Mechanical Engineering Principles

FRICTION 179

Efficiencyη =

useful work done

actual work done

which is usually expressed as a percentage

i.e. η=

Wp

W(μπd+p)×πd

(πd−μp)

=

p(πd−μp)

(μπd+p)×πd

Dividing throughout byπdgives:

η=

p

(

1 −

μp

πd

)

(μπd+p)

=

p( 1 −tanλtanθ)

πd

(

μ+

p

πd

)

=

p( 1 −tanλtanθ)

πd(tanλ+tanθ)

However,

tan(λ+θ)=

tanλ+tanθ

( 1 −tanλtanθ)

from compound angle formulae

Hence, η=

p

πd

1

tan(λ+θ)

but

p

πd

=tanθ

hence,efficiency,

η=

tanθ

tan(λ+θ)

( 15. 33 )

From equations (15.31) and (15.32),

thework lost in friction

=

W(μπd+p)

(πd−μp)

×πd–Wp ( 15. 34 )

Problem 12. The coefficient of friction on

the sliding surface of a screw jack is 0.2. If

the pitch equals 1 cm, andD 1 =4cmand

D 2 =5 cm, calculate the efficiency of the

screw jack.

Working in millimetres,

d=

(D 1 +D 2 )

2

=

( 40 + 50 )

2

=45 mm,

p=1cm=10 mm,

tanθ=

p

πd

=

10

π× 45

= 0. 0707 ,

from which, θ=tan−^1 ( 0. 0707 )= 4. 05 °,

and tanλ=μ= 0. 2 ,

from which, λ=tan−^1 ( 0. 2 )= 11. 31 °

From equation (15.33),

efficiencyη=

tanθ

tan(λ+θ)

=

0. 0707

tan( 11. 31 + 4. 05 )°

=

0. 0707

0. 2747

= 0. 257

i.e. η= 25 .7%

Now try the following exercises

Exercise 77 Further problem on the effi-

ciency of a screw jack

1. The coefficient of friction on the sliding
   surface of a screw jack whose thread is
   similar to Figure 15.15, is 0.24. If the
   pitch equals 12 mm, andD 1 =42 mm
   andD 2 =56 mm, calculate the efficiency
   of the screw jack. [24.06%]

Exercise 78 Short answer questions on

friction

1. The.........of frictional force depends
   on the.........of surfaces in contact.


2. The.........of frictional force depends
   on the size of the.........to the sur-
   faces in contact.


3. The ......... of frictional force is
   always .........to the direction of
   motion.


4. The coefficient of friction between sur-
   faces should be a .........value for
   materials concerned with bearings.