Two persons are holding a rope of negligible weight tightly at its ends so that it is horizontal. A

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4-1.Newton's Laws of Motion

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Two persons are holding a rope of negligible weight tightly at its ends so that it is horizontal. A $15\, kg$ weight is attached to the rope at the mid-point, which now no longer remains horizontal. The minimum tension required to completely straighten the rope is

A$15\, kg$

B$\frac{15}{2}kg$

C$5\, kg$

Dinfinitely large

Solution

The tension required to bring the string into horizontal position can be found out using the
following formulae.
$2 T \sin x=m g$
The following equation is true for body which attains equilibrium. Also, when the body will
be in equilibrium the angle formed will be $0.$
Therefore,
$T=m g / 2 \sin x,$ where $x=0$
$T=m g / 2 \sin 0,$ where the value of $\sin =0$
Then, the amount of tension required is infinite.

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Two persons are holding a rope of negligible weight tightly at its ends so that it is horizontal. A $15\, kg$ weight is attached to the rope at the mid-point, which now no longer remains horizontal. The minimum tension required to completely straighten the rope is

Solution

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