$\text{Ma}-2\text{T}=0$
$\Rightarrow\text{Ma = 2T}\Rightarrow\text{T} = \frac{\text{Ma}}{2}.$
$\text{T + Ma}-\text{Mg}=0$
$\Rightarrow\frac{\text{Ma}}{2}+\text{ma = Mg}.$ $\Big(\text{because = T}=\frac{\text{Ma}}{2}\Big)$
$\Rightarrow3\text{Ma = 2Mg}\Rightarrow\text{a}=\frac{2\text{g}}{3}$
- acceleration of mass M is $\frac{2\text{g}}{3}.$
- Tension $\text{T}=\frac{\text{Ma}}{2}=\frac{\text{M}}{2}=\frac{2\text{g}}{3}=\frac{\text{Mg}}{3}$
- Let, R1 = resultant of tensions = force exerted by the clamp on the pulley
$\text{R}^1=\sqrt{\text{T}^2+\text{T}^2}=\sqrt{2}\text{T}$
$\therefore\text{R}=\sqrt{2}\text{T}=\sqrt{2}\frac{\text{Mg}}{3}=\frac{\sqrt{2}\text{Mg}}{3}$
Again, $\tan\theta=\frac{\text{T}}{\text{T}}=1\Rightarrow\theta=45^{\circ}.$
So, it is $\frac{\sqrt{2}\text{Mg}}{3}$ at an angle of 45° with horizontal.
