Question

Two identical ladders are arranged as shown in the figure. Mass of each ladder is $M$ and length $L$. The system is in equilibrium. Find direction and magnitude of friction force acting at A or B.

• A. $f=(\frac{M+m}{2})g\tan \theta$ horizontally outwards
• B. $f=(\frac{M+m}{2})g\cot \theta$ horizontally inwards
• C. $f=(\frac{M+m}{2})g\cos \theta$ horizontally outwards
• D. None of these.

Answer 1

The correct option is B $f=(\frac{M+m}{2})g\cot \theta$ horizontally inwards

Drawing the FBD of both rods
As the rods are in equilibrium

$\Sigma F_X=0$

$\Sigma F_y=0$

${\tau }_{net}=0$
For right rod

$N_2+\frac{mg}{2}+Mg=N$ (1)

$N_1=f$ (2)
For left rod

$N_2+N=Mg+\frac{mg}{2}$ (3)

From $\left(1\right)and\left(3\right)N_1=f$

$=N_2=0=N=Mg+\frac{mg}{2}$

Balancing torque about $O$,

$Mg\frac{L}{2}\cos \theta +fL\sin \theta =NL\cos \theta$

$\frac{Mg\cos \theta }{2}+f\sin \theta =(Mg+\frac{mg}{2})\cos \theta$

$f=(\frac{Mg}{2}+\frac{mg}{2})\cot \theta =f=(\frac{M+m}{2})g\cot \theta$