Question

find the value of T1 and T2 from the figure

Answer 1

Answer:

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Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

As the figure has not specified the mass,

Let us assume that the given body has a mass "m".

$\huge{\bold{\underline{Explanation:-}}}$

Let's assume that the system is in equilibrium.

$\large{\bold{Resolving \: The \: Tension(T_2) \: into \: components.}}$

$\large{\bold{After \: resolving, \: it's \: one \: component \: will \: be \: (T_2 \sin 30 \degree) \: vertically \: upwards.}}$

$\large{\bold{and, \: the \: other \: component \: will \: be \: (T_2 \cos 30 \degree) horizontally.}}$

As we have assumed the system in equilibrium.

The Force (F) will be equal to the weight of the given block.

$\large{ \therefore F = W}$

$\large{\boxed{ F = mg \: -----(1)}}$

Now,

CASE-1

As the system is in equilibrium,

therefore,

$\large{T_2 \sin 30 \degree = F}$

From equation (1)

$\large{ \implies T_2 \sin 30 \degree = mg}$

As sin 30° = 1/2

$\large{ \implies T_2 \times \dfrac{1}{2} = mg}$

$\large{ \implies T_2 = 2 \times mg \: -----(2)}$

$\huge{\boxed{\boxed{T_2 = 2mg}}}$

CASE-2

As the system is in equilibrium,

therefore,

$\large{T_1 = T_2 \cos 30 \degree}$

From equation (2) Substituting the value of $\large{T_2}$ in the above equation,

$\large{ \implies T_1 = 2mg . \cos 30 \degree}$

as cos 30° = √3/2

$\large{ \implies T_1 = 2mg \times \dfrac{ \sqrt{3}}{2}}$

$\large{ \implies T_1 = \cancel{2}mg \times \dfrac{ \sqrt{3}}{ \cancel{2}}}$

$\huge{\boxed{\boxed{T_1 = \sqrt{3} mg}}}$

So, the $\large{\bold{T_1}}$ is √3 mg ,

and the $\large{\bold{T_2}}$ is 2mg.

#refer the attachment for figure.