A block and tackle pulley system is used to lift a load having four number of pulleys. It lifts a load 450 kg through a vertical height of 20 m. The effort applied is 250 kg.
(i) Draw a diagram of the system.
(ii) Find the work done by the effort.
(Take g = 10 ms-2 )
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}dL = 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 loadDistance moved by effort
\bold{Mechanical \ advantage = \frac{Load}{Effort}}Mechanical advantage=EffortLoad
\bold{Efficiency = \frac{Mechanical \ Advantage}{Velocity \ Ratio} \times 100}Efficiency=Velocity RatioMechanical Advantage×100
Solution:
Velocity Ratio:
\bold{Velocity \ Ratio = \frac{Distance \ moved \ by \ effort}{Distance \ moved \ by \ load}}Velocity Ratio=Distance moved by loadDistance moved by effort
\bold{Velocity \ Ratio = \frac{d_E}{d_L}}Velocity Ratio=dLdE
\bold{4 = \frac{d_E}{20}}4=20dE
\bold{\therefore d_E = Distance \ moved \ by \ effort = 80 \ m}∴dE=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=EffortLoad
\bold{Mechanical \ advantage = \frac{150}{50}}Mechanical advantage=50150
\bold{\therefore Mechanical \ advantage = 3}∴Mechanical advantage=3
Efficiency:
\bold{Efficiency = \frac{Mechanical \ Advantage}{Velocity \ Ratio} \times 100}Efficiency=Velocity RatioMechanical Advantage×100
\bold{Efficiency = \frac{3}{4} \times 100}Efficiency=43×100
\bold{Efficiency = 0.75 \times 100}Efficiency=0.75×100
\bold{ \therefore Efficiency = 75 \%}∴Efficiency=75%