Engineering Mechanics

Chapter 10 : Principles of Lifting Machines „„„„„ 181

Mechanical advantage
Substituting the value of W in the law of the machine,
P = mW + C = 0.04 × 2000 + 7.5 = 87.5 N

∴

2000

M.A. 22.9

87.5

W

P

== = Ans.

Efficiency of the machine

We know that M.A. 22.9 0.5725 57.25%

V.R. 40

η= = = =^ Ans.

Maximum efficiency of the machine
We know that *maximum efficiency of the machine,
11
Max. 0.625 62.5%
m V.R. 0.04 40

η= = = =

××

Ans.

Relation between F and W
We know that effort lost in friction,

(effort ) – (0.04 7.5) –

V.R. V.R.

WW

FP==+W

11

0.04 – 7.5 0.04 – 7.5

V.R. 40

=+=+WW⎛⎞⎛⎞⎜⎟⎜⎟
⎝⎠⎝⎠
= W (0.04 – 0.025) + 7.5 = 0.015 W + 7.5 Ans.
Value of F when W is 2 kN
Substituting the value of W equal to 2 kN or 2000 N in the above equation,
F = (0.015 × 2000) + 7.5 = 37.5 N Ans.
Example 10.9. The law of a certain lifting machine is :
W
P8
50

=+

The velocity ratio of the machine is 100. Find the maximum possible mechanical advantage

and the maximum possible efficiency of the machine. Determine the effort required to overcome the

machine friction, while lifting a load of 600 N. Also calculate the efficiency of the machine at this

load.

Solution. Given: Law of lifting machine 80.02 8;

50

W

PW=+= + Velocity ratio (V.R.)

= 100 and load (W) = 600 N.
Maximum possible mechanical advantage
Comparing the given law of the machine with the standard relation for the law of the machine
(i.e. P = mW + C) we find that in the given law of the machine, m = 0.02. We know that maximum
possible mechanical advantage

(^) Max M.A.^1150
m 0.02
== =^ Ans.
Maximum possible efficiency
We know that maximum possible efficiency
111
0.5 50%
m V.R. 0.02 100 2
== ===
××^
Ans.

• In this example, the value of m = 0.04 and C = 7.5 as the law of the machine is given by the relation
  P = mW + C.