91Ó°ÊÓ

Mass,${}^{m=1.8×{10}^{7}kg}$

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

Inclined angle,${}^{\mathrm{θ}=24°}$

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:

$F_{Net}=\underset{}{∑ma}$

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),

$\underset{}{∑F_y=m{a}_y}$

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 ,

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

And,

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

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,

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

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

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

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