In the figure, at the free end of the light string, a force F is supplied to keep the suspended mass of 18 kg at rest. Then the force exerted by the ceiling on the system (assume that the string segments are vertical and the pulleys are light and smooth) is:(`g=10(m)/(s^2)`)

Answer

240

Explanation

FBD of 18 kg. block
`implies T = 180 N`. (as it is at rest). For the lower pulley:
Clearly : `3T_(1) = T implies T_(1) = 60 N`
and for the upper pulley:
`T_(2) = 4T_(1) = 240 N`
where `T_(2)` is the force exerted by ceiling on the pulley.