Question

A rope AB, 2.7 m long is connected at A and B to two points on the same level, 2.4 m apart. A load of 136 N is suspended from a point C on the rope 90 cm from A. What load connected to a point D on the rope, 60 cm from B, will be necessary to keep the portion CD level?

Answer 1

$\frac{F_c}{l_1}=\frac{F_d}{l_2}$

$F_d=\frac{F_c\times l_2}{l1}=\frac{136\times 90}{2340} = 5.23N$