# Example 3.9: Two blocks joined by a bar, with friction

In the mechanism of the figure, two identical blocks of mass $m$ are joined by a bar of negligible mass, pinned at their centers $G_1$ and $G_2$. Block 1 slides on a horizontal surface and block 2 on a vertical wall. There is dry friction with coefficient $\mu$ at the block contacts; friction at the pins is negligible. The bar is at an angle $\varphi$ from the horizontal. Find:

1. The horizontal force $F$ needed to start the motion from rest, in each direction.
2. Its numerical value for $\varphi = 45^\circ$ and $\mu = 0.2$.

<img src="figs/L03_blocks.png" alt="Two blocks joined by a bar" width="440px">

(Adapted from problem 7-1 of {cite:p}`cardona2001teoria`.)

:::{dropdown} Solution
**Bar.** Massless and pinned at its ends: it is a two-force member. Let $T$ be the force it transmits, positive in **tension**. It pulls block 1 towards $G_2$, with direction $(\cos\varphi, -\sin\varphi)$, and block 2 towards $G_1$, with direction $(-\cos\varphi, \sin\varphi)$.

**Block 2.** It receives its weight $mg$, the bar force, the normal force from the wall $N_2$ (horizontal, to the right, since the wall is on its left) and friction $\mu N_2$ (vertical, opposite to the motion). From horizontal equilibrium:

$$ N_2 - T\cos\varphi = 0 \quad\rightarrow\quad N_2 = T \cos\varphi $$

**Block 1.** It receives its weight, the bar force, the normal force from the floor $N_1$, friction $\mu N_1$ (horizontal, opposite to the motion) and the force $F$. From vertical equilibrium:

$$ N_1 - mg - T\sin\varphi = 0 \quad\rightarrow\quad N_1 = mg + T \sin\varphi $$

At the start of motion, friction at both contacts equals $\mu N$, with the sense opposite to the motion about to happen. There are two cases.

**Case (a): $F$ to the left; block 1 moves left and block 2 moves up.** Friction on block 2 points down and on block 1, to the right:

$$
\begin{aligned}
\text{Block 2, vertical:} &\quad T\sin\varphi - mg - \mu T\cos\varphi = 0 \quad\rightarrow\quad T = \frac{mg}{\sin\varphi - \mu\cos\varphi} \\
\text{Block 1, horizontal:} &\quad -F + \mu N_1 + T\cos\varphi = 0 \quad\rightarrow\quad F = \mu mg + T\left( \cos\varphi + \mu\sin\varphi \right)
\end{aligned}
$$

Substituting $T$ and dividing by $\cos\varphi$:

$$ F_{\leftarrow} = mg \left[ \mu + \frac{1 + \mu\tan\varphi}{\tan\varphi - \mu} \right] = mg\, \frac{1 - \mu^2 + 2\mu\tan\varphi}{\tan\varphi - \mu} $$

This result only makes sense if $\tan\varphi > \mu$. If $\tan\varphi \le \mu$ ($\varphi \le \arctan\mu = 11.3^\circ$ for $\mu = 0.2$), the force of the bar on block 2 falls inside its friction cone: the mechanism **locks**, and no force $F$, however large, can lift block 2.

**Case (b): block 1 moves right and block 2 moves down.** Now the friction forces change sense:

$$
\begin{aligned}
\text{Block 2, vertical:} &\quad T\sin\varphi - mg + \mu T\cos\varphi = 0 \quad\rightarrow\quad T = \frac{mg}{\sin\varphi + \mu\cos\varphi} \\
\text{Block 1, horizontal:} &\quad F_x - \mu N_1 + T\cos\varphi = 0
\end{aligned}
$$

with $F_x$ the applied force, positive to the right. Proceeding in the same way:

$$ F_x = -mg\, \frac{1 - \mu^2 - 2\mu\tan\varphi}{\tan\varphi + \mu} $$

If $F_x < 0$, the weight of block 2 alone is enough to set the system in motion: to keep it at rest, it must be **held** with a force to the left of magnitude at least $|F_x|$.

**Numerical values** ($\varphi = 45^\circ$, $\mu = 0.2$, $\tan\varphi = 1$):

$$
F_{\leftarrow} = mg\,\frac{1 - 0.04 + 0.4}{1 - 0.2} = 1.70\, mg, \qquad
F_x = -mg\,\frac{1 - 0.04 - 0.4}{1 + 0.2} = -0.467\, mg
$$

That is, with no applied force block 2 falls. The system stays at rest if a force to the left $F$ between $0.467\,mg$ and $1.70\,mg$ is applied; above that range, block 2 moves up, and below it, it moves down. Without friction ($\mu = 0$), both limits would coincide at $F = mg/\tan\varphi = mg$: friction creates an **equilibrium band** instead of a single value.
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