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A uniform ladder of mass m and length L rests against the wall. The coefficients of static friction between the floor and the ladder and between the wall and the ladder are equal to each other (μ). What is the maximum value of angle θ that the ladder can make with the wall without sliding.

My way to solve this: since ladder is not moving

Torque on the ladder $$ -\frac{mgL}{2}\sin\theta-\mu N_{f}L\cos\theta+N_{f}L\sin\theta=0$$ Horizontal forces $$ N_{w}-f_{f}=0$$ Vertical forces $$ N_{f}+f_{w}-mg=0$$

where $N_{f}$ and $N_{w}$ is normal forcé due to floor and Wall, $f_{w}$ and $f_{f}$ friction due to Wall and floor. Now I solve for $N_{f}$ the first equation, to get: $$N_{f}=\frac{mg\tan\theta}{2(\tan\theta-\mu)}$$ Sincé the angle is the situation where the ladder is about to start sliding, friction forces will have their máximum value, so: $$f_{f}=\mu N_{f}$$ from 2º equation: $$N_{w}=f_{f}$$ and for same reason like befor $$f_{w}=\mu N_{w}$$

substituting all this values in the 3º equation I get: $$\frac{\tan\theta}{2(\tan\theta-\mu)}(1+\mu^{2})=1$$ and solving this for $\tan\theta$: $$\tan\theta=\frac{2\mu}{1+\mu^{2}}$$ but aparently this is wrong. Could someone check this and point where is the mistake. Ty

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  • $\begingroup$ About which point are you taking the torque? Also is the ladder leaning to the left or right in your problem. This is so we can check the equations. $\endgroup$ – ja72 Aug 3 '14 at 22:57
  • $\begingroup$ Related: physics.stackexchange.com/q/110680/392 $\endgroup$ – ja72 Aug 3 '14 at 23:23
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You can take the last two equations to solve for the normal forces and use them in the torque equation

$$ \left. \begin{align} N_f & = \frac{m g}{1+\mu^2} \\ N_w &= \frac{\mu m g}{1+\mu^2} \end{align} \right\} m g \ell \left(\frac{1}{1+\mu^2}-\frac{1}{2} \right)\sin\theta - m g \ell \frac{\mu}{1+\mu^2} \cos\theta =0 $$

$$ \left(\frac{1}{1+\mu^2}-\frac{1}{2}\right)\sin\theta = \frac{\mu}{1+\mu^2} \cos\theta $$

$$ \tan \theta = \frac{2 \mu}{2-(1+\mu^2)} = \frac{2 \mu}{1-\mu^2} $$

So it seems you did an algebraic error somewhere (not shown) to get $1+\mu^2$ in the denominator.

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Take different co-officient of friction between wall and ladder and between floor and ladder..

THen assuming equal values find it out..

I think you will get around 56*

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  • 1
    $\begingroup$ $\mu$ does not appear to be given a value, so the final answer must be given as a function of it, so it is entirely unclear where 56* comes from. Note also that "assuming equal values find it out.." is entirely unhelpful because OP got stuck while trying to "find it out." $\endgroup$ – Kyle Kanos Feb 9 '15 at 18:11

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