Question

please anybody explain me​

Answer 1

Assume the pulley is massless and frictionless, and the strings are light and inextensible.

Fig. 1 shows the FBD of the block, from which tension in the string T is given by,

$\small\longrightarrow\sf{T=mg\quad\dots(1)}$

as the system is in equilibrium.

Fig. 2 shows the FBD of the pulley, from which we get the tension in the string connecting pulley and the string ABC, is twice the tension in the string that passes over the pulley.

Fig. 3 shows the point of contact of the string ABC and the pulley. The point is acted upon by three forces and is at equilibrium so we can use Lami's theorem.

Lami's theorem states that when three forces acting at a point keep it in equilibrium, then each force is proportional to the sine of the angle between the other two forces, i.e., the ratio of each force to the sine of angle between the other two forces is a constant.

Here the three forces are 2T downwards, T₁ along BC and T₂ along BA. Angle between T₁ and T₂ is 90°, 2T and T₁ is 90°+θ, 2T and T₂ is 180°-θ. We are asked to find the value of T₁, i.e., the tension in the string BC.

Then by Lami's theorem,

$\small\text{ \longrightarrow\sf{\dfrac{T_1}{\sin(180^\textdegree-\theta)}=\dfrac{2T}{\sin90^\textdegree}} }$

$\small\text{ \longrightarrow\sf{\dfrac{T_1}{\sin\theta}=\dfrac{2T}{1}} }$

$\small\text{ \longrightarrow\sf{\dfrac{T_1}{\sin\theta}=2T} }$

$\small\longrightarrow\sf{T_1=2T\sin\theta}$

From (1),

$\small\text{ \longrightarrow\sf{\underline{\underline{T_1=2mg\sin\theta}}} }$

Hence (3) is the answer.