(^202) „„„„„ A Textbook of Engineering Mechanics
∴ Distance moved by the load
24
13
2
TT
r
TT
=π× × ...(ii)
∴
Distance moved by the effort
V.R.
Distance moved by the load

1 3
(^2424)
13
2
2
llT T
TTrT T
r
TT
π ⎛⎞
==×⎜⎟
π× × ⎝⎠
Now M.A.
W
P
= ...as usual
and efficiency, M.A.
V.R.
η= ...as usual
Example 11.15. In a double purchase crab winch, teeth of pinions are 20 and 25 and that
of spur wheels are 50 and 60. Length of the handle is 0.5 metre and radius of the load drum is 0.25
metre. If efficiency of the machine is 60%, find the effort required to lift a load of 720 N.
Solution. Given: No. of teeth of pinion (T 2 ) = 20 and (T 4 ) = 25; No. of teeth of spur
wheel (T 1 ) = 501 and (T 3 ) = 60; Length of the handle (l) = 0.5 m; Radius of the load drum (r) = 0.25
m; Efficiency (η) = 60% = 0.6 and load to be lifted (W) = 720 N.
Let P= Effort required in newton to lift the load.
We know that velocity ratio
1 3
24
1 0.5 50 60
V.R. 12
0.25 20 25
T T
rT T
⎛⎞⎛⎞
=×=⎜⎟⎜⎟×=
⎝⎠⎝⎠
and M.A. W^720
PP

∴ Efficiency
720
M.A. 60
0.6
V.R. 12
P
P

or^60 100 N
0.6
P==^ Ans.
Example 11.16. A double purchase crab used in a laboratory has the following dimensions :
Diameter of load drum = 160 mm
Length of handle = 360 mm
No. of teeth on pinions = 20 and 30
No. of teeth on spur wheels = 75 and 90
When tested, it was found that an effort of 90 N was required to lift a load of 1800 N and an
effort of 135 N was required to lift a load of 3150 N. Determine :
(a) Law of the machine,
(b) Probable effort to lift a load of 4500 N,
(c) Efficiency of the machine in the above case,
(d) Maximum efficiency of the machine.