Question

A mass of `3kg` is suspended by a rope of length `2m` from the ceilling A force of `40N` in the horizontal direction is applied atmidpoint `P` of the rope the as shown. What is the angle the rope makes with the vertical in equilibrium and the tension in part of string attached to the ceiling ? (Neglect the mass of the rope, `g =10m//s^(2))` .

Answer 1

Resolving the tension `T_(1)` into two mutually perpendicular compounds we have
`T_(1) cos 0 =W =30 N , T_(1) sin theta =40N`
`:. Tan theta = (4)/(3) (or) theta = tan^(-1) ((4)/(5)) =53^(@)`
The tension in part of string attached to the ceiling
`T_(1) = sqrt(W^(2) +F^(2)) = sqrt(30^(2) +40^(2)) =50 N`
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