Mechanical Engineering Principles

(Dana P.) #1
SIMPLE MACHINES 199

Dividing both numerator and denominator by Fl
gives:


Fl
aFl+b

=

1

a+

b
Fl

When the load is large,Flis large and


b
Fl

is small

compared with a. The force ratio then becomes


approximately equal to


1
a

and is called thelimiting

force ratio,i.e


limiting ratio=

1
a

Thelimiting efficiencyof a simple machine is
defined as the ratio of the limiting force ratio to the
movement ratio, i.e.


Limiting efficiency

=

1
a×movement ratio

×100%

whereais the constant for the law of the machine:


Fe=aFl+ b

Due to friction and inertia, the limiting efficiency of
simple machines is usually well below 100%.


Problem 1. A simple machine raises a load
of 160 kg through a distance of 1.6 m. The
effort applied to the machine is 200 N and
moves through a distance of 16 m. Takingg
as 9.8 m/s^2 , determine the force ratio,
movement ratio and efficiency of the
machine.

From equation (18.1),


force ratio=

load
effort

=

160 kg
200 N

=

160 × 9 .8N
200 N

= 7. 84

From equation (18.2),


movement ratio=

distance moved by the effort
distance moved by the load

=

16 m
1 .6m

= 10

From equation (18.3),

efficiency=

force ratio
movement ratio

×100%

=

7. 84
10

× 100 = 78 .4%

Problem 2. For the simple machine of
Problem 1, determine: (a) the distance
moved by the effort to move the load
through a distance of 0.9 m, (b) the effort
which would be required to raise a load of
200 kg, assuming the same efficiency, (c) the
efficiency if, due to lubrication, the effort to
raise the 160 kg load is reduced to 180 N.

(a) Since the movement ratio is 10, then from
equation (18.2),

distance moved by the effort

= 10 ×distance moved by the load
= 10 × 0. 9 =9m

(b) Since the force ratio is 7.84, then from equa-
tion (18.1),

effort=

load
7. 84

=

200 × 9. 8
7. 84

=250 N

(c) The new force ratio is given by

load
effort

=

160 × 9. 8
180

= 8. 711

Hencethe new efficiency after lubrication

=

8. 711
10

× 100 = 87 .11%

Problem 3. In a test on a simple machine,
the effort/load graph was a straight line of
the formFe=aFl+b. Two values lying on
the graph were atFe=10 N,Fl=30 N,
and atFe=74 N,Fl=350 N. The
movement ratio of the machine was 17.
Determine: (a) the limiting force ratio,
(b) the limiting efficiency of the machine.

(a) The equationFe=aFl+bis of the form
y=mx+c,wheremis the gradient of the
graph. The slope of the line passing through
Free download pdf