A cubical box is filled with sand and weighs $\mathbf{890}\mathit{N}$. We wish to “roll” the box by pushing horizontally on one of the upper edges. (a) What minimum force is required? (b) What minimum coefficient of static friction between box and floor is required? (c) If there is a more efficient way to roll the box, find the smallest possible force that would have to be applied directly to the box to roll it. (Hint:At the onset of tipping, where is the normal force located?)

Weight of box is,$W=890N$ .

Using the first equilibrium condition given below, we can find the relation between F(external force) and$\mathbf{}{\mathit{F}}_{\mathbf{f}}$(friction force).Then, using the second equilibrium condition, we can find the ${\mathit{F}}_{\mathbf{N}}$(contact force) in terms of W (Weight of the object). Using the third equilibrium condition and the above two relations, we can find $\mathit{F}$and$\mathbf{}\mathit{\mu }$ (coefficient of friction).The smallest possible force that would have to be applied directly to the box to roll it can be calculated by using equilibrium conditions for torque when$\mathit{F}$is acting in the upward right direction.

The free Body Diagram can be drawn as,

Using the condition of equilibrium in the x direction, we can write,

$\sum F_x=0$

Substitute the values in the above expression, and we get,

$\begin{array}{c}F-F_f=0\\ F=F_f\end{array}$ (1)

Using the condition of equilibrium in the x direction, we can write,

$\sum F_y=0$

Substitute the values in the above expression, and we get,

$\begin{array}{c}{F}_N-W=0\\ F_N=W\end{array}$ (2)

Now the net torque is zero, in this case, then we can write is as,

$\sum \tau =0$

Substitute the values in the above expression, and we get,

$\begin{array}{c}F\left(L\right)-W\left(\frac{L}{2}\right)=0\\ F=\frac{W}{2}\end{array}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}F=890/2\\ =445N\end{array}$

Thus, the magnitude of the minimum force required is$445N$ .

Substitute the values in equations 1 and 2, and we get,

$F_N=890N$

And

$F_f=445N$

The coefficient of friction can be calculated from the formula as,

$\mu =\frac{F_f}{F}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}\mu =\frac{445}{890}\\ =0.5\end{array}$

Thus, the minimum coefficient of friction force between box and floor is$0.5$.

Free Body Diagram

The box can be rolled with a smaller applied force if the force points upwards and to the right.Let$\theta$be the angle the force makes with horizontal.The torque equation then becomes:

$\begin{array}{c}FL\cos \theta +FL\sin \theta -\frac{WL}{2}=0\\ F=\frac{W}{2(\cos \theta +\sin \theta )}\end{array}$

We are choosing $\theta =45°$so that$\cos \theta +\sin \theta$will be the maximum, and we will be able to calculate the smallest possible force.

Substitute the values in the above expression, and we get,

$\begin{array}{c}F=\frac{890}{2(\cos 45°+\sin 45°)}\\ =315\text{\hspace{0.17em}N}\end{array}$

Thus, the smallest possible force is $315N$.