√32
23
14
√34
From the free body diagram
ΣF=0
2+mgsin30° = μmgcos30° (i)
10=mgsin30°+μmgcos30° (ii)
Adding eq(i) and (ii), we get
12=2μmgcos30°
6=μmgcos30°.. (iii)
4=mgsin30° (iv)
From eqn(iii) and (iv), we get
32=μ × cot30°
32=μ × √3
μ = 3/(2√3) = √3/2