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See Fig 5.8 .A mass of 6 kg is suspended...

See Fig 5.8 .A mass of 6 kg is suspended by a rope of length 2m from the ceiling . A force of 50 N in the horizontal direction is applied at the mid-point P of the rope, as shown. What is the angle the rope makes with the vertical in equilibrium ? (Take g = 10 m `s^(-2)`). Neglect the mass of the rope.

Text Solution

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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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