat rigid surface, inclined at an angle to the horizontal. An inclined plane spreads the amount of work needed to move

an object over a larger distance so that less force is needed at any particular moment.

Diagram 6 below gives an arrangement of single moving pulleys.Copy the diagram

If the effort applied at the free end of the string is E,(i) show the direction and magnitude of the forces exerted by the four strings marked (1) to (4).(ii) what is the load which can be lifted by the effort ?(iii) calculate the mechanical advantage of the system of pulleys.(ii) Suppose the load lifted = LBut effort applied = E (given)Let the effort Arm = xThen the load Arm = x/4Now, by the principle of machineLoad X load Arm = Effort x Effort Armi.e L (x/4) = E(x)Therefore , L = 4E(iii) M.A is defined as the ratio of the load lifted by the machine to the effort applied on the machinei.e M.A = L/E = 4E/ E = 4


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Hence, M.A = 4The diagram alongside shows a system of 5 pulleys.

1.Copy the diagram and complete it by drawing a string around the pulleys. Mark the position of load and effort.2. If the load is raised by 1 m, through what distance will the effort move? Ans: 1 The complete diagram of a string around the pulleys is shown as follows with the marked positions of effort and load.

2. Let the distance moved by the effort = DDistance moved by the load = d = 1 m (given)Let the load = W


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and the effort applied = E By the principle of machines E. D = W . dTherefore, D/d = W/EBut W/E = 5Hence D = (W X d)/E = 5 x 1 m D = 5 m PROB. 3. Ans: 1 The complete diagram of a string around the pulleys is shown as

follows with the marked positions of effort and load. A resistance of 1500 N is

overcome by a machine of V.R 6 and efficiency 80%. Find

i) Mechanical advantage

ii) Effort required to overcome resistance. (2mks)

Ans.

i) h = ,

M.A.= h ×V.R. = ×6=4.8

ii)

.

PROB. 4. The diagram below shows a lever in use.

i) To which class of lever does it belong?

ii) If AB = 1 m, AF = 0.4 m, find its mechanical advantage.

iii) Calculate the value of E.

Ans.

i) Class I

ii) 1.5

iii) 10


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II) A single fixed pulley does not provide any mechanical advantage but still it is used very often. why?

The single fixed pulley inverts the direction of the force. An upward force on the load is changed to a downward pull.This is very convenient to lift load. So it is used very often.(c) A block and tackle system has the velocity ratio 3. A man can exert a pull of 200 kgf. What is the maximum load he can raise with this pulley system if its efficiency is 60%?

i)

ii) Mechanical advantage

iii) Effort required to overcome resistance. (2mks)

Ans.

i) h = ,

M.A.= h ×V.R. = ×6=4.8

ii)

.

PROB. 2. The diagram below shows a lever in use.


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i) To which class of lever does it belong?

ii) If AB = 1 m, AF = 0.4 m, find its mechanical advantage.

iii) Calculate the value of E.

Ans.

i) Class I

ii) 1.5

iii) 10

PROB. 5. A force of 5 kgf is required to cut a metal sheet. A shears used for cutting the metal sheet

has its blades 5 cm long, while its handle is 10 cm long. What effort is needed to cut the sheet? Ans.

2.5

Q.1. Let the distance moved by the effort = D

Distance moved by the load = d = 1 m (given)

Let the load = W

and the effort applied = E

By the principle of machines E. D = W . d

Therefore, D/d = W/E

But W/E = 5

Hence D = (W X d)/E = 5 x 1 m

D = 5 m

Fig. shows a

block and tackle system of pulleys used to lift a load


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i) How many strands of tackle are supporting the load?

ii) Draw arrows to represent tension in each strand.

iii) What is the mechanical advantage of the system?

iv) When load is pulled up by a distance 1 m, how far does the effort end move?


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