Engineering Mechanics

(^200) „„„„„ A Textbook of Engineering Mechanics
Solution. Given: Length of lever (l) = 700 mm; No. of pinion teeth (T 2 ) = 12; No. of spur
geer teeth (T 1 ) = 96 and dia of load axle = 200 mm or radius (r) = 200/2 = 100 mm.
(i) Law of the machine
When P 1 = 60 N, W 1 = 1800 N and when P 2 = 120 N, W 2 = 3960 N.
Substituting the values of P and W in the law of the machine i.e., P = mW + C
60 = (m × 1800) + C ...(i)
and 120 = (m × 3960) + C ...(ii)
Subtracting equation (i) from equation (ii)
60 = m × 2160
or^601
2160 36
m==
Now substituting this value of m in equation (i),
1
60 1800 50
36
=× +=+⎛⎞⎜⎟CC
⎝⎠
∴ C = 60 – 50 = 10
and now substituting the value of m = 1/36 and C = 10 in the law of machine,
1
10
36
PW=+ Ans.
(ii) Efficiencies of the machine in both the cases
We know that velocity ratio
1
2
700 96
V.R. 56
100 12
l T
rT
=× = × =
and mechanical advantage in the first case
1
1
1800
M.A. 30
60
W
P
== =
∴ Efficiency 1
M.A 30
0.536 53.6%
V.R. 56
η= = = =^ Ans.
Similarly, mechanical advangate in the second case,
2
2
3960
M.A. 33
120
W
P
== =
∴ Efficiency 2
M.A. 33
0.589 58.9%
V.R. 56
η= = = = Ans.