Pls solve asap plsss

Dear Student,

       
   
(a) From the free body diagram

       

R = 4g cos 30°
$$\begin{gathered} \Rightarrow \mathrm{R}=4\times 10\times \frac{\sqrt{3}}{2} \\ =20\sqrt{3} \mathrm{N} \left(1\right) \end{gathered}$$

μ₂R + m₁a − p − m₁g sin θ = 0
μ₂R + 4a − p − 4g sin 30° = 0              
⇒ 0.3 $\times$ (40) cos 30° + 4a − p − 40 sin 30° = 20             (2)

      

R₁ = 2g cos 30° $=10\sqrt{3}$                        (3)
p + 2a − μ₁R₁ − 2g sin 30° = 0               (4)

From Equation (2),
$6\sqrt{3}+4a-p-20=0$

From Equation (4),
$$\begin{gathered} p+2a+2\sqrt{3}-10=10 \\ 6\sqrt{3}+6a+30+2\sqrt{3}=0 \\ \Rightarrow 6a=30-8\sqrt{3} \\ =30-13.85=16.15 \\ \Rightarrow a=\frac{16.15}{6} \\ =2.69=2.7 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$


Regards