A uniform beam AB hinged at A is kept horizontal by supporting and setting a 100 N weight by using a string atied at B and passing over a smooth pulley at C.

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written 7.9 years ago by

teamques10 ★ 68k

FBD:

Pulley:

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Block:

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Beam:

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$θ = \tan^{-1} (\dfrac 36)=26.56^0 $

From FBD of block,

$∑f_y=0⟹R+T=100 \\ R=100-T -------(1) $

From FBD of Beam,

$∑f_x=0⟹ H_A-T \cos 26.56=0 \\ H_A=T \cos 26.56 -----(2) \\ ∑f_y=0⟹V_A-22.5-R+T \sin 26.56=0 \\ V_A-22.5-100+T+T \sin 26.56=0 \space\space from (1) \\ ∴V_A+1.447T=122.5 -----(3) \\ M_A=0⟹ -22.5×1-R×4+T \sin 26.56×10=0 \\ -22.5-4(100-T)+4.47T=0 \\ 422.5=8.47T \\ =49.88 N \\ R= 50.11 N ------- using (1) \\ H_A=44.62 N using (2) \\ H_A=50.323 N \space\space\space using (3) $

From FBD of Pully,

$∑f_y=0⟹H_B= -T \cos 26.56 \\ ∴H_B=-44.62N \\ ∑f_y=0⟹V_B= T+T \sin 26.56 \\ ∴V_B=72.183 N \\ R_A=\sqrt{44.62^2+50.32^2 }=67.25 N \\ θ = \tan^{-1} (\dfrac {50.32}{44.62})=48.43^\circ \\ R_A=67.25 N\lt48.43^0 \\ R_B= \sqrt{44.62^2+72.183^2} =84.86 N \\ θ= \tan^{-1} (\dfrac {72.183}{44.62})=58.28^0 \\ R_B=84.86N\gt58.28^0$

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