Subject: Science

We use various tools to perform different types of work in our daily life. The tools or simple devices used for making our work easier, faster and more convenient are simple machines.

Those tools or devices which make our work easier and convenient in the direction of the force is called simple machine.

Purpose of using simple machine

Simple machines are used to make our work easier in the following ways:

- By multiplying force.
- By changing the directions of force.
- By transferring force from one point to another.
- By increasing the rate of doing work.

**Mechanical advantage**

The simple machine requires force to do work. The resistive force to be overcome is called load and the force applied to overcome the load is called effort.

The ratio of the load to the effort in a simple machine is called mechanical advantages or actual advantage of the machine.

Therefore,

Mechanical advantage has no unit. It is the simple ratio of two forces and affected by friction.

If a machine overcomes a load ‘L’ and the distance travelled by the load is ‘Ld’. Similarly, the effort applied in the machine is ‘E’ and the distance travelled by effort is ‘Ed’, then

Velocity ratio of simple machine is the ratio of distance travelled by an effort to the distance travelled by the load in the machine. As velocity ratio or ideal mechanical advantage is a simple ratio of two distances, it also does not have the unit. The friction is not involved in it.

If a machine overcomes a load ‘L’ and the distance travelled by the load is ‘Ld’, the work done by the load is L × Ld. It is also called output work or useful work. Therefore,

Output work = L × Ld

Likewise, the effort applied to overcome the load is E and the distance covered by effort is Ed, the work done by effort is E × Ed. It is also called input work. Therefore,

Input work = E × Ed

The efficiency of a simple machine is defined as the ratio of useful work done by machine (output work) to the total work put into the machine (input work).

For ideal or perfect machine, work output will be equal to the work input. Ideal machinesare those imaginary machines which are frictionless. In practice, the work output of a machine is always less than work input due to the effect of friction. If the frictional force in the machine increases the efficiency decreases. Because machines are frictionless in practice, the efficiency of a machine can never be 100%.

We know that,

η=\(\frac{output \;work} {input\; work}\) ×100%

or,η=\(\frac{work\;done\;by\;load} {work\;done\;by\;effort}\)×100%

or,η=\(\frac{L×Ld}{E×Ed}\)×100%

But the mechanical advantages =\(\frac{L}{E}\) and velocity ratio = \(\frac{E_d}{L_d}\)

η=\(\frac{MA}{VR}\) ×100%

Thus, the relation shows that efficiency is the ratio of mechanical advantage to velocity ratio in percentage.

The principle of simple machine states that “if there is no friction in a simple machine, work output and work input are found equal in that machine”

Mathematically,

Output work = Input work

Or, L × Ld = E × Ed

- Those tools or devices which make our work easier and convenient in the direction of force is called simple machine.
- Simple machine helps us by magnifying force, accelerating work and by changing the direction of force.
- Velocity ratio of simple machine is the ratio of distance traveled by effort to the distance traveled by load in the machine.
- The efficiency of a simple machine is defined as the ratio of useful work done by machine (output work) to the total work put into machine (input work).
- The ratio of the load to the effort in a simple machine is called mechanical advantages or actual advantage of the machine.

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

Simple machine is a device which is mechanically simple and helps to make our work easier and convenient. They are of six types:

- Lever
- Pulley
- Wheel and axle
- Inclined plane
- Screw
- Wedge

- By multiplying force
- By changing the direction of applied force
- By gaining speed to work faster
- By doing jobs which would be dangerous and unsafe.

It is defined as the ratio of load to the effort applied.

Mechanical advantage (M.A) =

Mechanical advantage (M.A) =

During calculation, friction is not counted but there is always friction inside a machine. So, actual value of MA is not same as the calculated value of it.

It is defined as the ratio of distance moved by the effort to the distance moved by the load .

Velocity ratio(VR) =

Velocity ratio(VR) =

It is because friction is not involved in VR.

