A 100 kg weight is suspended from the centre of a centre of a rope . In equilibrium the two halves of the rope subtend an angle of 120 degree with each other. Find the tension in the rope.
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A 100 kg weight is suspended from the centre of a centre of a rope . In equilibrium the two halves of the rope subtend an angle of 120 degree with each other. Find the tension in the rope.
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Consider the first attachment as the given figure.
Now,
Let's make the Free body diagram of the given condition .
Refer to attachment 2 for the free body diagram.
As, the given condition is equilibrium so,
So,
The Tension in the string/rope would be 1000 Newtons.
_____________________________
Explanation:
Mass of the body (m) = 100 kg.
The two halves of the rope subtend 120° angle.
The given condition is in equilibrium.
Find:-}}}}
†ToFind:−
The tension in the rope.
Solution:-}}}}
†Solution-
Consider the first attachment as the given figure.
Now,
Let the tension in one half of the string be T.
Let's make the Free body diagram of the given condition .
Refer to attachment 2 for the free body diagram.
As, the given condition is equilibrium so,
\begin{gathered}\colon\implies \sf \; mg = T\:cos\:60\degree+T\:cos\:60\degree\\\\\colon\implies\sf\;100\times\cancel{10}=\cancel2T\:cos\:60\degree\\\\\colon\implies\sf\;100\times5=T\times \dfrac{1}{2}\\\\\colon\implies\sf\;T=500\times2\\\\\co
:⟹mg=Tcos60°+Tcos60°
:⟹100×
10
=
2
Tcos60°
:⟹100×5=T×
2
1
:⟹T=500×2
:→
T=1000N
So,
The Tension in the string/rope would be 1000 Newtons.