(^168) „„„„„ A Textbook of Engineering Mechanics
If there would have been no friction between the screw and the nut, then φ will be zero. In such
a case, the value of effort (P 0 ) necessary to raise the same load, will be given by the equation :
P 0 = W tan α ...[substituting φ = 0 in equation (i)]
∴^0
Ideal effort tan tan
Efficiency ( ) =
Actual effort tan ( ) tan ( )
P W
PW
αα
η===
α+φ α+φ
It shows that the efficiency of a screw jack is independent of the weight lifted or effort ap-
plied. The above equation for the efficiency of a screw jack may also be written as:
sin
cos sin cos ( )
sin ( ) cos sin ( )
cos ( )
α
α α× α+φ
η= =
α+φ α× α+φ
α+φ
or
sin cos ( )
1– 1–
cos sin ( )
α× α+φ
η=
α× α+φ
cos sin ( ) – sin cos ( )
cos sin ( )
αα+φ αα+φ

αα+φ
sin
cos sin ( )
φ

αα+φ [Q sin (A – B) = sin A cos B – cos A sin B]
2sin
1–
2cos sin( )
φ
η=
αα+φ [Multiplying and dividing by 2]
2sin
sin (2 ) – sin
φ

α+φφ
...[Q 2 cos A sin B = sin (A + B) + sin (A – B)]
Now for the efficiency to be maximum, the term (1 – η) should be the least. Or in other words,
the value of sin (2α + φ) should be the greatest. This is only possible, when
2 α + φ = 90° or 2 α = 90° – φ
∴ 45 –
2
φ
α= °
It shows that the maximum efficiency of a screw jack is also independent of the weight lifted
or effort applied.
Example 9.13. A load of 2.5 kN is to be raised by a screw jack with mean diameter of
75 mm and pitch of 12 mm. Find the efficiency of the screw jack, if the coefficient of friction between
the screw and nut is 0.075.
Solution. Given: Load (W) = 2.5 kN ; Mean diameter of the screw (d) = 75 mm; Pitch of the
screen (p) = 12 mm and coefficient of friction between the screw and nut (μ) = 0.075 = tan φ.
We know that
12
tan 0.051
75
p
d
α= = =
ππ×
and efficiency of the screw jack,
tan tan 0.051
tan ( ) tan tan 0.051 0.075
1– tan tan 1–(0.051 0.075)
αα
η= = =
α+φ α+ φ +
αφ ×
0.051
0.1265

= 0.403 = 40.3% Ans.