Chapter 8, Problem 96A

To determine

The force required to just tip the box.

Answer to Problem 96A

The force required to just tip the box is $F=\frac{\left(0.86\right)M}{\left(h-0.25\text{m}\right)}$ , where M is the mass of box.

Explanation of Solution

Given:

The distance of rope from the rope is $h$ .

The width of the box is $0.25m$ .

The height of the box is $H=0.50m$ .

The coefficient of friction is $\mu =0.35$ .

Formula used:

The torque is the product of the lever arm and the applied force. The expression for the torque is,

$\tau =Fr$

Here, $F$ is the applied force, $r$ is the lever arm.

Calculation:

Consider a line that passes through centre of mass of box. Also, it is perpendicular to the box’s surface that is considered as the axis of rotation.

Consider the distance between the centre and bottom of the box be $x$ .

The free body diagram of the box is shown below.

The value of the distance $x$ is,

$\begin{array}{c}x=\frac{H}{2}\\ =\frac{0.50m}{2}\\ =0.25\text{m}\end{array}$

The frictional force acting between the box surface and the floor is denoted by $F_{\text{f}}$ .

The weight of the box, whose mass is M , is

$\begin{array}{c}{F}_{\text{g}}=Mg\\ =9.8\times M\\ =9.8M\text{N}\end{array}$

The torque $\left({\tau }_{\text{f}}\right)$ produced due to the frictional force $\left(F_{\text{f}}\right)$ is,

$\begin{array}{l}{\tau }_{\text{f}}=F_{\text{f}}x\\ {\tau }_{\text{f}}=F_{\text{f}}\left(\frac{H}{2}\right)\\ {\tau }_{\text{f}}=\mu F_{\text{g}}\left(\frac{H}{2}\right)\end{array}$

Consider the force applied on the rope as $F$ . The distance between the centre of mass of the box and rope is $r$ .

The torque $\left(\tau \right)$ produced by the force is,

$\begin{array}{l}\tau =Fr\\ \tau =F\left(h-\frac{H}{2}\right)\end{array}$

For rotational equilibrium of the box,

$\begin{array}{l}{\tau }_{\text{f}}=\tau \\ \mu F_{\text{g}}\left(\frac{H}{2}\right)=F\left(h-\frac{H}{2}\right)\\ F=\frac{\mu F_{\text{g}}\left(\frac{H}{2}\right)}{\left(h-\frac{H}{2}\right)}\end{array}$

Substitute the values in the above equation,

$\begin{array}{c}F=\frac{\left(0.35\right)\left(9.8M\right)\left(\frac{0.50\text{m}}{2}\right)}{\left(h-\frac{0.50\text{m}}{2}\right)}\\ F=\frac{\left(0.8575\right)M}{\left(h-0.25\right)}\\ \approx \frac{0.86M}{\left(h-0.25\right)}N\end{array}$

The denominator of the equation will be zero when the box is pulled at the height of center of mass. This means that box can be pulled with any amount of force and it will not tip the box.