91影视

Inclination angle is$胃=26掳$

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

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

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

$蚁=1.0\text{鈥�}c{m}^3$

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

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

The cross-sectional area of the bolt is$\text{A}=6.4c{m}^2\left(\frac{1{\text{鈥尘}}^2}{{\text{10}}^{\text{4}}{\text{鈥塩尘}}^{\text{2}}}\right)=6.4脳{10}^{鈭�4}{\text{鈥尘}}^2$

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

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:

${\text{F}}_{\text{friction}}={\text{渭}}_{\text{static}}脳{\text{F}}_{\text{N}}$

$\text{F=ma}$

$\text{Shearstress=}\frac{{\text{F}}_{\text{Minimumforcerequiredtostabilizetheslab}}}{\text{苍耻尘产别谤辞蹿产辞濒迟蝉脳础谤别补辞蹿产辞濒迟}}$

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,

$\begin{array}{c}\text{V}=\text{LTW}\\ \text{鈥夆赌夆赌夆赌�}=43\text{鈥�}m脳2.5\text{鈥�}m脳12\text{鈥�}m\\ =1290\text{鈥�}{m}^3\end{array}$

$\begin{array}{c}\text{F}=\text{蚁Vgsin胃}\\ =(3.2脳{10}^3\text{鈥�}kg脳1290\text{鈥�}{m}^3)脳(9.8\text{鈥�}m/s^2脳sin26掳)\\ =1.77脳{10}^7\text{鈥�}N\end{array}$

${\text{F}}_{\text{friction}}={\text{渭}}_{\text{static}}脳{\text{F}}_{\text{N}}$

From the free boy diagram, we can say that,

${\text{F}}_{\text{N}}=\text{尘驳鈥塩辞蝉}\left(胃\right)$

${\text{F}}_{\text{friction}}={\text{渭}}_{\text{static}}\text{脳尘驳肠辞蝉胃}$

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

As the bolts are required to stabilize the slab,

The minimum force required to stabilize the slab will be,

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

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