Question 793872
The 3 feet of pipe to one side of the fulcrum balance the 3ft of pipe to the other side.
The weight (x kg) put on the small end, has to balance the 18 feet of pipe extending from 3 to 21 feet away from the fulcrum.
That 18-foot stretch of pipe is 18/24=3/4 of the entire pipe and weighs
(3/4)(20kg)=15kg
The weight could be considered applied at the midpoint, (3ft+21ft)/2=12ft from the fulcrum.
The torque produced by that weight and the torque produced by the balancing weight on the short end (at 3 ft from the fulcrum) should have the same magnitude, so
(3ft)(x kg) = (12ft)(15kg) --> x kg = 15kg (12ft/3ft = 4(15kg) = 60kg
You need a 60 kg weight applied on the small end to balance the pipe.