Answer
$\theta=2^{\circ}$
Work Step by Step
Basically, there are three forces acting i.e weight W, friction f and Normal force N. Let's first resolve the weight into perpendicular and horizontal components:
Weight in perpendicular direction $W_p=W cos{\theta}$
Weight in horizontal direction $W_h=W sin{\theta}$
The net force in both horizontal and perpendicular direction is zero, hence
$N-W_p=0$
$N- Wcos{\theta}=0$
or $cos{\theta}=\frac{N}{W}$ ........eq(1)
Similarly for the horizontal component,
$f-W_h=0$
$f-Wsin{\theta}=0$
$sin{\theta}=\frac{f}{W}$
$sin{\theta}=\frac{\mu_sN}{W}$ ...........eq (2)
Dividing eq(2) by eq(1), we get
$tan{\theta}=\mu_s$
Now putting the value of $\mu_s$ and solving:
$tan{\theta}=0.04$
$\theta=tan^{-1}(0.04)$
$\theta=2.3^{\circ}$
After rounding off,
$\theta=2^{\circ}$
