Question

a body of weight √3N suspend vertically using a rope is pulled horizontally such that rope makes an angle300 to the vertical.then the tension in that rope is​

Answer 1

Answer:

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Given,

m = 0.03 kg

Using the law of conservation of mechanical energy,

Work done by the horizontal force = Gain in P. E. of mass m

or, F × 1 sin 30° = mg × ( 1 - 1 cos 30°)

or, F = mg × (1 -1 cos 30°)/1 sin 30°

= 0.03 g × (1 - √3/2)/(1/2)

= 0.03 g × (2 - √3/2)/(1/2)

= 0.03 g × (2 - √3)

= 0.03 g × 9.8 × (2 - √3)

= 0.079 N Answer

See diagram.

Let the tension force in the string be T.  Let the force required be F.   The static equilibrium of pendulum (mass) at an angle Ф requires:

 F = T SinФ                     W =  m g = T CosФ       =>  F = m g Tan Ф

As Ф increases, TanФ increases and so Force also increases.  So the force required maximum at Ф = 30°.

        F = 0.03 kg * Tan 30° * g   = √3/100  kg wt    or √3 g /100  Newtons

===============

If we want the average force along the horizontal direction (x) to move the mass to 30 deg., then we find that by integration.





Answer 2

The tension in the string is 2N

GIVEN

Weight of body = √3 N

Angel made after pulling the rope = 30°

TO FIND

The tension in the rope.

SOLUTION

We can simply solve the above problem as follows-

Let T be the tension on the rope.

So,

Horizontal component of T = Tcos30°

Vertical component of T = Tsin30°

Weight of body(acting downward) = √3 N

Since, The system is in equilibrium,

Tcos30° = √3

We know that,

Cos30° = √3/2

So,

$T \frac{ \sqrt{3} }{2} = \sqrt{3}$

$T = \frac{ \sqrt{3} \times 2 }{ \sqrt{3} }$

T = 2

Hence, The tension in the string is 2N

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