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Byju's Answer

Standard VIII

Physics

Free Body Diagram

A mass of 6...

Question

A mass of $6$ $k g$ is suspended by a rope of length $2$ $m$ from the ceiling. A force of $50$ $N$ horizontally is applied at the mid point $P$ of the rope. Calculate the angle of rope makes with the vertical. Neglect the mass of rope. $(g = 9.8$ $m s^{- 2})$

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Solution

$T cos θ = 60$
$T sin θ = 50$
$t a n θ = \frac{5}{6}$
$θ = t a n^{- 1} (\frac{5}{6})$

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Similar questions

Q. 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.

Q. A mass of

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is suspended by a rope of length

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equals:
(Take

g = 10 {ms}^{- 2}

and the rope to be massless)

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(Take

g = 10 m s^{- 2}

, neglect the mass of the rope)

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

Q. A block of mass

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is suspended by a massless rope of length

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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 horizontally is applied at the mid point P of the rope. Calculate the angle of rope makes with the vertical. Neglect the mass of rope. (g=9.8 ms−2)

Tcosθ=60Tsinθ=50tanθ=56θ=tan−1(56)

A mass of $6$ $k g$ is suspended by a rope of length $2$ $m$ from the ceiling. A force of $50$ $N$ horizontally is applied at the mid point $P$ of the rope. Calculate the angle of rope makes with the vertical. Neglect the mass of rope. $(g = 9.8$ $m s^{- 2})$

$T cos θ = 60$
$T sin θ = 50$
$t a n θ = \frac{5}{6}$
$θ = t a n^{- 1} (\frac{5}{6})$