A block wedge with a potential for planar failure is observed on a high wall face. The rock block base is $2m3m$ and it is inclined by $30$ degrees from the horizontal. Two cable anchors are installed horizontally to support this block wedge on the basis of a factor of safety of $2.$ The following information is also given:

Cohesive strength of the failure plane and coefficient of friction $-30$kpa&$0\mathrm{.}5,$ respectively.

Uplift and horizontal pressures due to water $-100$kpa & $60$kpa, respectively.

Rock Density $-2\mathrm{.}8$ tons $\frac{?}{m3}$

i$.$ Determine the tension on each cable bolt. $[10]$

ii$.$ Quantify the effect that artificial water drainage $($pumping$)$ would have on the stability of the wedge. $[5]$

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

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:

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

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

Where:

Where: $F$ is the factor of safety

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

$V$ 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.