A pulley system with velocity ratio of 4 is used to lift a load of150 kgf
through a vertical height of 20 m. The effort required is 50 kgf in the
downward direction. Calculate:
(i) The distance moved by the effort,
(ii) The work done by the effort and
(iii) The efficiency of the pulley system
Answers
Answer:
Distance moved by effort is 80 m
Total no. of pulleys is 4
Mechanical advantage is 3
Efficiency is 75%
Given:
Velocity ratio = 4
Load = 150 kgf
Vertical height = \bold{d_L}d
L
= 20 m
Effort = F = 50 kgf
To find:
Distance moved by effort = ?
Total no. of pulleys = ?
Mechanical advantage = ?
Efficiency = ?
Formula used:
\bold{Velocity \ Ratio = \frac{Distance \ moved \ by \ effort}{Distance \ moved \ by \ load}}Velocity Ratio=
Distance moved by load
Distance moved by effort
\bold{Mechanical \ advantage = \frac{Load}{Effort}}Mechanical advantage=
Effort
Load
\bold{Efficiency = \frac{Mechanical \ Advantage}{Velocity \ Ratio} \times 100}Efficiency=
Velocity Ratio
Mechanical Advantage
×100
Solution:
Velocity Ratio:
\bold{Velocity \ Ratio = \frac{Distance \ moved \ by \ effort}{Distance \ moved \ by \ load}}Velocity Ratio=
Distance moved by load
Distance moved by effort
\bold{Velocity \ Ratio = \frac{d_E}{d_L}}Velocity Ratio=
d
L
d
E
\bold{4 = \frac{d_E}{20}}4=
20
d
E
\bold{\therefore d_E = Distance \ moved \ by \ effort = 80 \ m}∴d
E
=Distance moved by effort=80 m
Number of pulley:
Number of pulley = Velocity Ratio
Number of pulley = 4
Mechanical advantage:
\bold{Mechanical \ advantage = \frac{Load}{Effort}}Mechanical advantage=
Effort
Load
\bold{Mechanical \ advantage = \frac{150}{50}}Mechanical advantage=
50
150
\bold{\therefore Mechanical \ advantage = 3}∴Mechanical advantage=3
Efficiency:
\bold{Efficiency = \frac{Mechanical \ Advantage}{Velocity \ Ratio} \times 100}Efficiency=
Velocity Ratio
Mechanical Advantage
×100
\bold{Efficiency = \frac{3}{4} \times 100}Efficiency=
4
3
×100
\bold{Efficiency = 0.75 \times 100}Efficiency=0.75×100
\bold{ \therefore Efficiency = 75 \%}∴Efficiency=75%