Engineering Mechanics

Chapter 11 : Simple Lifting Machines „„„„„ 197

We know that distance moved by effort in one revolution of wheel
=πD
If the worm is single threaded (i.e., for one revolution of wheel C, the screw S pushes the worm
wheel through one tooth) then the worm wheel and pulley A will move through.
1
revolution
T

=

∴ Net shortening of the string

2 r

T

π

=

Since the load drum is supported by two strings, therefore the load will be lifted through a
distance

12

2

rr

TT

ππ

=× = ...(ii)

∴

Distance moved by the effort

V.R.

Distance moved by the load

=

DDT

r r

T

π

==

π

...(iii)

Now M.A. =

W

P

...as usual

and efficiency,

M.A.

=

V.R.

η ...as usual

Note: In general, if the worm is n threaded then

V.R. DT

nr

=

Example 11.11. A worm geared pulley block has its effort wheel of 200 mm diameter. The
worm is single threaded and the worm wheel has 60 teeth. The load drum is of 80 mm diameter. Find
the efficiency of the block, if an effort of 75 N is required to lift a load of 9 kN.

Solution. Given: Diameter of effort wheel = 200 mm; No. of teeth in worm wheel (T) = 60;
Diameter of load drum = 80 mm or radius (r) = 40 mm; Load to be lifted (W) = 9 kN = 9000 N and
effort (P) = 75 N.

We know that velocity ratio of a worm geared pulley block,

200 60

V.R. 300

40

DT

r

×

== =

and

9000

M.A. 120

75

W

P

== =

∴ Efficiency,

M.A. 120

0.4 40%

V.R. 300

η= = = = Ans.

Example 11.12. A double threaded worm geared pulley block has effort wheel of 400 mm
diameter and load drum of 100 mm. If the machine is 35% efficient, find the load it can lift by an
effort of 80 N. The worm wheel has 50 teeth.

Solution. Given: No. of threads (n) = 2; Diameter of effort wheel = 400 mm; Diameter of
load drum = 100 mm or radius (r) = 50 mm; Efficiency (η) = 35% = 0.35; Effort (P) = 80 N
and no of teeth in worm wheel (T) = 50.

Let W= Load which can be lifted by the machine

T 2

T