Question

A pulley system with a velocity ratio of 4 is used
to lift a load of 150 kgf through a vertical height
of 20 m. The effort required is 50 kgf in
downward direction. Calculate :
(a) the distance moved by the effort,
(b) the work done by the effort,
(c) the mechanical advantage,
(d) the efficiency of the pulley system, and
(e) the total number of pulleys and the number
of pulleys in each block.
(g = 10 N kg-1)​

Answer 1

A pulley system with a velocity ratio of 4 is used to lift a load of 150 kgf through a vertical height of 20m. The effort required is 50 kgf in downward direction.

(a) The distance moved by the effort.

Here velocity ratio is given by,

$V.R=\frac{\text{distance moved by effort}}{\text{distance moved by load}}$

Here, distance moved by applied load is 20m and velocity ratio is 4.

∴ 4 = distance moved by effort/20

⇒ distance moved by effort = 4 × 20 = 80 m

Therefore the distance moved by effort is 80 m.

(b) The work done by the effort.

We know, work done is the product of displacement and force applied along the displacement.

∴ Work done by the effort = effort × distance moved by effort

= 50 kgf × 80 m

= 40000 J  [∵ 1 kgf m = 10 J ]

Therefore the work done by the effort is 40000 J.

(c) the mechanical advantage.

mechanical advantage is the ratio of load to effort.

$M.A=\frac{L}{E}=\frac{\text{150 kgf}}{\text{50 kgf}}=3$

Therefore the mechanical advantage is 3.

(d) the efficiency of the pulley system.

efficiency of the pulley system is given by,

$\eta=\frac{M.A}{V.R}\times100\\\\=\frac{3}{4}\times100=75$

Therefore the efficiency of the pulley system is 75%.

(e) The total number of pulleys and the number of pulleys in each block.

total number of pulleys = velocity ratio = 4 [ an even number ]

∴ number of pulleys in each block is 2.i.e., there are two pulleys in fixed block and 2 pulleys in movable block.

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