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 ….

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: mcq

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