2. Simple Lifting Machines
Chapter 11
Simple Lifting Machines

3. Learning Objectives Introduction Types of Lifting Machines
Simple Wheel and Axle
Differential Wheel and Axle
Weston’s Differential Pulley Block
Geared Pulley Block
Worm and Worm Wheel
Worm Geared Pulley Block
Single Purchase Crab Winch
Double Purchase Crab Winch
Simple Pulley
First System of Pulleys
Second System of Pulleys
Third System of Pulleys
Simple Screw Jack
Differential Screw Jack
Worm Geared Screw Jack

4. Double Purchase Crab Winch
A double purchase crab winch is an improved form of a single purchase crab winch, in which the velocity ratio is intensified with the help of one more spur wheel and a pinion. In a double purchase crab winch, there are two spur wheels of teeth T1 and T2 and T3 as well as two pinions of teeth T2 and T4.
The arrangement of spur wheels and pinions are such that the spur wheel with T1 gears with the pinion of teeth T2. Similarly, the spur wheel with teeth T3 gears with the pinion of the teeth T4, The effort is applied to a handle as shown in Fig
Fig Double purchase crab winch.

5. Let T1 and T3 = No. of teeth of spur wheels,
T2 and T4 = No. of teeth of the pinions
l = Length of the handle,
r = Radius of the load drum, W = Load lifted, and
P = Effort applied to lift the load, at the end of the handle.
We know that distance moved by the effort in one revolution of the handle,
= 2pl (i)
No. of revolutions made by the pinion 4 = 1
and no. of revolutions made by the wheel 3

6. Example
In a double purchase crab winch, teeth of pinions are 20 and 25 and that of spur wheels are 50 and 60. Length of the handle is 0.5 metre and radius of the load drum is 0.25 metre. If efficiency of the machine is 60%, find the effort required to lift a load of 720 N.
Solution
Given: No. of teeth of pinion (T2) = 20 and (T4) = 25; No. of teeth of spur wheel (T1) = 501 and (T3) = 60; Length of the handle (l) = 0.5 m; Radius of the load drum (r) = 0.25 m; Efficiency (h) = 60% = 0.6 and load to be lifted (W) = 720 N.
 Let P = Effort required in newton to lift the load.
 We know that velocity ratio

7. Simple Pulley
A simple pulley is a wheel of metal or wood, with a groove around its circumference, to receive rope or chain. The pulley rotates freely about its axle, which passes through its centre and is perpendicular to its surface plane. This axle is supported by a metal or a wooden frame, called block as show in Fig Following assumptions are made in the study of pulley system, which are quite reasonable from the practical point of view :
Fig Simple pulley

8. First system of pulleys. Second system of pulleys.
The weight of the pulley block is small as compared to the weight to be lifted, and thus may be neglected in calculations.
The friction between the pulley surface and the string is negligible, and thus the tension in the two sides of the rope, passing round the pulley, may be taken to be equal.
A little consideration will show, that in a simple pulley, its mechanical advantage as well as velocity ratio is 1 under the assumed conditions mentioned above. The only advantage of a simple pulley is that the effort can be applied Fig (a), (b) and (c). Simple pulleys are generally used in certain mechanical advantage and efficiency.
Though there are many typed of pulleys used by engineers, yet the following system of pulleys are commonly used :
First system of pulleys.
Second system of pulleys.
Third system of pulleys.
Fig Force applied in different directions.

9. First System of Pulleys
In Fig is shown the first system of pulleys. In this system, the pulleys are so arranged that there are as many strings as there are pulleys. The end of each string is fastened to a rigid ceiling; while the other end passing round the bottom periphery of the pulley, is fastened to the next higher pulley. The load is attached to the bottom-most pulley ; whereas the effort is applied to the far end of the string passing round the last pulley. Another pulley (no. 5) is used just to change the direction of the effort. The velocity ratio of the system may be obtained by considering a unit motion of the load. In this case, let the weight W be raised by x metres. Since the loads is supported on both sides of the string, thus this slackness of x metres will have to be taken up by the pulley 2. If the relative position of the pulley 2, with respect to the pulley 1, is to remain undisturbed, then the pulley 2 should move upwards through a distance of 2x metres.
Fig First system of pulleys

10. Now this upward movement of pulley 2 through a distance of 2x metres will cause a total slackness of 2 × (2x) = 22x metres in the string, which has to be taken up by the pulley 3. Thus the pulley 3 should move upwards through a distance of 22x metres, thus causing a slackness of 2 × (22x) = 23x in the string passing round the pulley 4. Thus the pulley 4 should move upwards through a distance of 23x metres causing a slackness of 2 × (23x) = 24x meters, which must be taken up by the free end of the string to which the effort is applied. Thus the effort must move through a distanece of 24x metres.

