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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.

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Mumbai University > FE > Sem 1 > Engineering Mechanics

Marks : 08

Years : MAY 2015

1 Answer
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FBD of the ladder can be drawn as follows

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$∑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 \cos 60 -600 \times 3 \cos 60 + 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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