If the beam $AD$ is loaded as shown, determine the horizontal force $P$ which must be applied to the wedge in order to remove it from under the beam. The coefficients of static friction at the wedge's top and bottom surfaces are ${}_{CA}=0\mathrm{.}25$ and ${}_{CB}=0\mathrm{.}35,$ respectively. If $P=0,$ is the wedge selflocking? Neglect the weight and size of the wedge and the thickness of the beam.

Use the following values for the image above;

$L_1=2\mathrm{.}4m$

$L_2=3\mathrm{.}9m$

$=14$ degree

$H=4\mathrm{.}5k\frac{N}{m}$

The horizontal force $P$ needed to remove the wedge is $kN.$