Question

the system shown in the figure is in equilibrium.find the tension in each string:T1,T2,T3,T4,TT5



those who answers correctly I will mark as brainliest

Answer 1

Consider the point A.

The tensions $\sf{T_1}$ and $\sf{T_2}$ each make an angle $\sf{30^o}$ with the vertical since the triangle ABC is assumed to be equilateral.

We can so write these tensions in vector form as,

$\longrightarrow \vec{\sf{T_1}}=\sf{T_1\left(\sin30^o\ \left(-\hat i\right)+\cos30^o\ \hat j\right)}$

$\longrightarrow \vec{\sf{T_1}}=\sf{T_1\left(-\dfrac{1}{2}\ \hat i+\dfrac{\sqrt3}{2}\ \hat j\right)}$

$\longrightarrow \vec{\sf{T_1}}=\sf{-\dfrac{T_1}{2}\ \hat i+\dfrac{T_1\sqrt3}{2}\ \hat j}$

and,

$\longrightarrow \vec{\sf{T_2}}=\sf{T_2\left(\sin30^o\ \hat i+\cos30^o\ \hat j\right)}$

$\longrightarrow \vec{\sf{T_2}}=\sf{T_2\left(\dfrac{1}{2}\ \hat i+\dfrac{\sqrt3}{2}\ \hat j\right)}$

$\longrightarrow \vec{\sf{T_2}}=\sf{\dfrac{T_2}{2}\ \hat i+\dfrac{T_2\sqrt3}{2}\ \hat j}$

The weight of the block is acting downwards so,

$\longrightarrow \vec{\sf{W}}=\sf{-200\ \hat j\ N}$

As the system is in equilibrium, the net force at A can be equated to zero.

$\longrightarrow\vec{\sf{T_1}}+\vec{\sf{T_2}}+\vec{\sf{W}}=0$

$\longrightarrow\sf{-\dfrac{T_1}{2}\ \hat i+\dfrac{T_1\sqrt3}{2}\ \hat j+\dfrac{T_2}{2}\ \hat i+\dfrac{T_2\sqrt3}{2}\ \hat j-200\ \hat j=0}$

$\longrightarrow\sf{-\left(\dfrac{T_1-T_2}{2}\right)\ \hat i+\left(\dfrac{(T_1+T_2)\sqrt3}{2}-200\right)\ \hat j=0}$

As each component would be equal to zero, we get,

$\sf{\longrightarrow T_1=T_2}$

$\sf{\longrightarrow T_1+T_2=\dfrac{400}{\sqrt3}}$

On solving them we get,

$\sf{\longrightarrow\underline{\underline{T_1=T_2=\dfrac{200}{\sqrt3}\ N}}}$

Consider the point B.

Here the tension $\sf{T_1}$ is in opposite direction to that at point A.

So when we write it in vector form the sign of the components will be changed. Thus,

$\longrightarrow \vec{\sf{T_1}}=\sf{\dfrac{T_1}{2}\ \hat i-\dfrac{T_1\sqrt3}{2}\ \hat j}$

Putting value of $\sf{T_1,}$

$\longrightarrow \vec{\sf{T_1}}=\sf{\dfrac{100}{\sqrt3}\ \hat i-100\ \hat j}$

The tension $\sf{T_3}$ is acting horizontally rightwards.

$\longrightarrow \vec{\sf{T_3}}=\sf{T_3\ \hat i}$

The tension $\sf{T_4}$ is making an angle $\sf{60^o}$ with the vertical.

$\longrightarrow \vec{\sf{T_4}}=\sf{T_4\left(\sin60^o\ \left(-\hat i\right)+\cos60^o\ \hat j\right)}$

$\longrightarrow \vec{\sf{T_4}}=\sf{T_4\left(-\dfrac{\sqrt3}{2}\ \hat i+\dfrac{1}{2}\ \hat j\right)}$

$\longrightarrow \vec{\sf{T_4}}=\sf{-\dfrac{T_4\sqrt3}{2}\ \hat i+\dfrac{T_4}{2}\ \hat j}$

As the system is in equilibrium,

$\longrightarrow\vec{\sf{T_1}}+\vec{\sf{T_3}}+\vec{\sf{T_4}}=\sf{0}$

$\longrightarrow\sf{\dfrac{100}{\sqrt3}\ \hat i-100\ \hat j+T_3\ \hat i-\dfrac{T_4\sqrt3}{2}\ \hat i+\dfrac{T_4}{2}\ \hat j=0}$

$\longrightarrow\sf{\left(\dfrac{100}{\sqrt3}+T_3-\dfrac{T_4\sqrt3}{2}\right)\ \hat i+\left(\dfrac{T_4}{2}-100\right)\ \hat j=0}$

Equating each component to zero and on solving we get,

$\sf{\longrightarrow \underline{\underline{T_3=\dfrac{200}{\sqrt3}\ N}}}$

$\sf{\longrightarrow \underline{\underline{T_4=200\ N}}}$

Consider the point C.

The tension $\sf{T_2}$ is in opposite direction to that at A. Thus,

$\longrightarrow \vec{\sf{T_2}}=\sf{-\dfrac{T_2}{2}\ \hat i-\dfrac{T_2\sqrt3}{2}\ \hat j}$

$\longrightarrow \vec{\sf{T_2}}=\sf{-\dfrac{100}{\sqrt3}\ \hat i-100\ \hat j}$

Also the tension $\sf{T_3}$ is acting horizontally leftwards.

$\longrightarrow \vec{\sf{T_3}}=\sf{-T_3\ \hat i}$

$\longrightarrow \vec{\sf{T_3}}=\sf{-\dfrac{200}{\sqrt3}\ \hat i}$

The tension $\sf{T_5}$ makes an angle $\sf{60^o}$ with the vertical.

$\longrightarrow\vec{\sf{T_5}}=\sf{T_5(\sin60^o\ \hat i+\cos60^o\ \hat j)}$

$\longrightarrow\vec{\sf{T_5}}=\sf{T_5\left(\dfrac{\sqrt3}{2}\ \hat i+\dfrac{1}{2}\ \hat j\right)}$

$\longrightarrow\vec{\sf{T_5}}=\sf{\dfrac{T_5\sqrt3}{2}\ \hat i+\dfrac{T_5}{2}\ \hat j}$

As the system is in equilibrium,

$\longrightarrow\vec{\sf{T_2}}+\vec{\sf{T_3}}+\vec{\sf{T_5}}=\sf{0}$

$\longrightarrow\sf{-\dfrac{100}{\sqrt3}\ \hat i-100\ \hat j-\dfrac{200}{\sqrt3}\ \hat i+\dfrac{T_5\sqrt3}{2}\ \hat i+\dfrac{T_5}{2}\ \hat j=0}$

$\longrightarrow\sf{\left(\dfrac{T_5\sqrt3}{2}-100\sqrt3\right)\ \hat i+\left(\dfrac{T_5}{2}-100\right)\ \hat j=0}$

On equating each component to zero we get,

$\sf{\longrightarrow\underline{\underline{T_5=200\ N}}}$

Answer 2

Explanation:

u only done that question right, so if ur process is right then the answer is also true, so plz keep ur process..