A uniform cube of side length $\mathbf{8}\mathbf{.0}\mathbf{\text{\hspace{0.17em}cm}}$ rests on a horizontal floor.The coefficient of static friction between cube and floor is m. A horizontal pull $\overrightarrow{\mathbf{P}}$is applied perpendicular to one of the vertical faces of the cube, at a distance $\mathbf{7}\mathbf{.0}\mathbf{\text{\hspace{0.17em}cm}}$above the floor on the vertical midline of the cube face. The magnitude of $\overrightarrow{\mathbf{P}}$is gradually increased. During that increase, for what values of$\mathbf{\text{μ}}$ will the cube eventually (a) begin to slide and (b) begin to tip? (Hint:At the onset of tipping, where is the normal force located?)

The side length of the cube is$8.0\text{\hspace{0.17em}cm}$.

The vertical distance at which the horizontal pull is applied on the cube side is$7.0\text{\hspace{0.17em}cm}$ .

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

$\begin{array}{l}{f}_s=\mu N\\ \Sigma {\tau }_{net}=0\end{array}$

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,

$\overrightarrow{f_s}=\overrightarrow{P}$

According to the definition of static friction force,

$\overrightarrow{f_s}=\mu \overrightarrow{N}$

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,

$\begin{array}{c}N=mg\\ \overrightarrow{f_s}=\overrightarrow{P}=\mu mg\end{array}$

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

$\overrightarrow{{\tau }_{net}}=0$

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.

$\begin{array}{l}\mu mgh=mg\left(\frac{L}{2}\right)\\ \mu =\frac{L}{2h}\\ \mu =\frac{\left(8.0\text{\hspace{0.17em}cm}\right)}{2\left(7.0\text{\hspace{0.17em}cm}\right)}\\ \mu =0.57\end{array}$

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

Thus,

$\mu <0.57$

The cube is about to tip then the coefficient of static frictionis greater than the derived value of$\mu$.

Thus,

$\mu >0.57$