Q: A mass of 10 kg is suspended by a rope of length 4 m , from the ceiling . A force F is applied horizontally at the mid point of the rope such that top half of the rope makes an angle of 45° with the vertical . Then F equals (Take g = 10 m/s2 and rope to be massless )
(a) 75 N
(b) 100 N
(c) 90 N
(d) 70 N
Click to See Answer :
Ans: (b)
Sol: 
Here AB = 2 m & BC = 2 m
Resolving Components ,
T cos45° = m g …(i)
T sin45° = F …(ii)
On dividing (ii) by (i)
$\large tan45^o = \frac{F}{mg}$
$\large 1 = \frac{F}{mg}$
F = m g = 10 × 10 = 100 N