A ladder AB of length 5 m and weight (W) 600 N is resting against a wall. Assuming frictionless contact at the floor (B) and the wall (A), the magnitude of the force P (in newton) required to maintain equilibrium of the ladder is

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Solution

Drawing FBD of ladder AB
$tan θ = \frac{3}{4}$
$(\sum F)_{y} = 0$
$\Rightarrow 600 N = N_{B}$

Taking moment about A
$- 600 \times 2.5 \times cos θ + N_{B} \times 5 \times cos θ - P \times 5 \times sin θ = 0$
$\Rightarrow - 600 \times 2.5 \times \frac{4}{5} + 600 \times 5 \times \frac{4}{5} - P \times 5 \times \frac{3}{5} = 0$
$\Rightarrow P = 400 N$

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A ladder AB of length 5 m and weight (W) 600 N is resting against a wall. Assuming frictionless contact at the floor (B) and the wall (A), the magnitude of the force P (in newton) required to maintain equilibrium of the ladder is

Drawing FBD of ladder AB tanθ=34 (∑F)y=0 ⇒600N=NB Taking moment about A −600×2.5×cosθ+NB×5×cosθ−P×5×sinθ=0 ⇒−600×2.5×45+600×5×45−P×5×35=0 ⇒P=400N