A mass of 6 kg is suspended by a rope of length 2 m from the ceiling. A force of 50 N in the horizontal direction is applied at the midpoint P of the rope.What is the angle the rope makes with the vertical in equilibrium?

Question

A mass of 6 kg is suspended by a rope of length 2 m from the ceiling. A force of 50 N in the horizontal direction is applied at the midpoint P of the rope.What is the angle the rope makes with the vertical in equilibrium? <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/d4dacb37b7e2dabd47b23f978c92e625-.png" /> (A)  sin -1((3/5)) (B)  sin -1((5/6)) (C)  tan -1((4/5)) (D)  tan -1((5/6))

Tardigrade Admin · Accepted Answer

At equilibrium, T1 sin θ=50 N T1 cos θ=T2 ⇒ tan θ=(50 N /T2)=(50 N /60 N ) or θ= tan -1((5/6)) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/a2c631c0b5da8c84c58156e8e3100fe5-.png" />

Q. A mass of $6 k g$ is suspended by a rope of length $2 m$ from the ceiling. A force of $50 N$ in the horizontal direction is applied at the midpoint $P$ of the rope.What is the angle the rope makes with the vertical in equilibrium?

$sin^{- 1} (\frac{3}{5})$

$sin^{- 1} (\frac{5}{6})$

$tan^{- 1} (\frac{4}{5})$

$tan^{- 1} (\frac{5}{6})$

At equilibrium, $T_{1} sin θ = 50 N$
$T_{1} cos θ = T_{2}$
$\Rightarrow tan θ = \frac{50 N}{T _{2}} = \frac{50 N}{60 N}$
or $θ = tan^{- 1} (\frac{5}{6})$

Q. A mass of 6kg is suspended by a rope of length 2m from the ceiling. A force of 50N in the horizontal direction is applied at the midpoint P of the rope.What is the angle the rope makes with the vertical in equilibrium?

sin−1(53​)

sin−1(65​)

tan−1(54​)

tan−1(65​)