11. Example
In a system of pulleys of the first type, there are three pulleys, and a weight of 320 N can be lifted by an effort of 50 N. Find the efficiency of the machine and the amount of friction.
Solution
Given : No. of pulleys (n) = 3 ; Weight lifted (W) = 320 N and effort (P) = 50 N.
Efficeincy of the machine
We know that velocity ratio of first system of pulleys.
V.R. = 2n = 23 = 8

12. Second System of Pulleys
In Fig (a) and (b) is shown second system of pulleys containing two blocks, one upper and the other lower, both carrying either equal number of pulleys or the upper block may have one pulley more than the lower one.
(a) Different no. of pulleys (b) Same no. of pulleys
Fig Second system of pulleys.

13. In both the cases, the upper block is fixed and the lower one is movable. There is obly one string, which passes round all the pulleys one end of which is fixed to the upper block (when both the blocks have the same no. of pulleys) or to the lower block (when the upper block has one pulley more than the lower one). The other end of the string is free and the effort is applied to this free end as shown in Fig (a) and (b). In both the cases the load is attached to lower block.
A little consideration will show, that for x displacement of the weight, the effort will move through a distance to nx, where n is the number of pulleys in both the blocks. Thus velocity ratio

14. Example
A weight of 1 kN is lifted by an effort of 125 N by second system of pulleys, having 5 pulleys in each block. Calculate the amount of effort wasted in friction and the frictional load.
Solution
Given: Weight lifted (W) = 1 kN = 1000 N ; Effort (P) = 125 N and no. of pulleys (n) = 2 × 5 = 10.
Amount of effort wasted in friction
 We know that velocity ratio
 V.R. = n = 10
and amount of effort wasted in friction,

15. Third System of Pulleys
In Fig , is shown a third system of pulleys. In this system, like the first system of pulleys, the pulleys are arranged in such a way that there are as many strings as there are pulleys. One end of each string is fixed to a block B –B, to which the load is attached. The other end of each string, passing round the upper periphery of the pulley, is fastened to the next lower pulley as shown in The velocity ratio, of the system, may be obtained by considering a unit motion of the load. In this case, let the weight W be raised by x metres. Since the llad is supported on all the strings, therefore all the strings will be slackened by x metres. Now consider the pulley 1, which is fixed to the ceiling. The slackness of string s1 equal to x metres will have to be taken up by the pulley 2, which should come down through a distance of 2x metres. But as the string s2 also slacks by x metres, therefore the string s1 will be pulled through a distance of (2x – x) = x metre.
Fig Third system of pulleys.

16. Now consider the pulley 2
Now consider the pulley 2. As the string s1 has been pulled through a distance x metres, therefore the string s2 will be pulled through a distance of 2x + x = 3x = (22 – 1)x. Similarly, in order to keep the relative position of the pulley 3 undisturbed, the string s3 will be pulled through a distance of (2 × 3x + x) = 7x = (23 – 1) x and the string s4 i.e., effort will be pulled through a distance of (2 × 7x + x) = 15x = (24 – 1) x.

17. Example
In a third system of pulleys, there are 4 pulleys. Find the effort required to lift a load of 1800 N, if efficiency of the machine is 75%. Calculate the amount of effort wasted in friction.
Solution
Given: No. of pulleys (n) = 4 ; Load lifted (W) = 1800 N and efficiency (h) = 75% = 0.75.
 Effort required to lift the load
Let P = Effort required in newton to lift the load. We know that velocity ratio of third system of pulleys.
  V.R. = 2n – 1 = 24 – 1 = 15

18. Simple Screw Jack
It consists of a screw, fitted in a nut, which forms the body of the jack. The principle, on which a screw jack works, is similar to that of an inclined plane. Fig shows a simple screw jack, which is rotated by the application of an effort at the end of the lever, for lifting the load. Now consider a single threaded simple screw jack. Let l = Length of the effort arm, p = Pitch of the screw, W = Load lifted, and P = Effort applied to lift the load at the end of teh lever. We know that distance moved by the effort in one revolution of screw, = 2pl ...(i)
Fig Simple screw jack.

