A certain vertical high wall has a wedge with a crack daylighting on top of the high wall. The crack is $4$m from the crest and is $5$m deep. Water column inside the crack is $3$m$.$ The height of the wedge is $7$m$.$ The plunge of the wedge is $500.$ The wedge measures $7$m across. Cohesion is $30$Pa$.$ Assume the density of the rock material to be $2\mathrm{.}5$ t$/$m$3.$ Determine number of $38$ ton cable anchors required to stabilise the wedge at FOS of $1\mathrm{.}5.$

Slope Stability Chart$\_$Rock masses containing major structural features: planar failure

The stability of a planar wedge or block, such as that shown in Figure $33,$ can be determined using the following equation:

$F=\frac{C(A)+(Wcos-U-Vsin-Tcos)tan}{Wsin+Vcos-Tsin},V=0\mathrm{.}5Y_w{Z}_{WV-\text{a}\text{r}\text{e}\text{a}}$

$U=0\mathrm{.}5Y_w(H-Z{)}_{U-\text{a}\text{r}\text{e}\text{a}}$

Where: $F$ is the factor of safety Where:

c is the cohesion on the failure plane

A is the area of the base of the wedge or block

$W$ is the weight of the wedge or block

$,$ is the angle of inclination of the failure plane

$U,$ is the uplift force on the failure plane due to water pressure

$Y_w$ is the unit weight of water $U$ and $V$ areas upon which $U$ and $V$ are generated

$V$ is the horizontal force on the wedge due to water pressure in a tension crack

$T$ is the tensile force in a cable anchor

$$ is the angle of inclination between the anchor and the normal to the failure plane

$,$ is the friction angle on the failure plane.