In Fig. 12-63, a rectangular slab of slate rests on a bedrock surface inclined at angle $\theta =26°$. The slab has length $L=43m$, thickness $T=2.5m$, and width,$W=12m$and $1.0c{m}^3$of it has a mass of $3.2g$. The coefficient of static friction between slab and bedrock is $0.39$. (a) Calculate the component of the gravitational force on the slab parallel to the bedrock surface. (b) Calculate the magnitude of the static frictional force on the slab. By comparing (a) and (b), you can see that the slab is in danger of sliding. This is prevented only by chance protrusions of bedrock. (c) To stabilize the slab, bolts are to be driven perpendicular to the bedrock surface (two bolts are shown). If each bolt has a cross-sectional area of $6.4{\text{\hspace{0.17em}cm}}^{\text{2}}$and will snap under a shearing stress of, $3.6\times {10}^8{\text{N/m}}^{\text{2}}$what is the minimum number of bolts needed? Assume that the bolts do not affect the normal force.

Inclination angle is$\theta =26°$

The length of the slab is$\text{L}=43m$

The thickness of the slab is$\text{T}=2.5m$

The width of the slab$\text{w}=12m$

$\rho =1.0c{m}^3$

The mass of the slab is$\text{m}=3.2g$

The coefficient of static friction is${\mu }_s=0.39$

The cross-sectional area of the bolt is$\text{A}=6.4c{m}^2\left(\frac{1{\text{\hspace{0.17em}m}}^2}{{\text{10}}^{\text{4}}{\text{\hspace{0.17em}cm}}^{\text{2}}}\right)=6.4\times {10}^{-4}{\text{\hspace{0.17em}m}}^2$

Shear Stress is$=3.6\times {10}^8{\text{N/m}}^{\text{2}}$

By drawing the free body diagram of the slab, we can determine the forces which are required.To stabilize the slab we use bolts. Using the formula for stress, and knowing how much force is required to stabilize the slab, we can calculate the number of bolts.

Formula:

${\text{F}}_{\text{friction}}={\text{μ}}_{\text{static}}\times {\text{F}}_{\text{N}}$

$\text{F=ma}$

$\text{Shearstress=}\frac{{\text{F}}_{\text{Minimumforcerequiredtostabilizetheslab}}}{\text{numberofbolts×Areaofbolt}}$

From the Free body diagram, the component of the gravitational force along the incline will be,

$\text{F}=\text{ma}$

The volume of the slab,

$\begin{array}{c}\text{V}=\text{LTW}\\ =43m\times 2.5m\times 12m\\ =1290m^3\end{array}$

$\begin{array}{c}\text{F}=\text{ρVgsinθ}\\ =(3.2\times {10}^{3}kg\times 1290m^3)\times (9.8m/s^2\times sin26°)\\ =1.77\times {10}^{7}N\end{array}$

${\text{F}}_{\text{friction}}={\text{μ}}_{\text{static}}\times {\text{F}}_{\text{N}}$

From the free boy diagram, we can say that,

${\text{F}}_{\text{N}}=\text{mg\hspace{0.17em}cos}\left(\theta \right)$

${\text{F}}_{\text{friction}}={\text{μ}}_{\text{static}}\text{×mgcosθ}$

$\begin{array}{c}{F}_{friction}=0.39\times ((3.2\times {10}^{3}kg\times 1290m^3)\times (9.8m/s^2\times cos26°))\\ =1.4\times {10}^{7}N\end{array}$

As the bolts are required to stabilize the slab,

The minimum force required to stabilize the slab will be,

$\begin{array}{c}{\text{F}}_{\text{minimum}}=\text{F}-{\text{F}}_{\text{friction}}\\ =1.77\times {10}^{7}N-1.4\times {10}^{7}N\\ =3.7\times {10}^{6}N\end{array}$

$\begin{array}{c}{\text{F}}_{\text{minimum}}=\text{F}-{\text{F}}_{\text{friction}}\\ =1.77\times {10}^{7}N-1.4\times {10}^{7}N\\ =3.7\times {10}^{6}N\end{array}$