Question

A mass of 4 kg is suspended by a rope of a length 4 m. From a celing . A Force of 20N in the horizontal direction is applied at the mid point of the rope as show in the figure . What is the angle which the rope makes the vertical in equilibrium ?
{ neglate mass of the rope , Take g = 10 ms-²}​

Answer 1

Answer:

For horizontal equilibrium-

T

3

=T

1

sinθ

For vertical equilibrium-

T

2

=T

1

cosθ

Taking ratios-

T

2

T

3

=tanθ

And for equilibrium of body T

2

=6kgwt=6×10=60N

⟹tanθ=

6

5

or θ=tan

−1

6

5

Answer 2

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the concept of Force has been used. We know that Force is the product of Tension and displacement in a certain direction. Here the condition of equilibrium we are given. Firstly we can find the horizontal equilibrium and then vertical equilibrium and then take their ratio to find the value of θ.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{\pink{\tan\theta\;=\;\bf{\dfrac{\sin\theta}{\cos\theta}}}}}$

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★ Solution :-

Given,

» Mass of the body = 4 Kg

» Length of rope = 4 m

» Force applied on midpoint of rope = 20 N

» Acceleration due to gravity = 10 m/sec²

____________________________________________

~ For forces acting on rope ::

We see that we are asked for equilibrium condition in the question.

This means the force applied on the body in horizontal direction will be equal to the product of Tension ans sin θ which is displacement.

So, for horizontal equilibrium :-

$\\\;\bf{\mapsto\;\;T_{3}\;=\;T_{1}\sin\theta}$

• We already know the value of T₃. This is because in figure we see that T₃ is force applied in horizontal direction so,

→ T₃ = 20 N

By applying this value, we get

$\\\;\bf{\mapsto\;\;T_{1}\sin\theta\;=\;\red{20\;\;N}}$

• From figure we can also see that, in equilibrium position the product of Tension and cos θ which is displacement will be equal to the weight of the substance which is suspended.

Also,

→ Weight = Mass × Acceleration due to gravity

→ 4 Kg Weight = 4 × 10 N

→ 4 Kg Weight = 40 N

So, for vertical equilibrium :-

$\\\;\bf{\mapsto\;\;T_{2}\;=\;T_{1}\cos\theta}$

• We know that T₂ is the weight applied to the rope. And in figure T₂ makes equilibrium with T sin θ which is in vertical direction,

→ T₂ = 40 N

By applying this value, we get

$\\\;\bf{\mapsto\;\;T_{2}\cos\theta\;=\;\blue{40\;\;N}}$

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~ For the required angle ::

Let the angle which rope makes with the vertical in equilibrium be θ . Then tan θ will be ratio of sin θ and cos θ.

We already know that,

$\\\;\sf{\rightarrow\;\;\tan\theta\;=\;\bf{\dfrac{\sin\theta}{\cos\theta}}}$

Now if we multiply both numerator and denominator by T₁ then values will bw same only of the fraction because a same number is being multiplied at numerator and denominator. So,

$\\\;\sf{\rightarrow\;\;\tan\theta\;=\;\bf{\dfrac{\sin\theta}{\cos\theta}\;\times\;\dfrac{T_{1}}{T_{1}}}}$

$\\\;\sf{\rightarrow\;\;\tan\theta\;=\;\bf{\dfrac{T_{1}\sin\theta}{T_{1}\cos\theta}}}$

From above we get,

$\\\;\sf{\rightarrow\;\;\tan\theta\;=\;\bf{\dfrac{T_{3}}{T_{2}}}}$

$\\\;\sf{\rightarrow\;\;\tan\theta\;=\;\bf{\dfrac{20\;\:N}{40\;\:N}}}$

$\\\;\sf{\rightarrow\;\;\tan\theta\;=\;\bf{\dfrac{20}{40}}}$

$\\\;\sf{\rightarrow\;\;\tan\theta\;=\;\bf{\dfrac{1}{2}}}$

$\;\\\sf{\rightarrow\;\;\theta\;=\;\bf{\green{\tan^{-1}\;\dfrac{1}{2}}}}$

Now if we simplify more, then

$\;\\\sf{\rightarrow\;\;\theta\;=\;\bf{\orange{26.5651^{\circ}}}}$

This is the required answer.

$\\\;\underline{\boxed{\tt{Hence,\;\;required\;\;angle\;\:\theta\;=\;\bf{\purple{\tan^{-1}\;\dfrac{1}{2}}}}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;Force\;=\;Mass\;\times\;Acceleration}$

$\\\;\sf{\leadsto\;\;Tension\;=\;\dfrac{Force}{Length}}$

$\\\;\sf{\leadsto\;\;Surface\;Energy\;=\;\dfrac{Work}{Area}}$

$\\\;\sf{\leadsto\;\;Stress\;=\;\dfrac{Force}{Area}}$

$\\\;\sf{\leadsto\;\;Strain\;=\;\dfrac{Change\;in\;Dimension}{Original\;Dimension}}$