Mechanical Engineering Principles

202 MECHANICAL ENGINEERING PRINCIPLES

(b) Ann-pulley system has a movement ratio of
n, hence this 6-pulley system has amovement
ratioof 6
(c) From equation (18.3),

efficiency=

force ratio

movement ratio

× 100

=

3. 75

6

× 100 = 62 .5%

Now try the following exercise

Exercise 90 Further problems on pulleys

1. A pulley system consists of four pulleys
   in an upper block and three pulleys in
   a lower block. Make a sketch of this
   arrangement showing how a movement
   ratio of 7 may be obtained. If the force
   ratio is 4.2, what is the efficiency of the
   pulley. [60%]


2. A three-pulley lifting system is used to
   raise a load of 4.5 kN. Determine the
   effort required to raise this load when
   losses are neglected. If the actual effort
   required is 1.6 kN, determine the effi-
   ciency of the pulley system at this load.
   [1.5 kN, 93.75%]

18.4 The screw-jack

Asimple screw-jackis shown in Figure 18.3 and

is a simple machine since it changes both the mag-

nitude and the line of action of a force.

Load

Effective radius

r

Bar

Screw

lead

L

Body

Ta b l e

Figure 18.3

The screw of the table of the jack is located in

a fixed nut in the body of the jack. As the table

is rotated by means of a bar, it raises or lowers a

load placed on the table. For a single-start thread, as

shown, for one complete revolution of the table, the

effort moves through a distance 2πr, and the load

moves through a distance equal to the lead of the

screw, say,L.

Movement ratio=

2 πr

L

( 18. 4 )

Problem 6. A screw-jack is being used to

support the axle of a car, the load on it being

2.4 kN. The screw jack has an effort of

effective radius 200 mm and a single-start

square thread, having a lead of 5 mm.

Determine the efficiency of the jack if an

effort of 60 N is required to raise the car

axle.

From equation (18.3),

efficiency=

force ratio

movement ratio

×100%

where force ratio=

load

effort

=

2400 N

60 N

= 40

From equation (18.4),

movement ratio=

2 πr

L

=

2 π( 200 )mm

5mm

= 251. 3

Hence, efficiency=

force ratio

movement ratio

× 100

=

40

251. 3

× 100 = 15 .9%

Now try the following exercise

Exercise 91 Further problems on the

screw-jack

1. Sketch a simple screw-jack. The single-
   start screw of such a jack has a lead
   of 6 mm and the effective length of the
   operating bar from the centre of the screw
   is 300 mm. Calculate the load which can
   be raised by an effort of 150 N if the
   efficiency at this load is 20%.
   [9.425 kN]