Question

A rope is hanging vertically with its two ends fixed to walls at same horizontal level (as shown). Tension at the bottom most point is 90N. If the mass per unit length of rope is 4 kgm-1, find radius of curvature (in meter) at point ‘C’ {Take}

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Answer 1

Answer:r=3

see the picture for the answer of the folowing

answer of the problem answer is 3

Answer 2

Given:

Tension (T$_L$) at the bottom-most point of the rope = 90N

The mass per unit length of the rope (μ) = 4kg/m

To Find:

The radius of curvature at point C (makes 30° angle with the horizontal)

Solution:

Let T be the tension at point C and m be the mass of the rope.

The FBD of point C is shown below in the attachment.

Since point C is in equilibrium

⇒The net force on it should be zero.

According to the diagram,

T sin θ = mg     - (1)

T cos θ = T$_L$     - (2)

Dividing equation (1) by (2) we get:

tan θ = mg / T$_L$    - (3)

Differentiating equation (3) we get:

$sec^2\theta d\theta = \frac{gdm}{T_L}$

Substituting dm = μRdθ

or $sec^2\theta d\theta = \frac{\mu gR d\theta }{T_L}$

or sec²θ = μRg / T$_L$

Taking θ = 30° and cross multiplying,

R = 90 X sec²30°  / 4 X 10

= 90 X 4 / 4 X 10 X 3

= 3m

Hence, the radius of curvature of the rope is 3m.