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
Comments
Hi sir. Good day. Question
Hi sir. Good day. Question lang po, hindi po ba may normal force pa rin na kasama kapag nag-slide up ung block sa inclined plane without tipping over? Sinubukan ko po kasi magslide ng block pataas sa incline and nakita ko may mga certain heights din na may tipping at wala kaya confused lang po ako kung bakit natanggal ung normal force sa problem. Naisip ko rin since the block is moving baka kaya neglected ung normal force.