A 2.0-m-high box with a 1.0-m-square base is moved across a rough floor as in Fig. 9–89. The uniform box weighs 250 N and has a coefficient of static friction with the floor of 0.60. What minimum force must be exerted on the box to make it slide? What is the maximum height h above the floor that this force can be applied without tipping the box over? Note that as the box tips, the normal force and the friction force will act at the lowest corner.

The length of the box is\(l = 2\;{\rm{m}}\).

The width of the box is\(b = 1\;{\rm{m}}\).

The weight of the box is\(W = 250\;{\rm{N}}\).

The coefficient of static friction between the box and the floor is \({\mu _{\rm{s}}} = 0.6\).

A free body diagram (FBD) is a simplified representation of a rigid body to visualize all the force vectors acting on it. The free body shown below depicts the forces on the uniform box.

The forces acting on the box are the following:

• Weight of the box engine,\(W = mg\)vertically downwards
• Normal force, Nperpendicular to the floor
• Applied force (pushing), F
• Frictional force,\({F_{{\rm{fr}}}}\)between the floor and the box

The uniform box is considered to be on the verge of both sliding and slipping yet in a state of equilibrium. The static friction will have a maximum value because the box is just about to slide.

Suppose the box is just about to slide and touch the floor, such that all the forces on the box due to the contact with the ground are at the lower right-hand corner.

The forces along the y-direction (parallel to the edge of the box) are given by the equilibrium equation.

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

This shows that the normal force balances the weight (mg) of the box.

The forces along the x-axis (perpendicular to the edge of the box) are given as follows:

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

From equations (i) and (ii),

\(\begin{array}{c}F = {\mu _{\rm{S}}}W\\ = \left( {0.6} \right)\left( {250\;{\rm{N}}} \right)\\ = 150\;{\rm{N}}\end{array}\)

At the critical moment when the box is just about to tip and slide, the net torque is equal to zero. Take the torques in the clockwise direction as positive.

Using the equation of equilibrium,

\(\begin{array}{r}\sum {\tau = 0} \\Fh - W\left( {\frac{b}{2}} \right) = 0\\h = \frac{W}{F}\left( {\frac{b}{2}} \right)\end{array}\)

Substituting numerical values in the above expression gives the height h above the floor, where the force should be applied without tipping the box.

\(\begin{array}{c}h = \frac{{250\;{\rm{N}}}}{{150\;{\rm{N}}}}\left( {\frac{{1\;{\rm{m}}}}{2}} \right)\\ = 0.83\;{\rm{m}}\end{array}\)