Order of Pulley

First Order Pulley: VR = 2ⁿ where n is number of movable pulleys, only one pulley is fixed.

Second Order Pulley: Three fixed and three movable pulleys are in the system. Single string passes over all the pulleys. VR = n where n is total number of pulleys.

Third Order Pulley: One pulley is fixed; one end of each string is connected to load. VR = 2ⁿ – 1 where n is total number of pulleys.

All the three orders of pulleys are shown in Figure 13.8.

Figure 13.8 (a) First Order Pulley, (b) Second Order Pulley, and (c) Third Order Pulley

Example 13.6: Four pulleys are arranged in first order in which one pulley is fixed. If the efficiency of the system is 95%, find the effort required to raise a load of 5,000 N. Also, find the friction of the machine.

Solution:

Number of movable pulleys, n = 3

VR for first order pulleys = 2ⁿ = 2³ = 8

Example 13.7: Six pulleys are arranged in second order, three pulleys in the upper block and three pulleys in lower block. It was found that a load of 1,200 N was raised by effort of 300 N. Calculate the efficiency and effort lost by friction in the machine.

Solution:

Ideal effort required to raise a load of 1,200 N, 

     Effort lost in friction = 300 – 200 = 100 N

Example 13.8: Four pulleys arranged in third order. A load of 1,500 N was raised by an effort of 300 N. Calculate the efficiency and effort lost in friction of the machine.

Solution:

VR for second order pulleys = 2ⁿ – 1 = 2⁴ – 1 = 15

Ideal effort required to raise a load of 1,200 N, 

    Effort lost in friction = 300 – 100 = 200 N