Tension and acceleration on inclined plane

For 5kg box, we will get:

$5×10 \sin 37°-T=5a \ ...(1)$

For 3kg box, we will get:

$T-3×10 \sin 53°=3a \ ...(2)$

Solving $(1)$ and $(2)$ , we get:

$a=\dfrac{3}{4}$

$T=\dfrac{105}{4}$

$\therefore a+T=\dfrac{108}{4}=\boxed{27}$

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Note: $a=\text{acceleration of box}$ and $T=\text{Tension}$

$\sin 37°=\dfrac{3}{5}$ and $\sin 53°=\dfrac{4}{5}$ .

The vertical component of the force exerted by $5 \ \text{kg}$ block = $\text{mg}\sin\theta = 5×10×\sin{(37)}^° = 30 \ \text{N}$ .
The vertical component of the force exerted by $3 \ \text{kg}$ block = $\text{mg}\sin\theta = 3×10×\sin{(53)}^° = 24 \ \text{N}$ .

So, the net force is $30 - 24 = 6 \ \text{N}$ .
Now, $\begin{aligned} \text{F} & = \text{ma}\\ \text{a} & = \dfrac{6}{5+3}\\ \text{a} & = 0.75 \ {\text{m/sec}}^2 \end{aligned}$

Now, since the vertical component of the force exerted by the $5 \ \text{g}$ block is more,
$\begin{aligned} \text{mg}\sin\theta - \text{T} & = \text{ma}\\ \\ 30 - \text{T} & = 5×0.75\\ \\ \text{T} & = 30 - 3.75\\ \\ \text{T} & = 26.25 \ \text{N} \end{aligned}$

So, $\begin{aligned} \text{T + a} & = 26.25 + 0.75\\ & = \color{#3D99F6}{\boxed{27}} \end{aligned}$

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