Page 179 - WCS - Electrical

Load
15000 =
=
MA =
2π ∗
Effort

Distance

7
8∗
V.R. =
=

moved
Distance
load
by
2∗
22∗
3008

8

Load
15000 =
=
MA =
8

8

Load
2π ∗
300
Load
= 15000 =
15000 =
work
Output

8

MA =
MA =
Load

Load
Distance == 2π
300
15000
100%
2π ∗ 300∗
Efficiency =
MA =
by

8
MA
Effort =
Load

work
Iutput
2π ∗
300
=
15000
Effort Effort Effort Effort moved by × effort = 15000 = = 2π ∗ 8 300 15000 ∗ moved 8 300 load
= =
MA =
by

load
Distance

moved
300
2π ∗
Effort
V.R. = Distance moved by effort 15000 ∗ 8∗ 7

effort

Distance stance
Di
work
by
Output
m
by effort
moved oved
WORKSHOP CALCULATION & SCIENCE - CITS 8 = 15000 ∗ 7815000 ∗ ∗ 8∗ 7
Distance

by

moved
load
by
Efficiency =
effort

Dis
V.R. =

V.R. =moved tance
effort

2∗
Distance

22∗
by
Load V.R. = Di Distance stance m moved oved moved work = 15000 = 15000 ∗ = 8 =7∗ 3008 πD 85000 ∗ D 7∗
1
load
Iutput
by
by load
V.R. =
Distance

MA = Distance moved moved by effort = 2∗ 22∗ = 3008∗ = 3008 =
2∗ 22
by

load
15000 ∗
Distance
moved

Effort V.R. = Dist Output m ance work by oved load by load 2π ∗ 300 2∗ 22∗ 3008 8∗ 7 2∗ 2 πd 30082∗ d
=
3008
2∗
22∗
Efficiency =
by

work
Iutput
work
=

100%
Efficiency = workOutput O Output workutput × 100% work Distance moved mov load by load
× 100%
Efficiency =
×
Ded loadby ed

Output

Distance mov Distance

Distance
Efficiency =
work
100%
load
by

Iutput =
Distance moved by effort Iutput Load× workIutput 100%× Distance moved × by load Distance =moved = moved by load by load
Efficiency workOutput

moved
Distance
= work
15000 ∗
mov

ded loadby d
Distance stance
Di
Iutput
by
move
×
100%
Efficiency =
V.R. = Effort × Distance work by effort 8∗ 7 = Distance moved by load load
=
moved
=
Distance moved by load I Output workutput work 2∗ 22∗ 3008 Distance moved by load 1 moved  by load
Distance
=
1

Efficiency =
2
of
Distance

Iutput
work
Efficiency =
2
Efficiency = workOutput O Output workutput work work πD = D = moved  πd loadby  − πd  
=

Output
D
Efficiency =
Distance

Output work Load Iutput = workIutput work moved by load πD πd = πD DπD 1 =
d
Efficiency workOutput
D
=
=
π
Efficiency = × 100% Iut = workput × Iutput work Distance moved by load = = πd = d π  1 DD − d 2  
Efficiency =
πd = d
d =
effort
Iutput work Effort Distance moved by = πd πD = D 2  πd d 
work
Iutput
d
=
Load× Distance moved by load Distance moved by load D πd D D d 2 
=
1
m
Load×
Load×
Distance
Distance
Output work Effort × Distance moved moved oved by by lo load ad 1 D d π  d − d 
D
by
effort

= Distance
Load×
=

m
Mechanical loadby oved
Advantage =by moved ance
Efficiency = = Ef Effort × Difort × = Load× Dist by by effort effort load 1  D 1 d d =  
2
moved oved
Distance stance
m
− πd
=
Velocity πD
Ratio
Iutput work Load× Distance moved by load by effort = = D = d 2 of  πd of  π 1   d 2 2 2  

Effort ×
effort
Distance

moved
1
by
1
moved
Distance


Effort ×

=
 1
= of d
1
2
Distanc
by effortby
= 1
Effort ×

moved

load
=

 πd 
Distance 21

=(Mechanical Advantage )/(Velocity Ratio)
2  
Load × Distance moved e moved moved oved Load load πd d 1 of   1 2 − πdπd Velocity 21 1    πd − πdd  2 − πd    − πd    by Load
of
 =
= 1
=
2
 
