Four forces and a couple acting on a plate as shown in figure. Determine the resultant force and locate it with respect to point A.

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written 7.9 years ago by

teamques10 ★ 68k

• modified 4.5 years ago

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engineering mechanics

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written 7.9 years ago by

teamques10 ★ 68k

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$∑f_x=200 \cos ⁡36.87+ 100=260N \\ ∑f_y =40-80-200 \sin⁡36.87= -160 N $

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$R=\sqrt{260^2+ (-160)^2 } \\ R= 305.286 N \\ θ= \tan^{-1} (\dfrac {260}{160}) =31.61^\circ \\ M_A=-100×200+40×200+80×300-40×10^3+200 \sin⁡36.87×600+200 \cos 36.87 × 200 \\ M_A=76 ×10^3 N.mm $

Let resultant act at perpendicular distance of x m from A.

$x= \dfrac {M_A}R = \dfrac {76 ×10^3}{305.286} =248.95mm $

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