When the cylinder starts to turn due to P, the normal force under horizontal surface is zero. See the free body diagram below.
$x = (1)(\sin 60^\circ)$
$x = \frac{1}{2}\sqrt{3} \, \text{ m}$
$\Sigma M_A = 0$
$(1 + x)P = 12x$
$(1 + \frac{1}{2}\sqrt{3})P = 12(\frac{1}{2}\sqrt{3})$
$1.866P = 10.392$
$P = 5.569 \, \text{ kN}$ answer
Comments
Hi sir. Goodmorning. Bakit po
Hi sir. Goodmorning. Bakit po pala natanggal ung normal force sa horizontal surface when the cylinder turns due to force "P" and ung normal force sa may inclined surface ay hinde? I was thinking about the cylinder would be on the same spot po pa rin kasi if turned by a force "P" since angle of friction=tan-1(0.3)= 17° is lesser than the angle of the inclined plane.