**Sliding up the incline**
$\Sigma F_y = 0$

$N = W \cos \theta = \frac{4}{5}W$

$f = \mu N = 0.30(\frac{4}{5}W) = \frac{6}{25}W$

$\Sigma F_x = 0$

$P = W \sin \theta + f$

$P = \frac{3}{5}W + \frac{6}{25}W$

$P = \frac{21}{25}W$

**Tipping over**

$\Sigma M_A = 0$

$Ph = 40(W \sin \theta) + 20(W \cos \theta)$

$\frac{21}{25}Wh = 40(\frac{3}{5}W) + 20(\frac{4}{5}W)$

$h = 47.62 \, \text{ cm}$ *answer*

Good day sir! Why is the normal force not considered when we summate moments about A? Is it because the normal force equals the vertical component of the weight with respect to the inclined plane? If so, why dont we just cancel the two(W

_{y}& N) so that whats left is Ph=W_{x}(40cm)? Please help sir.In tipping over, the mode of impending motion is rotation; about

Ain this problem. When the block is about to tip over, there will be noNbut the component of weight in they-direction is still there.