Question

Solve This Question. ​

Answer 1

Solution :

Let us draw a free body diagram of above system :

From the above figure PQW is 90°

PQY = 90° & $YQT_2\: = \: QT_{2}Y\: = \: 30^{\circ}$

Now,

$\angle{T_{2}QW} \: =\: 90^{\circ}\: +\: 30^{\circ}$

$\angle{T_{2}QW}\: =\: 120^{\circ}$

Now, PQR

$\angle{T_{2}QT_{1}} \: =\: 360^{\circ}\: - \: (120^{\circ}\: +\: 90^{\circ})$

$\angle{T_{2}QT_{1}} \: =\: 150^{\circ}$

By applying Lamis theorem in system :

$\dfrac{T_1}{sin\:\alpha} \: =\: \dfrac{T_2}{sin\:\beta}\: =\: \dfrac{W}{sin\:\gamma}$

$T_1\: =\: Tension \:in \:string\: one$

$T_2\: =\: Tension\: in \:string\: second$

$W\: =\: Weight$

$\dfrac{T_1}{sin\:120} \: =\: \dfrac{T_2}{sin\:90}\: =\: \dfrac{70}{sin\:150}$

W = Mg = 70 kg × 10 m/sec = 70 kg.m/sec

W = 70 N

Take System 1

$\dfrac{T_1}{sin\:120}\: =\: \dfrac{7}{sin\:150}$

$\dfrac{T_1}{\dfrac{\sqrt{3}}{2}} \: =\:\dfrac{70}{\dfrac{1}{2}}$

$\dfrac{T_1}{\dfrac{\sqrt{3}}{\cancel{2}}} \: =\:\dfrac{70}{\dfrac{1}{\cancel{2}}}$

$\dfrac{T_1}{\sqrt{3}} \: =\:70$

$T_1\: =\: 70\sqrt{3}\: N$

Take system 2

$\dfrac{T_2}{sin\:90} \: =\:\dfrac{70}{sin\:150}$

$\dfrac{T_2}{1} \: =\:\dfrac{70}{\dfrac{1}{2}}$

$\dfrac{T_2}{1} \: =\:\dfrac{70}{\dfrac{1}{2}}$

$\dfrac{T_2}{1} \: =\:2\times70$

$T_2\: =\: 140\: N$

Answer 2

Answer:

• The Tension (T₁) in PQ is 70 √{3} N
• The Tension (T₂) in QR is 140 N

Given:

1. ∠ R = 30°
2. ∠ PQS = 90°
3. Let the Tension in PQ be T₁
4. Let the Tension in QR be T₂

Explanation:

$\rule{300}{1.5}$

Firstly finding the ∠ PQR and ∠ RQS,

Now, we can see that,

" ∠ R and ∠ PQR form linear pair ",

Therefore,

⇒ ∠ R + ∠ PQR = 180°

Substituting the values,

⇒ 30° + ∠ PQR = 180°

⇒ ∠ PQR = 180° - 30°

⇒ ∠ PQR = 150°

⇒ ∠ PQR = 150°

∴ We got the ∠ PQR angle.

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$\rule{300}{1.5}$

Now, we know that all angles add up to 360°

Therefore,

⇒ ∠ PQR + ∠ PQS + ∠ RQS = 360°

Substituting the values,

⇒ 150° + 90° + ∠ RQS = 360°

⇒ 240° + ∠ RQS = 360°

⇒ ∠ RQS = 360° - 240°

⇒ ∠ RQS = 120°

⇒ ∠ RQS = 120°

∴ We got the ∠ RQS angle.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From Lami's theorem's we know,

$\large \bigstar\;\underline{\boxed{\sf \dfrac{F_{1}}{\sin \alpha} = \dfrac{F_{2}}{\sin \beta} = \dfrac{F_{3}}{\sin \gamma}}}$

Here, we need to know something,

1] F₁ will be the weight of the 7 Kg block, therefore, alpha will be opposite to it i.e. 150°.

2] F₂ will be the Tension in PQ, therefore, Beta will be opposite to it i.e. 120°.

3] F₃ will be the Tension in QR, therefore, gamma will be opposite to it i.e. 90°.

Substituting the values,

$\displaystyle \longrightarrow \sf\dfrac{7\times 10}{\sin 150^{\circ}} = \dfrac{T_{1}}{\sin 120^{\circ}} = \dfrac{T_{2}}{\sin 90^{\circ}}\\\\\\\longrightarrow\sf \dfrac{70}{\sin 150^{\circ}} = \dfrac{T_{1}}{\sin 120^{\circ}} = \dfrac{T_{2}}{\sin 90^{\circ}}$

Now,

Case-1

$\displaystyle\longrightarrow\sf \dfrac{70}{\sin 150^{\circ}} = \dfrac{T_{1}}{\sin 120^{\circ}}\\\\\\\longrightarrow\sf \dfrac{70}{1/2}=\dfrac{T_{1}}{\sqrt{3}/2}\\\\\\\longrightarrow\sf \dfrac{2\times 70}{1}=\dfrac{T_{1}\times 2}{\sqrt{3}}\\\\\\\longrightarrow\sf T_{1}=\dfrac{2\times 70\times \sqrt{3}}{2}\\\\\\\longrightarrow\sf T_{1}=70\sqrt{3}\\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf T_{1}=70\sqrt{3}\;N}}}}$

∴ The Tension (T₁) in PQ is 70 √{3} N.

Case-2

$\displaystyle\longrightarrow\sf \dfrac{70}{\sin 150^{\circ}} = \dfrac{T_{2}}{\sin 90^{\circ}}\\\\\\\longrightarrow\sf \dfrac{70}{1/2}=\dfrac{T_{2}}{1}\\\\\\\longrightarrow\sf \dfrac{2\times 70}{1}=\dfrac{T_{2}}{1}\\\\\\\longrightarrow\sf T_{2}=\dfrac{2\times 70\times 1}{1}\\\\\\\longrightarrow\sf T_{2}=140\\\\\\\longrightarrow \large{\underline{\boxed{\red{\sf T_{2}=140\;N}}}}$

∴ The Tension (T₂) in QR is 140 N.

Refer the attachment for diagram.

$\rule{300}{1.5}$