19. Example Solution and distance moved by the load = p ...(ii)
A screw jack has a thread of 10 mm pitch. What effort applied at the end of a handle 400 mm long will be required to lift a load of 2 kN, if the efficiency at this load is 45%.
Solution
Given: Pitch of thread (p) = 10 mm; Length of the handle (l) = 400 mm; Load lifted (W) = 2 kN = 2000 N and efficiency (n) = 45% = 0.45.
 Let P = Effort required to lift the load.

20. Differential Screw Jack
It is an improved form of a simple screw Jack in which the velocity ratio is intensified with the help of a differential screw. In Fig is shown a jack, with a differential screw. The principle on which this machine works, is the same as that of any other differential machine, i.e., action of one part of the machine is subtracted from the action of another part. In this machine, the differential screw is in two parts, A and B. Part A is threaded both on inside and outside ; whereas the part B is threaded on the outside only. The external threads of A gear with the threads of the nut C, which forms the body of the differential screw jack. The internal threads of A gear with the external threads of the screw B. Thus the part A behaves as a screw for the nut C and as a nut for the screw B.

21. Fig. 11.15. Differential screw jack
The screw B does not rotate, but moves in vertical direction only, and carries the load. When the effort is applied at the lever, the screw A rises up and simultaneously the screw B goes down. Thus the lift of the load is algebraic sum of the motions of the screw A and screw B. Let p1 = Pitch of screw A, p2 = Pitch of the screw B, l = Length of the lever arm, W = Load lifted, and P = Effort applied, at the end of the lever to lift the load. We know that distance moved by the effort in one revoluton of the lever arm, = 2pl ...(i)

22. Example \Upward distance moved by the screw A = p1
 and downward distance moved by the screw B
= p2
 Distance through which the load is lifted
= p1 – p (ii)
Example
In a differential screw jack, the screw threads have pitch of 10 mm and 7 mm. If the efficiency of the machine is 28%, find the effort required at the end of an arm 360 mm long to lift a load of 5 kN.

23. Solution
Given: Pitch of the screw jack (p1) = 10 mm and (p2) = 7 mm ; Efficiency (h) = 28% = 0.28; Arm length of screw jack (l) = 360 mm and load lifted (W) = 5 kN = 5000 N.
Let P = Effort required to lift the load. We know that velocity ratio of a differential screw Jack.

24. Worm Geared Screw Jack
It is a further improved form of differential screw jack, in which the velocity ratio is further intensified with the help of a geared screw jack. In Fig is shown a jack with worm geared screw. It is also an improved form of screw jack, in which the screw is lifted with the help of worm and worm wheel, instead of effort at the end of a lever. Now consider a worm geared screw jack. Let l = Radius of the effort wheel (or length of the handle). p = Pitch of the screw, W = Load lifted, P = Effort applied to lift the load, and T = No. of teeth on the worm wheel.
Fig Worm geared screw jack.

25. We know that distance moved by effort in one revolution of wheel (or handle) = 2pl ...(i) If the worm is single threaded (i.e. for one revolution of the wheel A, the screw S pushes the worm wheel through one teeth) then the worm wheel move through 1/T revolution. Therefore distance moved by the load

26. Example
A worm geared screw jack has the following particulars : Length of handle = 300 mm No. of teeth in the worm wheel = 50 Pitch of screw = 10 mm Effort applied= 100 N Load lifted= 100 kN If the worm is double threaded, find the efficiency of the jack.
Solution
Given: Length of handle (l) = 300 mm ; No. of teeth in the worm wheel (T) = 50 ; Pitch of screw (p) = 10 mm ; Effort (P) = 100 N ; Load lifted (W) = 100 kN = N and no. of threads (n) = 2.
 We know that velocity ratio of a worm geared screw Jack,