91影视

Mass,${}^{m=1.8脳{10}^{7}kg}$

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

Inclined angle,${}^{胃=24掳}$

The problem is based on Newton鈥檚 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:

$F_{Net}=\underset{}{鈭�ma}$

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

Free body diagram of Block:

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

$\underset{}{鈭�F_y=m{a}_y}$

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

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

Relation between static frictional force and normal force is ,

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

And,

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

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

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

Therefore, the block will not slide.

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

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

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

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

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