Worked Exercise — Box with a Spring: Equilibrium Zone

Problem

A box of mass $m = 10\;\text{kg}$ rests on the floor and is connected to a horizontal spring fixed to the wall. The spring has constant $k = 100\;\text{N/m}$ and natural length $\ell_0 = 3\;\text{m}$. The coefficient of static friction between box and floor is $\mu = 0{,}5$. Over what range of distances $\ell$ from the wall does the box stay put?

Solution

Step 1 — Forces involved. The horizontal forces on the box are the spring force $F_m = k\,\abs{\ell-\ell_0}$ and the static friction $F_a$, which opposes the impending motion and satisfies $\abs{F_a}\leq\mu R$.

Step 2 — Normal reaction and maximum friction. Vertically the floor’s reaction balances the weight:

$R = mg = 10\cdot 9{,}81 = 98{,}1\;\text{N}$

$F_a^{\text{max}} = \mu R = 0{,}5\cdot 98{,}1 = 49{,}05\;\text{N}$

Step 3 — Stretched spring ($\ell > \ell_0$). The spring pulls the box toward the wall and friction must oppose it; equilibrium holds as long as the elastic force does not exceed the maximum friction:

$k(\ell-\ell_0) \leq F_a^{\text{max}} \;\Longleftrightarrow\; \ell \leq \ell_0 + \frac{F_a^{\text{max}}}{k} = 3 + \frac{49{,}05}{100} = 3{,}49\;\text{m}$

Step 4 — Compressed spring ($\ell < \ell_0$). The spring pushes the box away from the wall and again friction must suffice:

$k(\ell_0-\ell) \leq F_a^{\text{max}} \;\Longleftrightarrow\; \ell \geq \ell_0 - \frac{F_a^{\text{max}}}{k} = 3 - \frac{49{,}05}{100} = 2{,}51\;\text{m}$

Step 5 — Equilibrium range. There is a whole “quiet zone” in which the box stays still thanks to the interplay between spring and friction:

$\ev{\ell\in[2{,}51;\;3{,}49]\;\text{m}}$

Outside this range the spring wins and the box starts to move.

Links

Topics: Statica ed equilibrio Concepts: Equilibrio statico · Forza elastica e legge di Hooke · Attrito statico · Forza normale Objects: Molla · Blocco