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A ladder of 4 m length weighing 200N is placed as shown in fig. 13 ?b=0.25 & ?A=0.35. Calculate the minimum horizontal force to be applied at A to prevent slipping.

written 4.6 years ago by teamques10 ★ 70k

modified 4.0 years ago by krithikkm200 • 10

engineering mechanics

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written 4.6 years ago by teamques10 ★ 70k

$\sum M_A=0$

$-200(2\cos60^0)+0.25R_B(4\cos60^0)-600(3\cos60^0)+0.25R_B(4\sin60^0)=0$

$R_b = 277.49 N$

$\sum F_y = 0$

$\\ .\cdot. R_A-200-600+0.25R_B= 0$

$\therefore R_A =277.49N$

$\sum F_x=P+0.35R_A-R_B=0$

$ P+0.35(730.63)-277.49=0$

$\\ \\ \\\\ .\cdot. P=21.77N$

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