m=60 kg, ladder length =6.5 m, height of the wall =6 m
Therefore torque due to the weight of the body
a) τ=600×6.5/2sinθ=i
⇒τ=600×6.5/2×√[1−(6/6.5)2]
⇒τ=735 N−m.
b) R2=mg=60×9.8
R−1=μR2⇒6.5R1cosθ=60gsinθ×6.5/2
⇒R1=60 gtanθ=60 g×(2.5/12) [because tanθ=2.5/6]
⇒R1=(25/2)g=122.5 N