Question

A pulley system with a velocity ratio 4 is used to lift a load of 150kgf through a vertical height of 20 m. the effort required is 50kgf in downward direction.
calculate:
distance moved by effort
total no. of pulleys
mechanical advantage
efficiency

Answer 1

The mechanical advantage is 3.5 , Efficiency is 87.5 , Work done by effort is 1875 Joules  and the 3.75 m

Explanation:

Given data:

• V-R = 4
• Load = 150 Kg
• dL = 20 m
• Effort = 50 Kg

dE = ?

Solution:

V = R = dE / dL

4 = dE / 20

dE = 80 m

Since V = R ( no. of pulley is a system of n pulleys.)

Number of pulleys = 4

Mechanical advantage = 175/50 = 3.5

Efficiency = (mechanical advantage/velocity ratio)×100  = (3.5/4)×100 = 87.5

Work done by effort = 50×10×(15/4)= 1875 Joules

Distance moved by effort  = 15/4 = 3.75 m

Thus the mechanical advantage is 3.5 , Efficiency is 87.5 , Work done by effort is 1875 Joules  and the 3.75 m

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250 N force is required to raise 75 kg mass from a pulley. If rope is pulled 12 m then the load is lifted to 3 m, the efficiency of pulley system will be?

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Answer 2

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 = $\bold{d_L}$ = 20 m

Effort = F = 50 kgf

To find:

Distance moved by effort = ?

Total no. of pulleys = ?

Mechanical advantage = ?

Efficiency = ?

Formula used:

$\bold{Velocity \ Ratio = \frac{Distance \ moved \ by \ effort}{Distance \ moved \ by \ load}}$

$\bold{Mechanical \ advantage = \frac{Load}{Effort}}$

$\bold{Efficiency = \frac{Mechanical \ Advantage}{Velocity \ Ratio} \times 100}$

Solution:

• Velocity Ratio:

$\bold{Velocity \ Ratio = \frac{Distance \ moved \ by \ effort}{Distance \ moved \ by \ load}}$

$\bold{Velocity \ Ratio = \frac{d_E}{d_L}}$

$\bold{4 = \frac{d_E}{20}}$

$\bold{\therefore d_E = Distance \ moved \ by \ effort = 80 \ m}$

• Number of pulley:

Number of pulley = Velocity Ratio

Number of pulley = 4

• Mechanical advantage:

$\bold{Mechanical \ advantage = \frac{Load}{Effort}}$

$\bold{Mechanical \ advantage = \frac{150}{50}}$

$\bold{\therefore Mechanical \ advantage = 3}$

• Efficiency:

$\bold{Efficiency = \frac{Mechanical \ Advantage}{Velocity \ Ratio} \times 100}$

$\bold{Efficiency = \frac{3}{4} \times 100}$

$\bold{Efficiency = 0.75 \times 100}$

$\bold{ \therefore Efficiency = 75 \%}$