91Ó°ÊÓ

The side length of the cube is$8.0\text{ c³¾}$.

The vertical distance at which the horizontal pull is applied on the cube side is$7.0\text{ c³¾}$ .

You can draw the free body diagram and use the static friction and static equilibrium concept. The formulas used are given below.

$\begin{array}{l}{f}_s=\mathrm{μ}N\\ \mathrm{Î\pounds }{\mathrm{Ï„}}_{net}=0\end{array}$

The cube just tends to move hence there is static friction between the base of the cube and the ground. According to the free body diagram,$P$is applied force at a distance$h$from the$O$point.

According to Newton’s second law of motion,

$\stackrel{→}{f_s}=\stackrel{→}{P}$

According to the definition of static friction force,

$\stackrel{→}{f_s}=\mathrm{μ}\stackrel{→}{N}$

The normal force is acting in an upward direction and the gravitational force acts in the downward direction at the center of the cube. Hence by using Newton’s second law,

$\begin{array}{c}N=mg\\ \stackrel{→}{f_s}=\stackrel{→}{P}=\mathrm{μ}mg\end{array}$

The cube is about to move, hence we can apply the equilibrium condition.

$\stackrel{→}{{\mathrm{τ}}_{net}}=0$

The cube can be rotated at the point. The applied force$P$is the horizontal line of action and its moment of the arm isthe vertical distance from O. The gravitational forceis the vertical line of action at the center of the cube and its moment of the arm is half of the side of the cube.

$\begin{array}{l}\mathrm{μ}mgh=mg\left(\frac{L}{2}\right)\\ \mathrm{μ}=\frac{L}{2h}\\ \mathrm{μ}=\frac{\left(8.0\text{ c³¾}\right)}{2\left(7.0\text{ c³¾}\right)}\\ \mathrm{μ}=0.57\end{array}$

A cube is about to slide only when the coefficient of static frictionis less than the derived value of$\mathrm{μ}$.

Thus,

$\mathrm{μ}<0.57$

The cube is about to tip then the coefficient of static frictionis greater than the derived value of$\mathrm{μ}$.

Thus,

$\mathrm{μ}>0.57$