Question

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.

Short answer

a) Component of the gravitational force, along the slab$=1.77\times {10}^{7}N$

b) Force of static friction$=1.4\times {10}^7\text{\hspace{0.17em}N}$

c) Number of bolts$\text{n}=16$

Step-by-step solution

Listing the given quantities

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

Understanding the concept of stress and strain

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:

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

$\text{F=ma}$

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

 Step 3: Free body diagram of slab

(a) Calculation ofComponent of the gravitational force, along the slab

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,

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

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

(b) Calculation offorce of static friction

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

From the free boy diagram, we can say that,

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

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

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

(c) Calculation ofnumber of bolts

As the bolts are required to stabilize the slab,

The minimum force required to stabilize the slab will be,

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

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

The minimum number of bolts required,