First slide

Equilibrium of a particle

Question

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

Easy

A

100N
B

90N
C

75N
D

70N

Solution

$T \cos 45^{o} = m g \Rightarrow \frac{T}{\sqrt{2}} = 100 . . . . . . (1)$
$T \sin 45^{o} = F \Rightarrow \frac{T}{\sqrt{2}} = F \Rightarrow F = 100 N [f r o m e q u a t i o n (1)]$

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