Solution:

Given,

Load (L) = 2400 N,
Load distance (LD) = 50 cm,
Effort distance (ED) = 90 cm,
Effort (E) = ?,

We have, from the principle of a simple machine,

L × LD = E × ED
Or, 2400 × 50 = E × 90
Or, \(E=\frac{2400\times50}{90}\)
∴ E = 1333.33 N

Hence the required effort is 1333.33 N.