Ratio = 

 2 πd
Load DistanceLoad
by
by load
m
Distance
π of
d  πd
= =
− d −
 
move
Distance
Effort

effort
1
Load

=
load
× e
× moved

 1 
Load
2  
2 2
D
MA = istance
moved
From the above expression, it can be found by cross multiplication.

 
2
d − d
π d
= π

=

Ef d
by
= 2
m
by effort
Load× Distance moved by ×loa = Distanc Distance × by by d moved oved Effort effort by load D 1 π  1 − d 1  1 1 − d  Distance  moved by Load
Distance stance
Effort =Difort

d
2
moved
Load
load
by
 1
=
= Effort = Distance moved by effort moved by effort d 1 2  2  =  π  d  − d 
Distance
Effort
×
1 1 
2  2
Mechanical Advantage = Efficiency x Velocity Ratio
effort
Effort × Distance move by d = Mechanical Advantage = by effort 1  2  2  = π   d − dπ   d  − d 2   2  = 2   πD 
1

Distance
moved
Effort
2
1 
 1
d − d 
d 
2 
1200 Ratioelocity
V
d
-
= 1
d 2 
= =
1

2 

Examples = = M Mechanical Advantageechanical Advantage = 1 1 = 1 2 of  πd 1 − πd   π  d − d  π π d − π  π 1  1 d  2  


 d −



Mechanical
=
2 
= 
300 Velocity RatioVelocity
=
Advantage =
Mechanical

Load Distance moved by load Advantage = Velocity Ratio 1 Ratio =  π =1  d − d 2    by Load
Distance=
Velocity 1
2 moved
Mechanical
Advantage =
=

= × A load of 1200 kg is lifted using a simple machine having a velocity ratio of 5 . Find out effiency of machine, if


2
Ratio  1
2
Velocity
2 
 2
1
Load
Effort Distance moved by effort MA = Velocity Ratio = 2 π d − d =   = = Ve Ratio = = Ratio = 2 Di Distancmove stance Distance  moved by Load
d mov
π e
d  by ed Loadby
-d
effort applied is 300 kg Load
moved

Distance

by

Load
 Load
=moved
Di

by
Velocity Ratiolocity stance

πD/
=
Load
Load
=
by Load
m
Distance stance
moved oved
Velocity Di
by
=
Ratio =

Mechanical
Solution: MA
= MA =
MA =
by
Distance

Load
moved

Load
2 

moved
by

Distance
1
Velocity
=
= Mechanical Advantage = MA = Load Effort Effort Advantage π  d − Velocity Ratio = Ratio = Distance moved 2 by Load Load Load

πD
1
d 
Effort =
MA

Load
by
Distance
Load
moved
=
Velocity
Ratio
πD
MA =

RatioVelocity Effort Effort Effort =  2  = πD  1 = d- 2 πD ∗ πD 2
1200
d =
π
= πD

=
1200

 1
2
-
d d-
π d
π. 
d
d 
4

 1
=
=
1200 300 1200 1200 Velocity Ratio = Distance moved by Load = d-  2   π πD  1  =  2  1  1 2 - d 2   

d
π

π 

 d-
 d
=


=

=
300
2  

 
2
MA = Load 300 1200 300 = 5 300 Distance moved by Load π  1  1 d   2   2D 
d d
- d-
π

=4 =

2 
 

2 
Effort 300 =  d - d  π  π d  d - 1  1  - d d  =  1 2 
πD/ 1

Mechanical
Advantage
π

Velocity Ratio = 5 (given) MA = Load = πD = πD/  = = πD/  2  - d 2  2  π   d 1 d- - d 2  
d
πD/ 1
d
Mechanical Advantageechanical
M
π
Advantage
πD/
Ratio
Velocity
2
1200 Mechanical Advantage Effort π  1 - d 2  = πD/  2 = 2  2
Advantage
Efficiency if machine = Mechanical
Ratio
Velocity Ratiolocity
Ve
= Mechanical Advantage d  = πD ∗ 2 D 15

300 Velocity Ratio Velocity Ratio   = πD  1 ∗ d- 2 2  2 = =
2
= πD dπ. ∗
Velocity Ratio 75  1 2  = πD ∗   = πD 21  2 d 2 6
4

 1  ∗ 

=
π. d
-
4
d 

2
 1
2

=
=
π. 

