The accelerating matchbox

Problem

The accelerating matchbox. A matchbox of height $h$, width $b$ and uniform density stands on a table that is made to accelerate horizontally with acceleration $a$. The coefficient of static friction between box and table is $\mu_s$. At what value of $a$ does the box start to slide? At what value does it instead start to tip over? Which of the two phenomena occurs first if $\mu_s > b/h$? Reworked from (Povey 2015, § 4.3).

Solution

We describe the box in the (non-inertial) reference frame of the table: besides the weight $mg$ (downward) and the normal $N$, an inertial force $ma$ appears, horizontal, applied at the centre of mass (at half height, $h/2$, and half width, $b/2$). Static friction $f\le\mu_s N$ must keep the box still.

Sliding threshold. Vertically $N = mg$. The box does not slide as long as friction manages to balance the inertial force: $ma \le \mu_s N = \mu_s mg$. It starts to slide when $\ev{a_\text{slide} = \mu_s\,g}$

Tipping threshold. The box tips over rotating about the “rear” edge (the one on the side opposite to the acceleration). Let us calculate the moments about that edge:

• the weight $mg$ has lever arm $b/2$ and gives a stabilising moment $mg\cdot\dfrac{b}{2}$;
• the inertial force $ma$ has lever arm $h/2$ (height of the centre of mass) and gives a tipping moment $ma\cdot\dfrac{h}{2}$.

Tipping starts when the tipping moment exceeds the stabilising one: $ma\cdot\frac{h}{2} > mg\cdot\frac{b}{2} \;\Rightarrow\; \ev{a_\text{tip} = \dfrac{b}{h}\,g}$

Which happens first. The phenomenon with the lower acceleration threshold occurs first. Let us compare: $a_\text{slide} = \mu_s g \qquad\text{vs}\qquad a_\text{tip} = \frac{b}{h}g$ If $\mu_s > b/h$, then $a_\text{slide} > a_\text{tip}$: the tipping threshold is lower, so the box tips over before sliding.

$\ev{\mu_s > b/h \;\Rightarrow\; \text{tips over before sliding}}$

Interpretation via the height of the centre of mass. The box slides (rather than tipping) if $a_\text{slide} < a_\text{tip}$, i.e. $\mu_s g < \dfrac{b}{h}g \Rightarrow h < \dfrac{b}{\mu_s}$. Since the centre of mass of a uniform body sits at $y = h/2$, the condition becomes: the box slides as long as the height of the centre of mass does not exceed $\ev{y_\text{max} = \dfrac{b}{2\mu_s}}$ Beyond this height (a box too tall and narrow, or too much friction) tipping prevails. That is why tall, thin objects fall over when you push the surface they rest on, while low, wide ones slide.

Links

Topics: Dinamica · Attrito Concepts: Attrito statico · Seconda legge di Newton · Forza normale · Forza peso Skills: Applicazione delle leggi di Newton · Diagramma di corpo libero · Impostazione simbolica Methods: Metodo del diagramma di corpo libero Objects: Blocco