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A ladder of 4 m length weighing 200 N is placed as shown in fig . $\mu_B=0.25$ & $\mu_A =0.35$ Calculate the minimum horizontal force to be applied at A to prevent slipping
written 8.4 years ago by gravatar for teamques10 teamques10 ★ 68k •   modified 4.5 years ago

enter image description here
engineering mechanics
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written 8.4 years ago by gravatar for teamques10 teamques10 ★ 68k •   modified 8.4 years ago

Angle of plane, $θ = \tan^{-1}(4/12) = 18.4$

Solution:

FBD of the ladder can be drawn as follows

enter image description here

$∑ Fx = 0 \\ P + 0.35 Na – Nb = 0 … i \\ ∑Fy = 0 \\ -200 – 600 + Na + 0.25Nb = 0 \\ Na + 0.25Nb = 800 … ii \\ ∑M = 0 (\text {about point A }) \\ -200 \times 2\cos60 -600 \times 3 \cos60 + Nb \times 4 \sin 60 + 0.25 Nb \times 4 \cos60 = 0 \\ Nb = 277.5 N $

Put in ii to get $Na = 730.6 N$

Put in (i) to get $P = 21.8N$
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