Question

Two identical ladders are arranged as shown in figure. Mass of each ladder is $M$ and length is $L$. A load of mass $m$ has been attached at the touching point $\left(\text{O}\right)$ of the ladders, as shown in figure. The system is in equilibrium. Find the magnitude of static frictional force acting at either point $A$ and $B$.

• A. $\left(\frac{M+m}{2}\right)g\tan \theta$
• B. $\left(\frac{M+m}{2}\right)g\cot \theta$
• C. $\left(\frac{M+m}{2}\right)g\cos \theta$
• D. $\left(\frac{M+m}{2}\right)g\sin \theta$

Answer 1

The correct option is B $\left(\frac{M+m}{2}\right)g\cot \theta$

Let $N$ be the normal force acting at point $A$ and $B$ (same by symmetry)

Let $N_1$ and $N_2$ be the horizontal and vertical components of normal reaction exerted by each ladder on the other. (equal and opposite force will act on the two ladders at point $\text{O}$ according to Newton's 3rd law)

From symmetry of the ladder arrangement, both ladders will share the load $mg$ equally, hence $\frac{mg}{2}$ will be the load on each ladder.


Let $f$ be the friction acting between each ladder and the ground and due to symmetry, it will be same for both.
As the ladders are in equilibrium,
${\tau }_{\text{net}}=0,F_{\text{net}}=0$
For translational equilibrium,
$\sum F_x=0$
$\sum F_y=0$
For right ladder,
$N_2+\frac{mg}{2}+Mg=N...\left(i\right)$
and $N_1=f...\left(ii\right)$
For left ladder,
$N_2+N=Mg+\frac{mg}{2}...\left(iii\right)$
and $N_1=f$
From equation $\left(i\right)$ and $\left(iii\right)$,
$N-Mg-\frac{mg}{2}+N=Mg+\frac{mg}{2}$
$\Rightarrow 2N=2Mg+mg$
$\therefore N=Mg+\frac{mg}{2}...\left(iv\right)$

For rotational equilibrium, the net torque about $\text{O}$ should be zero on any ladder,
${\tau }_{\text{net}}=0...\left(v\right)$ for right ladder.
Taking anticlockwise sense of rotation for torque as $+ve$
$\&\tau =F\times r_{\perp }$
$\because$ Forces $N_1,N_2\&\frac{mg}{2}$ pass through reference point $\text{O}$, hence their torque about point $\text{O}$ is zero


Substituting torques with proper sign in eq $\left(v\right)$:
$-Mg\left(\frac{L}{2}\cos \theta \right)-f\left(L\sin \theta \right)+N\left(L\cos \theta \right)=0$
Putting value of $N$ from Eq $\left(iv\right)$,
$\Rightarrow \frac{Mg\cos \theta }{2}+f\sin \theta =(Mg+\frac{mg}{2})\cos \theta$
$\Rightarrow f\sin \theta =Mg\cos \theta -\frac{Mg\cos \theta }{2}+\frac{mg}{2}\cos \theta$
$\Rightarrow f\sin \theta =\left(\frac{M+m}{2}\right)g\cos \theta$
$\therefore f=\left(\frac{M+m}{2}\right)g\cot \theta$