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When we press an end of a stick/rod against a wall (rough enough to ensure that friction balances the weight) what happens is that owing to friction the point of normal force shifts to prevent toppling. I derived an expression to find the distance shift and it in turn depends on the length of rod. So does it mean that there has to be a restraint to how long a rod one can hold this way?enter image description here I derived an expression for the "x" using restraints (net torque=0) and got the below expression: $$x=\frac{mgl}{2F}$$ But is there an intuitive explanation for such a derivation. Does it really make sense that length of rod plays a role here?

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  • $\begingroup$ I would question the factor of (2) if (x) is the shift down from the line through the center of mass. $\endgroup$
    – R.W. Bird
    Commented Jul 29, 2021 at 19:24

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It makes sense, because the larger $l$ is, the larger the moment due to $mg$ about the point that static friction (which always balances $mg$) acts. This moment is trying to turn the rod counter-clockwise. As you've discovered, the location of the normal force must shift to balance this moment for static equilibrium. That is, $N$ provides a clockwise moment. You can't get it from $F$ because you have forced its line of action through the midpoint of the rod.

If the rod is long enough for a given force $F$, then you're right, $x$ will lie outside of the body and the rod will begin to tip counter-clockwise. However, one way to increase the moment from $N$ is to simply increase the pressing force $F$. This increases the magnitude of $N$ and hence its clockwise moment. You can also see it in your formula for $x$, which is inversely dependent on the applied load $F$.

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Let the maximum force you can exert on the left end of the rod be $F^\text{max}$ and the vertical width of the rod be $w$. Then your $\tau^\text{net} = 0$ equation yields $$l^\text{max} = \frac{wF^\text{max}}{mg}.$$ Any rods of length $l > l^\text{max}$ will fall (or at least move away from the horizontal).

It's helpful to check limiting cases here. When $m$ is very large, $l^\text{max} \rightarrow 0$. Think about holding a 100 lb dumbbell against a wall. Even were it quite short, would you be able to hold it in a strictly horizontal position, only by applying a horizontal force?

Now imagine that the dumbbell is 2 or 3 meters long, with a tiny vertical width. Wouldn't you need to push much harder to hold the dumbbell in place?

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To get an intuitive understanding of your formula, think of it in this rearrangement

$$F=\frac{mgl}{2x}$$

Any rod can be held up, but the force required increases as the length of the rod increases and the width decreases. If $2x$ is the width of the rod $w$, a force of at least

$$F=\frac{mgl}{w}$$

must be applied.

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The distance to the center of mass $\ell/2$ is a key factor as the moment arm of the weight that needs to be countered by the offset in axial reaction.

It is via this torque balance that you derived the equation

$$ \tau = ( mg) (\ell /2) = (F) (x) \tag{1} $$

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