# Example 3.10: Sliding door (rolling and sliding)

A sliding door of mass $m$ hangs from two rollers $A$ and $B$, of radius $r_c$, which roll on a horizontal rail. The rolling resistance coefficient of the rollers is $\rho$ and the dry friction coefficient between the rollers and the rail is $\mu$. The center of mass $G$ of the door is centered between the rollers, and the handle $P$ is a distance $h$ below the rail, at the same height as $G$. Compute the horizontal force $F$ to be applied at the handle to open the door (to the right) if:

1. Both rollers work properly and roll.
2. One of the rollers is jammed and slides instead of rolling.

Data: $m = 125$ kg, $l = 1$ m, $h = 2.25$ m, $r_c = 50$ mm, $\mu = 0.2$, $\rho = 5$ mm, $g = 9.81$ m/s².

<img src="figs/L03_door.png" alt="Sliding door" width="400px">

(Adapted from problem 7-17 of {cite:p}`cardona2001teoria`.)

:::{dropdown} Solution
The door is assumed to move slowly (no inertia effects) and friction at the roller axles is neglected.

**Resisting force of each roller.** The rail exerts on each roller a normal force $N$ and a tangential force $T$, opposite to the motion.

- If the roller **rolls**, its axle exerts no torque on it, so the moment of $T$ about the axle balances the rolling resistance moment: $T\, r_c = \rho N$, that is, $T = \dfrac{\rho}{r_c} N = 0.1\, N$.
- If the roller is **jammed**, it slides on the rail and the friction force is $T = \mu N = 0.2\, N$.

In both cases, $T$ acts at the height of the rail and is transmitted to the door through the roller axle.

**Equilibrium of the door.** With $\hat{i}$ to the right, the equations are:

$$
\begin{aligned}
\sum F_x = 0: &\quad F - T_A - T_B = 0 \\
\sum F_y = 0: &\quad N_A + N_B - mg = 0 \\
\sum M_G = 0: &\quad -l\,N_A + l\,N_B + h\left( T_A + T_B \right) = 0
\end{aligned}
$$

Force $F$ is at the height of $G$ and has no moment about it. The tangential forces, a height $h$ above, tend to tip the door, loading roller $A$ (the rear one) and unloading $B$:

$$ N_A - N_B = \frac{h}{l}\, F, \qquad N_A = \frac{1}{2}\left( mg + \frac{h}{l}F \right), \qquad N_B = \frac{1}{2}\left( mg - \frac{h}{l}F \right) $$

**1. Both rollers roll.** $T_A + T_B = \frac{\rho}{r_c}(N_A + N_B) = \frac{\rho}{r_c}\, mg$, which does not depend on how the load is shared:

$$ F = \frac{\rho}{r_c}\, mg = 0.1 \cdot 125 \cdot 9.81 = 122.6~\text{N} $$

**2a. Roller $A$ jammed.** $T_A = \mu N_A$ and $T_B = \frac{\rho}{r_c} N_B$:

$$
F = \frac{\mu}{2}\left( mg + \frac{h}{l}F \right) + \frac{\rho}{2r_c}\left( mg - \frac{h}{l}F \right)
\quad\rightarrow\quad
F = \frac{\left( \mu + \frac{\rho}{r_c} \right) \frac{mg}{2}}{1 - \left( \mu - \frac{\rho}{r_c} \right)\frac{h}{2l}}
= \frac{0.3 \cdot 613.1}{1 - 0.1 \cdot 1.125} = 207.3~\text{N}
$$

**2b. Roller $B$ jammed.** $T_A = \frac{\rho}{r_c} N_A$ and $T_B = \mu N_B$:

$$
F = \frac{\left( \mu + \frac{\rho}{r_c} \right) \frac{mg}{2}}{1 + \left( \mu - \frac{\rho}{r_c} \right)\frac{h}{2l}}
= \frac{0.3 \cdot 613.1}{1 + 0.1 \cdot 1.125} = 165.3~\text{N}
$$

It is worse if roller $A$ jams: when the door is pulled, the load transfer loads it, so it carries the larger normal force.

(With $g = 10$ m/s², as in the source, one gets $125$, $211$ and $169$ N.)

**Checking that the rollers roll in case 1.** A roller rolls without slipping if the required tangential force fits within the friction cone: $T = \frac{\rho}{r_c}N \le \mu N$, that is, $\frac{\rho}{r_c} = 0.1 \le \mu = 0.2$. ✓
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