Question

In a first order system of pulleys there are three movable pulleys. What is the effort required to raise a load of 6000 N ? Assume efficiency of the system to be 80%. 

If the same load is to be raised using 520 N, find the number of movable pulleys that are necessary. 

Assume a reduction of efficiency of 5% for each additional pulley used in the system.

Answer 1

VR = 2ⁿ , where n is the number of movable pulleys. 

VR = 23 = 8 

Now, MA = η × VR 

= 0.8 × 8 

= 6.4

\(\frac WP\) = 6.4

P = \(\frac W{6.4}\) = \(\frac {6000}{6.4}\)

P = 937.5 N 

In the second case, 

Effort = 520 N

Efficiency η = 0.80 – n₁ × 0.05 

where n₁ = number of additional pulleys required and equal to (n – 3). 

MA = η × VR

\(\frac WP\) = η × VR

W = P × η × 2ⁿ 

= P(0.8 – n1 × 0.05) × 2ⁿ 

= P[0.8 – (n – 3) × 0.05] 2ⁿ

By going for a trial and error solution, starting with one additional pulley i.e., totally with four pulleys, 

W = 520 [0.8 – (4 – 3) × 0.05] 2⁴ = 6240 N 

i.e., if four pulleys are used, a load of 6240 N can be raised with the help of 520 N effort.

∴ Number of movable pulleys required = 4