91Ó°ÊÓ

Given data

The side of the cube is\(l\).

The horizontal pull is\(F\).

The distance above the force is\(h\).

The coefficient of static friction is \({\mu _{\rm{s}}}\).

In this problem, the normal force will exert on the lower right corner of the block when the block is about to tilt.

In evaluating the coefficient of static friction, use the equations of equilibrium for forces in the x and y directions.

The following is the free body diagram.

The relation of forces in the y-direction can be written as follows:

\(\begin{array}{c}\sum {F_{\rm{y}}} = 0\\{F_{\rm{N}}} - mg = 0\\{F_{\rm{N}}} = mg\end{array}\)

Here, \({F_{\rm{N}}}\) is the normal force; mis the mass of the block; \(g\) is the gravitational acceleration.

The relation of forces in the x-direction can be written as follows:

\(\begin{array}{c}\sum {F_{\rm{x}}} = 0\\F - {F_{{\rm{fr}}}} = 0\\F = {\mu _{\rm{s}}}{F_{\rm{N}}}\\F = {\mu _{\rm{s}}}mg\end{array}\)

Here, \({F_{{\rm{fr}}}}\) is the frictional force.

The relation of net torque can be written as follows:

\(\begin{array}{c}\sum \tau = 0\\\left( {F \times h} \right) - \left( {mg \times \frac{l}{2}} \right) = 0\\Fh = \frac{{mgl}}{2}\end{array}\)

Plugging in the values in the above relation:

\(\begin{array}{c}\left( {{\mu _{\rm{s}}}mg} \right)h = \frac{{mgl}}{2}\\{\mu _{\rm{s}}} = \frac{l}{{2h}}\end{array}\)

When the coefficient of static friction is less than the obtained value, that is

\({\mu _{\rm{s}}} < \frac{l}{{2h}}\), then sliding will happen before tipping.

When the coefficient of static friction is greater than the obtained value, that is

\({\mu _{\rm{s}}} > \frac{l}{{2h}}\), then tipping will happen before sliding.