Figure 6-22 shows the cross section of a road cut into the side of a mountain. The solid line$\mathit{A}\mathit{A}\mathbf{\text{'}}$represents a weak bedding plane along which sliding is possible. Block B directly above the highway is separated from uphill rock by a large crack (called a joint), so that only friction between the block and the bedding plane prevents sliding. The mass of the block islocalid="1654084347613" ${}^{\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{}\mathbf{k}\mathbf{g}}$, the dip anglelocalid="1654084361257" ${}^{\mathbf{\theta }}$of the bedding plane islocalid="1654084374565" ${}^{\mathbf{24}\mathbf{°}}$, and the coefficient of static friction between block and plane islocalid="1654084400008" ${}^{\mathbf{0}\mathbf{.}\mathbf{63}}$. (a) Show that the block will not slide under these circumstances. (b) Next, water seeps into the joint and expands upon freezing, exerting on the block a forceparallel tolocalid="1654084460188" $\mathit{A}\mathit{A}\mathbf{\text{'}}$. What minimum value of force magnitudelocalid="1654084470850" $\mathit{F}$will trigger a slide down the plane?

Mass,${}^{m=1.8\times {10}^{7}kg}$

Coefficient of static friction,${}^{{\mu }_s=0.63}$

Inclined angle,${}^{\theta =24°}$

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Use the Newton's 2nd law of motion along vertical and horizontal direction.

Formula:

$F_{Net}=\underset{}{\sum ma}$

where, F is the net force, m is mass and a is an acceleration.

Free body diagram of Block:

By using Newton’s 2nd law along vertical direction (along y),

$\underset{}{\sum F_y=m{a}_y}$

Since block is not moving upward, ${}^{a_y=0}$

$$\begin{gathered} N-F_g\cos \left(24°\right)=0 \\ N=F_g\cos \left(24°\right) \end{gathered}$$

Relation between static frictional force and normal force is ,

$$\begin{gathered} f_s={\mu }_{s}N \\ ={\mu }_s{F}_g \\ =(0.63)\left(F_g\cos \left(24°\right)\right) \\ =1.02\times {10}^{8}N \end{gathered}$$

And,

$$\begin{gathered} F_g\sin \left(24°\right)=(1.8\times {10}^7)(9.81)\left(\sin \left(24°\right)\right) \\ =7.18\times {10}^{7}N \end{gathered}$$

In this case, upward force${}^{f_s}$is very greater than the downward force ${}^{F_g\sin \left(24°\right)}$

i.e.${}^{f_s>F_g\sin \left(24°\right)}$.

Therefore, the block will not slide.

Consider the force applied by the ice is F, then by using the Newton’s 2nd law of motion,

$\underset{}{\sum F_{\mathrm{x}}=m{a}_{\mathrm{x}}}$

${}^{a_x=0_{,}}$Since, block is not moving

$$\begin{gathered} f_s-F_g\sin \left(24°\right)-F=0 \\ F=f_s-F_g\sin \left(24°\right) \\ =\left(\begin{array}{c}1.02\times {10}^8\end{array}N\right)-\left(\begin{array}{c}7.18\times {10}^{7}N\end{array}\right) \\ =3.0\times {10}^{7}N \\ =3.0\times {10}^{7}N \end{gathered}$$

Therefore, the minimum value of force magnitude${}^F$, that will trigger a slide down the plane is${}^{3.0\times {10}^{7}N}$