Practically, no machine are frictionless. The friction decreases the mechanical advantage and efficiency of a machine whereas the VR of the machine is not affected by friction. So, value of MA is always less than value of VR in a simple machine.

The work done by the machine in overcoming the load is called output work.

i.e. Output work = Load x load distance

The work done on the machine by the effort is called input work.

i.e. Input work = Effort x effort distance

The principle of simple machine states that if there is no friction in a simple machine, work output and work input are found to be equal in the machine.

The percentage ratio of the output work to the input work is called efficiency of a simple machine. Mathematically, η = x 100%

A machine having 100% efficiency is called an ideal or a perfect machine.

Efficiency of a simple machine is always less than 100% because friction reduces the input work applied on the machine.

Friction is the main factor which reduces the efficiency of a simple machine. Efficiency of a simple machine can be increased by reducing the friction. But oiling and greasing the machine is a useful way to increase the efficiency of a machine.

The energy input to a machine is greater than its energy output. Source energy is always wasted to overcome the friction and even to move the parts of the machine itself. Hence, no machine is perfect.

We know that, Efficiency(η) = x 100% or, η = x 100% or, η = x 100% or, η = x 100% Hence, proved.

It means that 70% of the input energy is obtained as useful energy. Rest of the 30% is used in overcoming friction.

The grease does not affect the VR of a machine as grease reduces friction and the VR of a machine is not affected by friction.

Given, Load (L) = 300N

Load distance (Ld) = 10cm = 10/100m = 0.1m

Effort (E) = 100N

Effort distance (Ed) = 1m

Effort distance (Ed) = 1m

- Mechanical advantage (MA) = = = 3

- Velocity ratio (VR) = = = 10

- Efficiency (η) = x 100% = x 100% = 30%

Load (L) = 450N

Effort (E) = 150N

Load distance (Ld) = 25cm = 0.25m

Effort distance (Ed) = 1- 0.25 = 0.75m

We know that, Efficiency (η) = x 100% = x 100% = x 100% = 100%

Effort (E) = 150N

Load distance (Ld) = 25cm = 0.25m

Effort distance (Ed) = 1- 0.25 = 0.75m

We know that, Efficiency (η) = x 100% = x 100% = x 100% = 100%

- Name the given machine.
- Find the efficiency of the machine.

It is a crowbar used as first class lever of the simple machine.

Load (L) = 10N

Load distance (Ld) = 20 cm = 0.2m

Effort distance (E.d) = 70cm = 0.7m

Effort (E) = ?

Now, to balance it,

E x Ed = Ld

or, E x 0.7 = 0.2 x 10

or, E = 2/0.7

or, E = 20/7 N

Now,

Efficiency (η) = x 100% = x 100% = 100%

Load (L) = 10N

Load distance (Ld) = 20 cm = 0.2m

Effort distance (E.d) = 70cm = 0.7m

Effort (E) = ?

Now, to balance it,

E x Ed = Ld

or, E x 0.7 = 0.2 x 10

or, E = 2/0.7

or, E = 20/7 N

Now,

Efficiency (η) = x 100% = x 100% = 100%

Suppose unknown weight (W) is load and 40N and 20N act as an effort

To balance lever,

E x E.d = L x L.d

Or, E1 x Ed1 + E2 x Ed2 = W x 25cm

Or, 40 x 5 + 20 (20 +5) = W x 25cm

Or,700 = W x 25

Or, 700/25 = W

Or, 28 = W

Hence, the required weight is 28N.

To balance lever,

E x E.d = L x L.d

Or, E1 x Ed1 + E2 x Ed2 = W x 25cm

Or, 40 x 5 + 20 (20 +5) = W x 25cm

Or,700 = W x 25

Or, 700/25 = W

Or, 28 = W

Hence, the required weight is 28N.

Given,

Load (L) = 90N

Effort (E) = 60N

It is a single movable pulley, so, VR = 2,

Efficiency (η) = ?

