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I had this question in a mock exam and for some reason our lecturer hasn't given us the solutions and the exam is next week.

Two equal ladders are freely joint at one end and stand at rest with the other ends on rough horizontal ground. One is twice as heavy as the other. Show that if the coefficient of the critical friction between the ladders and surface is $\mu$, then the angle between the ladders cannot be greater than $2 \tan^{-1} (5 \mu / 3)$.

I have started as follows:

Let $N_{1}$ be the normal force at point $A$, the base of the first ladder with weight $W$. $N_2$ is the normal force of the ladder with weight $2W$. Ladders have lenght $2L$ and the angle between them I have called $2 \alpha$.

Now when calculating moments about B, I get the friction force working anti-clockwise with a force of $\mu N_1$ at a distance of $\sin \alpha 2L$. The weight acting anti-clockwise with a force of $W$ and a perpendicular distance $ \sin \alpha L$ and the normal force acting clockwise with a force $N_1$ and a perpendicular distance $\sin \alpha L$. Which yields:

$$ \mu N_1 \sin \alpha 2L + W \sin \alpha L - N_1 \sin \alpha 2L=0 $$

Now if the angle created by the ladder to the horizontal was $\theta$ then $N_1$ would be $W \sin \theta$ which would be $W \cos \alpha$, but when I substitute this into that equation and rearrange for $\tan$ I get $2\tan^{-1}(4\mu -4) = \alpha$.

I'm clearly missing something or making some fundamental mistake. Perhaps I can't think of each ladder as a separate system.

Any help would be greatly appreciated.

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2 Answers 2

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Notice that the number of unknowns is $3$, $N_1$ and $N_2$, and $f$, which is the friction force. The friction force on both ladders should be equal and opposite, otherwise CM of system would accelerate and there wouldn't be equilibrium. All you need to do is to find two equations, and get the value of $f$ in terms of the weight and other given values. Also, the contact force between the two ladders in contact at the top most point, is in an action-reaction pair. First, write equate net torque about point $A$(this is not necessary, but prudent, as it minimizes number of variables) and equate it to $0$. Then equate net force to $0$ along vertical. This gives you the two equations. You will find $N_1=\frac {5W}{4}$ and $N_2=\frac {7W}{4}$. Then, having this information, draw FBD of a single ladder keeping in mind two components of contact force at the top. Also, write torque equation from centre of mass of ladder. This will give you $f$ in terms of $W$ and angle. All that is left is to do $f\leq \mu N_1$. You will find that the maximum angle comes to be $2arctan(\frac {5\mu}{6})$, and not $2arctan(\frac {5\mu}{3})$.

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  • $\begingroup$ That’s great, thanks. One thing I’m not quite sure on is the reactionary forces at the top where the ladders are joint. When calculating torque about $A$, the base of one of the ladders, will there still just be 2 forces to consider? The force from the other ladder acting at the top and the weight? Also, does the force of the other ladder at the top act horizontally? Thank you again in advance $\endgroup$
    – Jbo
    May 15, 2021 at 8:17
  • $\begingroup$ No take two components for the force of the other ladder, a horizontal and a vertical component. So there would be three forces, namely, weight, and the two components of contact force from other ladder. $\endgroup$ May 15, 2021 at 8:35
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Considering the virtual work made by the involved forces, we have located at the ladder's barycenters respectively at $G_1=\frac l2(\sin(\frac{\alpha}{2}),\cos(\frac{\alpha}{2}))$ and $G_2=(l\sin(\frac{\alpha}{2})+\frac l2 \sin(\frac{\alpha}{2}),\frac l2\cos(\frac{\alpha}{2}))$. Assuming the left ladder pivoting without slip, and the right ladder end located at $B=(2l\sin(\frac{\alpha}{2}),0)$. A virtual displacement of $B$ due to a small $\alpha$ increment $\delta\alpha$ will produce

$$ \delta G_1(\alpha)\cdot(0,1)W_1+\delta G_2(\alpha)\cdot(0,1)W_2 = \delta B(\alpha)\cdot(1,0)H $$

or

$$ -\frac l2\sin(\frac{\alpha}{2})\frac{\delta\alpha}{2}W_1-\frac l2\sin(\frac{\alpha}{2})\frac{\delta\alpha}{2}W_2=2l\cos(\frac{\alpha}{2})\frac{\delta\alpha}{2}H $$

or

$$ -(W_1+W_2)\sin(\frac{\alpha}{2}) = 4\cos(\frac{\alpha}{2})H $$

where $H$ is the horizontal resistant force. Now considering $W_1 = - mg$ and $W_2 = -2mg$ we have $H = \mu 2m g$ and thus $\tan(\frac{\alpha}{2})\le \frac 83\mu$. Now making $W_1 = - 2mg$ and $W_2 = -mg$ we have analogously $\tan(\frac{\alpha}{2})\le \frac 43\mu$ so we follow with

$$ \tan\left(\frac{\alpha}{2}\right)\le \frac 43 \mu\Rightarrow \alpha \le 2\arctan\left(\frac 43 \mu\right) $$

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