Principles of Lifting Machines
10.18. MAXIMUM EFFICIENCY OF A LIFTING MACHINE We know that efficiency of a lifting machine,
Mechanical advantage
For *maximum efficiency, substituting the value of P = mW + C in the above equation,
Example 10.8. The law of a machine is given by the relation : P = 0.04 W + 7.5
where (P) is the effort required to lift a load (W), both expressed in newtons. What is the mechanical advantage and efficiency of the machine, when the load is 2 kN and velocity ratio is 40 ? What is the maximum efficiency of the machine ?
If (F) is the effort lost in friction, find the relation between F and W. Also find the value of F, when W is 2 kN.
Solution. Given: Law of machine P = 0.04 W + 7.5 ; Load (W) = 2 kN = 2000 N and velocity ratio (V.R.) = 40.
* We know that efficiency of a lifting machine, M.A.
η = V.R.
A little consideration will show that the efficiency will be maximum, when the mechanical advantage will be maximum.
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
We know that *maximum efficiency of the machine,
Max. 1 1 0.625 62.5%
V.R. 0.04 40
η = m = = =
× × Ans.
Relation between F and W
We know that effort lost in friction,
(effort) – (0.04 7.5) –
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 :
P W 8
=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 8 0.02 8;
50
P= W + = W + 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
We know that maximum possible efficiency
1 1 1
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) – 20 – 600 14 N
V.R. 100
F =P W = = Ans.
Efficiency of the machine
We know that mechanical advantage of the machine while lifting a load of 600 N.
M.A 600 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 1500 – 250 1250 0.08
m= = =
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 10.96 182.5
W
= 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,
1 1
1. 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]
2. 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%]
3. 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
1. What is a machine ? Explain the difference between a simple machine and a compound machine.
2. Define mechanical advantage of a machine.
3. What is an ideal machine ? 4. Define velocity ratio of a machine.
5. Derive the relation between mechanical advantage, velocity ratio and efficiency of a machine.
6. Explain how the efficiency of a simple machine is determined ?
7. What do you understand by the term ‘Reversibility’ of a machine ? Explain the difference between a reversible machine and a self-locking machine.
8. What is law of a machine ? Derive an equation for the same.
9. Obtain an equation for the maximum mechanical advantage and maximum efficiency of a machine.
10. What is meant by ‘friction in a machine’? In how many ways it can be expressed in terms of velocity ratio ?
OBJECTIVE TYPE QUESTIONS
1. The efficiency of a lifting machine is the ratio of (a) Its output to input
(b) Work done by it to the work done on it (c) Its mechanical advantage to its velocity ratio (d) All of the above.
2. If efficiency of a lifting machine is kept constant, its velocity ratio is directly proportional to its
(a) Mechanical advantage (b) Effort applied (c) Machine friction (d) All of the above
3. In an ideal machine, the mechanical advantage is ... velocity ratio
(a) Equal to (b) Less than (c) Greater than
4. A lifting machine having an efficiency less than 50% is known as (a) Reversible machine (b) Non-reversible machine (c) Ideal machine (d) None of the above
5. A weight of 1000 N can be lifted by an effort of 80 N. If the velocity ratio of the machine is 20, then the machine is
(a) Reversible (b) Non-reversible (c) Ideal
6. The maximum mechanical advantage of a lifting machine is
(a) 1 + m (b) 1 – m (c) 1/ m (d) m
7. The maximum efficiency of a lifting machine is
(a) 1/ m (b) V. R./ m (c) m/ V. R. (d) 1/ (m × V. R.)
ANSWERS
1. (d) 2. (a) 3.(a) 4. (b) 5. (a) 6.(c) 7.(d)
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Chapter 11 :
Simple Lifting Machines 185
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