Now,

Mechanical advantage (M.A) = = = 3/2

And,

Efficiency (η) = x 100% = x 100% = 75%

Load (L) = 90N

Effort (E) = 60N

It is a single movable pulley, so, VR = 2,

Efficiency (η) = ?

Now,

Mechanical advantage (M.A) = = = 3/2

And,

Efficiency (η) = x 100% = x 100% = 75%

Efficiency (η) = 90%

Load (L) = ?

Effort (E) = 400N

Velocity ratio (VR) = 2

According to formula,

η = x 100% = x 100%

or , 90% = x 100%

or, L = = 720N

Hence, 720 N can be lifted.

Load (L) = ?

Effort (E) = 400N

Velocity ratio (VR) = 2

According to formula,

η = x 100% = x 100%

or , 90% = x 100%

or, L = = 720N

Hence, 720 N can be lifted.

Effort (E) = 50N

Load (L) = 150N

Velocity Ratio (VR) = No. of pulley used = 4

Efficiency (η) = ?

We know,

η = x 100% = x 100%

= x 100% = 75%

Hence, the efficiency of the machine is 75%.

Load (L) = 150N

Velocity Ratio (VR) = No. of pulley used = 4

Efficiency (η) = ?

We know,

η = x 100% = x 100%

= x 100% = 75%

Hence, the efficiency of the machine is 75%.

- Calculate the MA and the efficiency of the machine.
- What are the input work and output work when the load is lifted by 5m?
- What is energy lost?

Given,

Velocity ratio (VR) = 4

Effort (E) = 36N

Total load (L) = 115 + 5N = 120N

MA = = 120/36 = 3.33

and efficiency(η) = x 100% = x 100% = 83.25%

When load is raised by 5m,

Load distance (L.d) = 5m

And VR = = Or, 4 =

∴E.d = 20m

So, work input = E x E.d = 36 x 20 = 720J

Work output = L x Ld = 120 x 5 = 600J

Total energy lost = Work input – Work output

= 720 – 600 = 120 J

Velocity ratio (VR) = 4

Effort (E) = 36N

Total load (L) = 115 + 5N = 120N

MA = = 120/36 = 3.33

and efficiency(η) = x 100% = x 100% = 83.25%

When load is raised by 5m,

Load distance (L.d) = 5m

And VR = = Or, 4 =

∴E.d = 20m

So, work input = E x E.d = 36 x 20 = 720J

Work output = L x Ld = 120 x 5 = 600J

Total energy lost = Work input – Work output

= 720 – 600 = 120 J

Given,

Load (L) = 24N

Effort (E) = 16N

Effort distance (E.d) = length of slope = 3m

Load distance (L.d) = height of slope = 1.5m

Now,

MA = = 24/16 = 3/2

VR = = 3/1.5 = 2

Efficiency (η) = x 100% = ¾ x 100% = 75%

Effort (E) = 16N

Effort distance (E.d) = length of slope = 3m

Load distance (L.d) = height of slope = 1.5m

Now,

MA = = 24/16 = 3/2

VR = = 3/1.5 = 2

Efficiency (η) = x 100% = ¾ x 100% = 75%

Given,

Pitch (P) = 1cm = 0.01m

Length of lever = 50cm = 0.5m

Effort (E) = ?

Load (L) = 500N

By using a condition,

MA = VR

or, =

or, =

or, E = = 1.59 N

Now,

MA = = 500/1.59 = 314.46

Hence, MA = 314.46 and E = 1.59N

Pitch (P) = 1cm = 0.01m

Length of lever = 50cm = 0.5m

Effort (E) = ?

Load (L) = 500N

By using a condition,

MA = VR

or, =

or, =

or, E = = 1.59 N

Now,

MA = = 500/1.59 = 314.46

Hence, MA = 314.46 and E = 1.59N

A 190 cm long crowbar of is provided abut a point 20 cm from its tip ; the force of 100 N is applied to its other end to displace a load of 600 N . Calculate MA,VR and η of the crowbar.

