Chapter 10 : Principles of Lifting Machines 183
∴ Law of the machine is given by the relation
P = 0.08 W + 22.5 Ans.Effort required when the load is 2 kN
Substituting the value of W equal to 2 kN or 2000 N in the above equation,
P = (0.08 × 2000) + 22.5 = 182.5 N Ans.Efficiency of the machine when the load is 2 kN
We know that M.A.^2000 10.96
182.5W
P== =and efficiency, M.A 10.96 0.609 60.9%
V.R. 18
η= = = =^ Ans.Maximum efficiency the machine can attain
We also know that maximum efficiency the machine can attain,
11
Max. 0.694 69.4%
m V.R. 0.08 18η====
××Ans.EXERCISE 10.2
- In a certain weight lifting machine, an effort of 15 N can lift a load of 300 N and an effort
of 20 N can lift a load of 500 N. Find the law of the machine. Also find the effort required
to lift a load of 880 N. [Ans. P = 0.025 W + 7.5 ; 29.5 N] - In a weight lifting machine, an effort of 40 N can lift a load of 1000 N and an effort of
55 N can lift a load of 1500 N. Find the law of the machine. Also find maximum
mechanical advantage and maximum efficiency of the machine. Take velocity ratio of the
machine as 48. [Ans. P = 0.03 W + 10 ; 33.3 ; 69.4%] - The following results were obtained from a test on a certain weight lifting machine having
a velocity ratio of 20 :
Load in N (W) 400 500 600 700 800 900 1000
Effort in N (P) 85 100 115 135 145 160 175
Plot the graph showing load and effort. From the graph, determine the law of the
machine. [Ans. P = 0.15 W + 2.5]
Hint. All the points, except 4, will lie on the straight line. Therefore ignore this point
and now study the law of the machine from any two remaining points.
QUESTIONS
- What is a machine? Explain the difference between a simple machine and a compound
machine. - Define mechanical advantage of a machine.
- What is an ideal machine?
- Define velocity ratio of a machine.
5. Derive the relation between mechanical advantage, velocity ratio and efficiency of a machine. - Explain how the efficiency of a simple machine is determined?