1. Tardigrade
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  3. Physics
  4. A mass of 10 kg is suspended by a rope of length 4 m, from the ceiling. A force F is applied horizontal!)' at the mid-point of the rope such that the top half of the rope makes an angle of 45° with the vertical. Then F equals: (Take g = 10 ms-2 and the rope to be massless)

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Q. A mass of $10\, kg$ is suspended by a rope of length $4\, m$, from the ceiling. A force $F$ is applied horizontal!)' at the mid-point of the rope such that the top half of the rope makes an angle of $45^°$ with the vertical. Then $F$ equals: (Take $g = 10\, ms^{-2}$ and the rope to be massless)

JEE MainJEE Main 2020Laws of Motion

Solution:

For equilibrium,
$T \,sin \,45^{\circ} = F \quad....\left(1\right)$
and $T \,cos \,45^{\circ } = 10g \quad ....\left(2\right)$
equation $\left(1\right)/\left(2\right)$
we get $F = 10g$
$= 100 \,N$

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