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A mass of 3kg is suspended by a rope of length 2m from the ceiling. A force of 40N in the' horizontal direction is applied at midpoint P of the rope 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) |
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Answer» Solution :In equilibrium `T_(2) =W = 30 N` Resolving the TENSION `T_(1)`into TWO mutually perpendicular COMPONENTS, we have `T_(1) cos_(x) = T_(2) = 30 N rArr T_(1)sin theta = 40 N` `tan theta = 4/3 rArr theta = 53^(@)` The tension in part of string attached to the ceiling `T_(1) = sqrt(W^(2) + F^(2)) = sqrt(30^(2) + 40^(2) N) = 50 N` |
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