1. Lever
2. Pulley
3. Wheel and axle
4. Inclined plane
5. Screw
6. 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.

The work done by the machine in overcoming the load is called output work.
i.e. Output work = Load x load 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.

Given, Load (L) = 300N
Load distance (Ld) = 10cm = 10/100m = 0.1m

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%

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 %

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.

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%

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%

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%

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%

Here,
First weight (L₁) = 30N
Second weight (E₁) = 20N
Third weight (E₂) = 10N
Distance between L₁ and fulcrum (Ld₁) = 30cm
Distance between L₂ and fulcrum (Ed₁) = 5cm
Distance between E₁ and fulcrum (Ed₂) = 20cm

Now,
Anticlockwise moment = L₁ \(\times\) Ld₁
= 30 \(\times\) 30 = 900N

Clockwisw moment = E₁ \(\times\) Ed₁ \(\times\) E₂ \(\times\) Ed₂
= 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 -
= E₁ \(\times\) Ed₁ + E₂ \(\times\) Ed₂ = L₁ \(\times\) Ld₂
= 20 \(\times\) 5 + 10 \(\times\) Ed₂ = 30 \(\times\) 30
= 100+10Ed₂ = 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.