Mechanical Engineering Principles

SIMPLE MACHINES 205

2. A compound gear train has a 30-tooth
   driver gearA, meshing with a 90-tooth
   follower gearB. Mounted on the same
   shaft asBand attached to it is a gear
   C with 60 teeth, meshing with a gear
   Don the output shaft having 120 teeth.
   Calculate the movement and force ratios if
   the overall efficiency of the gears is 72%.
   [6, 4.32]


3. A compound gear train is as shown in
   Figure 18.6. The movement ratio is 6 and
   the numbers of teeth on gearsA,Cand
   Dare 25, 100 and 60, respectively. Deter-
   mine the number of teeth on gearBand
   the force ratio when the efficiency is 60%.
   [250, 3.6]

18.6 Levers

Alevercan alter both the magnitude and the line
of action of a force and is thus classed as a simple
machine. There are three types or orders of levers,
as shown in Figure 18.7.

Fulcrum

ab

Load
Fl

Effort

Fe

(a)

Fulcrum

Effort

Load Fe

Fl

(b)

Fulcrum

Effort

Fe Load

Fl

(c)

Figure 18.7

A lever of the first order has the fulcrum

placed between the effort and the load, as shown

in Figure 18.7(a).

A lever of the second orderhas the load placed

between the effort and the fulcrum, as shown in

Figure 18.7(b).

A lever of the third orderhas the effort applied

between the load and the fulcrum, as shown in

Figure 18.7(c).

Problems on levers can largely be solved by

applying the principle of moments (see Chapter 5).

Thus for the lever shown in Figure 18.7(a), when

the lever is in equilibrium,

anticlockwise moment=clockwise moment

i.e. a×Fl=b×Fe

Thus,force ratio

=

Fl

Fe

=

b

a

=

distance of effort from fulcrum

distance of load from fulcrum

Problem 9. The load on a first-order lever,

similar to that shown in Figure 18.7(a), is

1.2 kN. Determine the effort, the force ratio

and the movement ratio when the distance

between the fulcrum and the load is 0.5 m

and the distance between the fulcrum and

effort is 1.5 m. Assume the lever is 100%

efficient.

Applying the principle of moments, for equilibrium:

anticlockwise moment=clockwise moment

i.e. 1200 N× 0 .5m=effort× 1 .5m

Hence, effort=

1200 × 0. 5

1. 5

=400 N

force ratio=

Fl

Fe

=

1200

400

= 3

Alternatively, force ratio=

b

a

=

1. 5

0. 5

= 3

This result shows that to lift a load of, say, 300 N,

an effort of 100 N is required.

Since, from equation (3),

efficiency=

force ratio

movement ratio

×100%