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)

Question

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) (A) 90 N (B) 75 N (C) 70 N (D) 100 N

Tardigrade Admin · Accepted Answer

For equilibrium, T sin 45° = F ....(1) and T cos 45° = 10g ....(2) equation (1)/(2) we get F = 10g = 100 N

Q. A mass of $10 k g$ 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 $4 5^{°}$ with the vertical. Then $F$ equals: (Take $g = 10 m s^{- 2}$ and the rope to be massless)

$90 N$

$75 N$

$70 N$

$100 N$

For equilibrium,
$T s in 4 5^{\circ} = F .... (1)$
and $T cos 4 5^{\circ} = 10 g .... (2)$
equation $(1) / (2)$
we get $F = 10 g$
$= 100 N$

Q. A mass of 10kg is suspended by a rope of length 4m, 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=10ms−2 and the rope to be massless)

90N

75N

70N

100N

For equilibrium, Tsin45∘=F....(1) and Tcos45∘=10g....(2) equation (1)/(2) we get F=10g =100N

$90 N$

$75 N$

$70 N$

$100 N$

For equilibrium,
$T s in 4 5^{\circ} = F .... (1)$
and $T cos 4 5^{\circ} = 10 g .... (2)$
equation $(1) / (2)$
we get $F = 10 g$
$= 100 N$