# Example 3.8: Bar clamp (friction locking)

The figure shows the friction-based mechanism by which a bar clamp locks and prevents its arm from sliding along the bar. Dry friction is assumed at both contact points, and the normal force $N$ is the same at both.

Determine the maximum width $e$ of the arm that ensures clamping without sliding, for a friction coefficient $\mu = 0.20$ between the bar and the arm. Data: $l = 80$ mm, $b = 20$ mm.

Hint: find the maximum value of $e$ from the force equilibrium in the case where the friction force is exactly at its maximum possible value.

<img src="figs/L03_clamp_dims.png" alt="Bar clamp: dimensions" width="340px">
<img src="figs/problemas_din_an/clamp_contact.png" alt="Bar clamp: contact points" width="340px">

(Exam of January 2023.)

:::{dropdown} Solution
We write the equilibrium of the moving arm of the clamp. When tightening, the workpiece exerts a horizontal force $F$ on the arm (to the right in the figure), at a distance $l$ above the bar. The slightly tilted arm touches the bar at two points: $A$, on the top face of the bar, and $B$, on the bottom face. At each of them there are a normal force $N$ and a friction force $R$, which opposes the sliding that $F$ would cause:

<img src="figs/problemas_din_an/clamp_fbd.png" alt="Free-body diagram of the arm" width="480px">

Since we look for the limit case, instead of Coulomb's inequality $R \le \mu N$ we use its maximum value, $R = \mu N$. The equations of statics are:

$$
\begin{aligned}
\sum F_x = 0: &\quad F - 2R = 0 \quad\rightarrow\quad R = \frac{F}{2} \quad\rightarrow\quad \mu N = \frac{F}{2} \quad\rightarrow\quad N = \frac{F}{2\mu} \\
\sum F_y = 0: &\quad N - N = 0 \quad \text{(gives no information)} \\
\sum M_{A} = 0: &\quad -l F + e N - R\, b = 0
\end{aligned}
$$

Substituting $R = F/2$ and $N = F/(2\mu)$ into the moment equation:

$$
-l F + e\, \frac{F}{2\mu} - \frac{F}{2}\, b = 0 \quad\rightarrow\quad -l + \frac{e}{2\mu} - \frac{b}{2} = 0
\quad\rightarrow\quad \boxed{e = 2\mu \left( l + \frac{b}{2} \right)}
$$

The force $F$ cancels out: the result **does not depend on how hard the clamp is tightened**. It is a self-locking mechanism: if the arm is narrow enough, it never slides, however large $F$ is. Numerically:

$$ e_{max} = 2 \cdot 0.20 \cdot \left( 80 + \frac{20}{2} \right) = 36~\text{mm} $$

For $e < 36$ mm, the moment created by $F$ forces $N$ to grow enough for the maximum friction $2\mu N$ to exceed $F$, and the arm does not slide.

**Check taking moments about $B$.** The result does not depend on the chosen point:

$$
\sum M_{B} = 0: \quad -(l+b) F + e N + R\, b = 0
\quad\rightarrow\quad
-(l+b) + \frac{e}{2\mu} + \frac{b}{2} = 0
\quad\rightarrow\quad
e = 2\mu \left( l + b - \frac{b}{2} \right) = 2\mu \left( l + \frac{b}{2} \right) \;✓
$$
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