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Standard XII

Physics

Tension in a String

A mass of 6 k...

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 mid-point P of the rope. as shown. What is the angle the rope makes with the vertical in equilibrium? $(T a k e g = 10 m s^{- 2})$ . Neglect the mass of the rope.

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Solution

For horizontal equilibrium-

$T_{3} = T_{1} sin θ$

For vertical equilibrium-

$T_{2} = T_{1} cos θ$

Taking ratios-

$\frac{T_{3}}{T_{2}} = tan θ$

And for equilibrium of body $T_{2} = 6 k g w t = 6 \times 10 = 60 N$

$⟹ tan θ = \frac{5}{6}$

or $θ = {tan}^{- 1} \frac{5}{6}$

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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 mid-point P of the rope. as shown. What is the angle the rope makes with the vertical in equilibrium? (Takeg=10ms−2). Neglect the mass of the rope.

For horizontal equilibrium-T3=T1sinθFor vertical equilibrium-T2=T1cosθTaking ratios-T3T2=tanθAnd for equilibrium of body T2=6kgwt=6×10=60N⟹tanθ=56 or θ=tan−156

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 mid-point P of the rope. as shown. What is the angle the rope makes with the vertical in equilibrium? $(T a k e g = 10 m s^{- 2})$ . Neglect the mass of the rope.

For horizontal equilibrium-

$T_{3} = T_{1} sin θ$

For vertical equilibrium-

$T_{2} = T_{1} cos θ$

Taking ratios-

$\frac{T_{3}}{T_{2}} = tan θ$

And for equilibrium of body $T_{2} = 6 k g w t = 6 \times 10 = 60 N$

$⟹ tan θ = \frac{5}{6}$

or $θ = {tan}^{- 1} \frac{5}{6}$