Engineering Mechanics

Chapter 11 : Simple Lifting Machines „„„„„ 191

We know that displacement of the effort in one revolution of the upper pulley block,

= πD ...(i)

This is also equal to the length of the chain pulled over the larger pulley. Since the smaller pulley also
turns with the larger one, therefore length of the chain released by the smaller pulley

= πd

∴ Net shortening of the chain = πD – πd = π (D – d)

This shortening of chain will be equally divided between the two portions of the chain,
supporting the load. Therefore distance through which the load will move up

1

(–) (–)

22

Dd Dd

π

=×π = ...(ii)

∴

Distance moved by the effort

V.R. =

Distance moved by the load

2

(–) –

2

DD

Dd Dd

π

==

π ... (iii)

Now M.A.

W

P

= ... as usual

and efficiency,

M.A.

V.R.

η= ...as usual

Notes: 1. Sometimes, the size of the upper pulleys are given in terms of radius (instead of diameters).

If R = Radius of the larger pulley, and

r = Radius of the smaller pulley.

Then

V.R.^2

R

Rr

=

2. Sometimes, the diameters of upper pulleys are not given and the relatives sizes are expressed in terms
   of number of teeth.
   If T 1 = No. of teeth of the larger pulley, and
   T 2 = No. of teeth of the smaller pulley

Then

1

12

2

V.R.

T

TT

=

Example 11.6. A weston differential pulley consists of a lower block and an upper block.
The upper block has two cogged grooves, one of which has a radius of 125 mm and the other has a
radius of 115 mm.

If efficiency of the machine is 80%, calculate the effort required to raise load of 1500 N.
Solution. Given: Radius of larger groove (R) = 125 mm; Radius of smaller groove (r) = 115
mm; Efficiency (η) = 80% = 0.8 and load to be raised = 1500 N.

Let P = Effort requried to raise the load.

We know that velocity ratio

22125

V.R. 25

–125–115

R

Rr

×

== =