A uniform ladder $24 m$ long weights $400 N$ It is placed against a vertical wall at an angle of $60^{\circ}$ with the ground. How far along the ladder can a $1600 N$ man climb before the ladder is on the verge of slipping? The angle of friction at all contact surfaces is $15^{\circ}$

12 m

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20 m

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20 m

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12 m

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Solution

The correct option is D $12 m$

$μ = t a n ϕ = t a n 15^{\circ} = 0.268$

$\sum F_{X} = 0$

$∴ 0.268 N_{A} = N_{B} . . . . . . . . . . . . (1)$

$\sum F_{Y} = 0$

$∴ N_{A} - 400 - 1600 + 0.268 N_{B} = 0$

Using Equation 1
$∴ N_{A} + 0.268 (0.268 N_{A}) = 2000$

$∴ N_{A} = 1865.98 N; N_{B} = 500.08 N$

$\sum M_{A} = 0$

$∴ 400 (12 C o s 60) + 1600 (X C o s 60) - 0.268 N_{B}$
Using
$N_{B} = 500.08 N$

$∴ X = 12 m$

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A uniform ladder 24m long weights 400N It is placed against a vertical wall at an angle of 60∘ with the ground. How far along the ladder can a 1600N man climb before the ladder is on the verge of slipping? The angle of friction at all contact surfaces is 15∘

The correct option is D 12m μ=tanϕ=tan15∘=0.268 ∑FX=0 ∴0.268NA=NB............(1) ∑FY=0 ∴NA−400−1600+0.268NB=0 Using Equation 1 ∴NA+0.268(0.268NA)=2000 ∴NA=1865.98N;NB=500.08N ∑MA=0 ∴400(12Cos60)+1600(XCos60)−0.268NB Using NB=500.08N ∴X=12m

A uniform ladder $24 m$ long weights $400 N$ It is placed against a vertical wall at an angle of $60^{\circ}$ with the ground. How far along the ladder can a $1600 N$ man climb before the ladder is on the verge of slipping? The angle of friction at all contact surfaces is $15^{\circ}$