Here, Load (L) = 600N

Effort (E) = 100 N

Load DIstance (Ld) = 0.20m

Length of Lever (l) = 190 cm = \(\frac{190}{100}\) = 1.9m

Effort DIstance (Ed) = length of lever - load distance = (1.9 - 0.2) = 1.7

Now,according to the formula ,

MA = \(\frac{L}{E}\)

= \(\frac{600}{100}\)

= 6

VR = \(\frac{ED}{LD}\)

= \(\frac{1.7 m}{0.20 m}\) = 8.5

\(\eta \)= \(\frac{MA}{VR}\) \(\times\) 100%

= \(\frac{1.7m}{0.20}\)

= \(\frac{6}{8.5}\)\(\times\) 100%

= 70.5 %

∴MA of the crowbar is 6, VR is 8.5 and \(\eta \) is 70.5 %

Draw a diagram of a lever suitable for raising a block of 600 N through a vertical distance of 6 cm with a velocity ratio of 10. GIven that the \(\eta \) of the machine is 80%. Calculate :**i. the mechanical advantage ii. the effort applied iii. the distance moved by theiv. input workv. output work**

Here, Load (L) = 600 N

Load distace (Ld) = 6 cm

Velocity ratio (VR) = 10

Efficiency \(\eta \) = 80%

According to the formula ,

i. \(\eta \) = \(\frac{MA}{VR}\) \(\times\) 100%

80% = \(\frac{MA}{10}\) \(\times\) 100%

or , 100 MA = 80 \(\times\) 10

∴ MA = \(\frac{ 80 \times 10 }{100}\) = 8

ii. MA = \(\frac{L}{E}\)

8 = \(\frac{600 N }{E}\)

or, E = \(\frac{600 N }{8}\)

= 75 N

iii. VR = \(\frac{Distance\;moved\;by\;the\;effort }{Distance\;moved\;by\;the\;load}\)

or, 10 = \(\frac{ ED }{6}\)

∴ Ed = 10 \(\times\) 6 = 60cm = 0.6m

iv. Input work = E \(\times\) Ed = 100 \(\times\) 0.6m = 60J

v. Output Work = L \(\times\) Ld = 600 \(\times\) 0.06m = 36J

Hence, in this lever MA is 8 , effort applied is 75 N , distance moved by the effort moved by the effort is 60cm , input work is 60J and output work is 36J.

A load of 600N is lifted by the five-pulley system up to 4 meters by applying an effort of 300 N. Calculate :**i. MAii. VRiii. Output workiv. Input workv. Efficiency**

Here , Load (L) = 600 N

Load distance (Ld) = 4m

No.of pulleys = 5

Effort (E) = 300N

(i) MA = ?

(ii) VR = ?

(iii) Output Work = ?

(iv) Efficiency = ?

According to the formua ,

i. MA = \(\frac{L}{E}\) = \(\frac{600N}{300N}\) = \(\frac{60}{30}\) = \(\frac{12}{6}\)

ii. VR = No.of pulleys used = 5

iii. Output Work = L \(\times\) Ld = 600N \(\times\) 4m = 2400J

For Effort Distance,

VR = \(\frac{Ed}{Ld}\)

5 = \(\frac{Ed}{4}\)

\(\therefore\) Ed = 20m

iv. Input Work = E \(\times\) Ed = 300N \(\times\) 20m = 6000J

v. \(\eta \) = \(\frac{Output\;work}{Input\;work}\)

= \(\frac{2400J}{6000J}\)

= \(\frac{24}{60}\) \(\times\) 100%

= 40%

Hence, MA is \(\frac{12}{6}\) VR is 5 , work output is 2400J, work input is 6000J and \(\eta \) is 40%

In a four-pulley system, a load of 400N is overcome. If MA is 3, calculate, the effort applied , VR and \(\eta \)?.

