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A mass of 6 kg is suspended by a rope of...

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

Text Solution

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Figure ( b) show force diagram of where as (c ) shows FBD of point P .
for equilbrium of W.
`T_(2) = 60 N`
At point P there are two component of `T_(1)` vertical component `T_(1) cos theta` which balances `T_(2)` and horizontal component `T_(1) sin theta ` which balances `50 N`
`tan theta = 0.833 ` or `theta = tan^(-1) ((5)/(6))`
`theta = 39^(@) 448`
` =39 .8^(@) = 40^(@)`
We see that result do not depend on length of rope of position of horizontal force .
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