4 5 4 4 = = πD/ π  d - d   = πD dπ.  1 ∗ d- 2D   π. 2  d d- Load d 150  
 -d
=0.8=80% 25




2

d
d
Mechanical Advantage = 5 4 5 5 = 5 2  = π.  1 - 2D =   Effect = 
2 2D
1
Load

=
A load of 75 kg is lifted using a simple machine having a velocity ratio of 4 . Find out effiency of machine, if effort
x
= d

5
Velocity Ratio MA = MA = Load Load 3 = πD ∗ 2 = d 1 2D d 2 d - = 2D 1 d - 1 2 d = 2 75 1 2D 15 15
Load

applied is 25 kg
d
- d
Effort
=
Load
MA =

-
2
=

d
15
Effort
MA

= d
d
15
4 Solution: MA = Effort Load Effort = Effort 4 π.  1 - d 2    D D 1 = d- D 15D 2 = 100 - × d 6 = 8
MA =
d
6
=
=
15

=
= =
75
5 Effort 75 Distance moved by effort = d = D d 15 D 15 15
d 6
=
6 =
=
75
2D
6
=
Velocity
=
75 25 = Ratio = 75 = Load == 150 d 8 6
d
Load = 75 25 25 = Distance moved by Load - d 2 = Effect = 6 x 150
1

Load 150Load
d
=0
MA = 25 = 25 Load = 15 = = 150 15
Load
150
Effort =3 3 3 2π l 15 = Effect 75 = Ef Effect = xfect 150 x x = = 8 x
25
Load
=
x
=

=
15 =
15 Effect
=
3
=
=
=
Effect
15 15
Velocity Ratio = 4 (given) 4 3 4 3 p D = 6 = 100 × 75 75 15 x 15
=
8
6
d
4
=
=
75
=15
3
×
75
75 4 Distance moved 4 by effort = 100 × 15 75 = 1515 × = = 8 × 150x8 15

15
=
6 8
10
100 = 60
=
=
15
Efficiency if machine
6
8
4
8
6
100
25
by Load
×
2ππ
=
Rati
=
Velocity
o =
15
Velocity D

effort =

=
Load
Distance

moved
by

effo
=
Velocity Ratiolocity
=
Ratio =
=(Mechanical Advantage )/(Velocity Ratio)rtby moved Distance Di Ratio = moved istance moved stance D Distance by by effort moved by effort = = 150 2 15 100 8 6 15 = 8 15 8
= Ve
Loadt
by effort Effec
1x
moved oved

m
Distance istance
=
by
by Load
Ratio =
3 Velocity Ratio = = Velocity Di Distance stance m moved oved effort Load p − p 8 150 15 15 8 8
by
Distance
moved

Distan
= = Velocity Ratio = 2π moved ce l by Load moved by Load = 15 8
Distance
150 150
8
x
4 = Dista 2π l2π moved nce l by Load = 75 × 15 = 15 150 15 15 8 = = 15

8
= 0.75 2π l p = = p p 2π l of screw 100 6 8 x = 150 x x 15 8 150 = 8
=
8
Pitch
=
x
=
F ×
p
=
Distance moved by effort p 2π l / Circumfere nce of screw 15 150x88x =
2
= 75%
p
= Velocity Ratio = Distance moved by effort 2ππ
150x8
Distance Velocity Loadby oved Ratio = = 8 150x8 8 150x8
= m
effort
moved oved
2ππ

Di
Distance stance
m
Working Principles of Simple Machines Loadby moved Distance by effort by by effort 2ππ 1 = p 2 2ππ 15 8 8 150x8

=moved
Ratio =
=
= Ve
Di
p − =
Velocity Ratiolocity stance
by 150
effort
moved

Velocity
moved oved =
= l2π Ratio = = Velocity Di Distance stance Distance effort Load 2ππ 1 2 p = 2 = 2ππ 8 150x8 8
m
by Load
by
1
Ratio =
by
moved

Distance
p − p
p −
1
= 1 Inclined Plane:Distance moved by Load p − 2 by Load x p − p 2 8
1 8
Distance
moved = p
=
Velocity
Ratio =
p
by

Load
Distance

moved
1
2
p
p −
- An inclined plane is a sloping surface used to raise or lower objects with less force.

Pitch
screw

of
=
Pi
f
screw
screw
2
Circumfere

nce
- It reduces the amount of force needed to lift a load by increasing the distance over which the force is applied.
screw
Distance moved F × / = F P F × /× itch / screw of 2ππ Pitch o tch of of Pitch of screw 150x8
by
= effort

= Velocity Ratio = = F × / 2 2 = Ci = Circumfere ncrcumfere nce o e f of screw 8
screw
/
F ×

screw
Distance moved by = Circumfere p Pitch 2 of Circumfere nce of screw
Load
Applied force Parallel to the Plane
2
1

screw
nce −
2 of p
/
F ×
2
Circumfere
Elfort x Distance moved by effort nce of screw
= F × / Pitch of screw
2 Circumfere nce of screw
166
CITS : WCS - Electrical - Exercise 17