Engineering Mechanics

(^182) „„„„„ A Textbook of Engineering Mechanics
Effort required to overcome the machine friction
We know that effort required to lift a load of 600 N
P = mW + 8 = (0.02 × 600) + 8 = 20 N
and effort required to overcome the machine friction, while lifting a load of 600 N,
(effort )
600
–20–14N
V.R. 100
W
FP=== Ans.
Efficiency of the machine
We know that mechanical advantage of the machine while lifting a load of 600 N.
600
M.A 30
20
W
P
== =
and efficiency, M.A.^30 0.3 30%
V.R. 100
η= = = =^ Ans.
Example 10.10. In an experiment of a weight lifting machine, with velocity ratio as 18, the
values of effort required to lift various loads were as given in the table below :
Load (W) in N 250 500 750 1000 1500 2500
Effort (P) in N 42.5 62.5 82.5 105 142.5 220
Plot a graph showing the relation between effort and load, and determine the law of the
machine. Find the effort required and efficiency of the machine, when the load is 2 kN. Also find the
maximum efficiency, this machine can attain.
Solution. Given: Velocity ratio (V. R.) = 18
Law of the machine
First of all, draw a suitable graph and plot the points 1, 2, 3, 4, 5, 6. From the geometry of the
points, we find that points 1, 2, 3, and 5 lie on the same straight line, whereas the point 4 lies above the
line and the point 6 below the line. Therefore let us ignore the points 4 and 6. Now draw a straight line
AB passing through the points 1, 2, 3 and 5 as shown in Fig. 10.3. Now let us measure the intercept
OA on y-y axis, which is equal to 22.5 N.
Fig. 10.3.
Now consider two points 1 and 5 (having maximum distance between them), which lie on the
straight line AB. From the geometry of these two points, we find that the slope of the line AB,
142.5 – 42.5 100
0.08
1500 – 250 1250
m===