Here,

No.of pulleys = 4

Load (L) = 400N

Mechanical advantage (MA) = 3**i. E= ?ii. VR=?iii. \(\eta \)=? **According to the formula,

i. MA = \(\frac{L}{E}\)

or, 3 = \(\frac{400}{E}\)

or, E = \(\frac{400}{3}\) = 133.33N

ii. VR = No.of pulleys used = 4

iii.\(\eta \)= \(\frac{MA}{VR}\) \(\times\) 100% = \(\frac{3}{4}\) = 75%

Hence, effort applied in it is 133.33N , VR is 4 and \(\eta \)is 75%

By studying the diagram, calculate :**i.Output Workii. Input Work iii. Mechanical advantage (MA)iv. Velocity ratio (VR) v. Efficieny (?)**

Here,

Load (L) = 700N

Effort (E) = 500N

Length of slope (L) = 8m

Height of slope (h) = 5m

i. Output Work = L \(\times\) Ld (h) = 700 \(\times\) 5m = 3500J

ii. Input Work = E \(\times\) Ed (l) = 500 \(\times\) 8m = 4000J

iii. MA = \(\frac{L}{E}\) = \(\frac{700}{500}\) = 1.4

iv. VR = \(\frac{l}{h}\) = \(\frac{8m}{5m}\) = 1.6

v. \(\eta \) = \(\frac{MA}{VR}\)\(\times\) 100%

In a wheel and axle, radius of the wheel is 20 cm and that of the axle is 5cm. If a load of 1000N is overcome by using an effort of 200N

on it, calculate MA, VR and \(\eta \).

Here,

Radius of wheel (R) = 20 cm

Radius of axle (r) = 5 cm

Load (L) = 1000N

Effort (E) = 200N

i. Mechanical advantage(MA)= ?

ii, Velocity Ratio(VR)= ?

iii. Efficiency \(\eta \) = ?

Accrding to the formua

i. MA = \(\frac{L}{E}\) = \(\frac{1000}{200}\) = 5

ii. VR = \(\frac{R}{r}\) = \(\frac{20}{5}\) = 4

iii. \(\eta \) = \(\frac{MA}{VR}\) \(\times\) 100%

= \(\frac{5}{4}\) \(\times\) 100% = 125%

Hence, in that wheel and axle MA is 5, VR is 4 and \(\eta \) is 125%

In the given figure, a lever is shown.Three weights 30N,20N and 10N are suspended in it, Now calculate :

i. Clockwise moment

ii. Anticlockwisw moment

iii. Can the lever be in the condition of baance in this situation ?

iv. What will be the location of 10N weight by keeping other loads unchanged to balance the lever ?

Here,

First weight (L_{1}) = 30N

Second weight (E_{1}) = 20N

Third weight (E_{2}) = 10N

Distance between L_{1} and fulcrum (Ld_{1}) = 30cm

Distance between L_{2} and fulcrum (Ed_{1}) = 5cm

Distance between E_{1} and fulcrum (Ed_{2}) = 20cm

Now,

Anticlockwise moment = L_{1} \(\times\) Ld_{1}

= 30 \(\times\) 30 = 900N

Clockwisw moment = E_{1} \(\times\) Ed_{1} \(\times\) E_{2} \(\times\) Ed_{2}

= 20 \(\times\) 5 + 10 \(\times\) 20

= 300N

iii. The lever will not be balanced, as the clockwise moment and anticlockwise moment are not equal.

iv. To balance the lever by changing the location of 10N weight, there should be -

= E_{1} \(\times\) Ed_{1} + E_{2} \(\times\) Ed_{2} = L_{1} \(\times\) Ld_{2}

= 20 \(\times\) 5 + 10 \(\times\) Ed_{2} = 30 \(\times\) 30

= 100+10Ed_{2} = 900

= \(\frac{900-100}{10}\) = \(\frac{800}{10}\) = 80cm

Thus, the lever can be belanced by keeping the 10N weight at a distance of 80cm from the lever.

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