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A mass of 5 kg is suspended by a rope of...

A mass of 5 kg is suspended by a rope of length of 2 m from a ceiling. A force of 50 N in horizontal direction is applied at the mid point of rope. What is the angle which the rope makes with the vertical in equilibrium ? (`g = 10 m s^(-2)`). Neglect the mass of the rope.

Text Solution

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We are given: mass, m = 5 kg
Horizontal force at mid point of rope = 50 N
We are to calculate, angle `theta`
Consider the equilibrium of weight,
`T_(2) = 5 xx 10=50 N` ….(i)
Now, consider equilibrium of mass at point P.
`T_(1)costheta= T_(2)= 50 N`
`T_(1)sintheta= 50 N` ......(ii)
Dividing equation (ii) by (i) we get
which gives `tantheta= (50)/(50)` or `theta= tan^(-1)1= 45^(@)`
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