A uniform plank of weight $W$ and total length $2L$ is placed as shown in figure with its ends in contact with the inclined planes. the angle.of friction is $15°$ . determine the maximum value of the angle $a$ at which slipping impends.

 

Appling equilibrium equations,we get

$\sum _{ }^{ }x=F_A\cos 60+N_A\sin 60+F_B\cos 45-N_B\sin 45=0$

Also we know htat

$F_A=0.268N_A$ and $F_B=0.268N_B$

Solving above equations we get $N_{A}0.158N_B$

$\sum _{ }^{ }Y=N_A\cos 60-F_A\sin 60+N_B\cos 45+F_B\sin 45=W$

Solving above equations we get

$N_B=0.966W$and $F=0.259W$

Taking moment about A and equating it to zero, we get

$\sum _{ }^{ }{M}_A=(w\times L\cos \alpha )-(N_B \cos 45°\times 2L\cos \alpha )+(N_B \cos 45°\times 2L\cos \alpha )-(F_B\sin 45°\times 2L\sin \cos \alpha )-(F_B\cos 45°\times 2L\sin \sin \alpha )=0$

By putting the values of known quantities in above equation we get $\alpha =36.2°$