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:
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
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).
We know that,
or,η=\(\frac{L×Ld}{E×Ed}\)×100%
But the mechanical advantages =\(\frac{L}{E}\) and velocity ratio = \(\frac{E_d}{L_d}\)
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
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,
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 the
iv. input work
v. 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. MA
ii. VR
iii. Output work
iv. Input work
v. 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 Work
ii. 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 (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.
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