Question

Two parallel 60-N forces are applied as shown to the corners A and C of a 200-mm square plate. Determine the moment of the couple formed by the two forces (a) by multiplying their magnitude by their perpendicular distance, (b) by resolving each force into horizontal and vertical components and adding the moments of the two resulting couples

Answer 1

We have that perpendicular distance and the two resulting couples

$m=-16.39m\\\\m=16.39Nm(cw)$

From the question we are told that

Two parallel 60-N forces

A and C of a 200-mm square plate.

Generally the equation for AC   is mathematically given as

$AC=\sqrt{0.2^2+(0.2^2)}\\\\AC=0.2\sqrt{2m}\\\\sin75=\frac{d}{AC}\\\\d=(0.2\sqrt{2})sin75\\\\d=0.2732m\\\\m=Fxd\\\\m=-60x0.2732m\\\\m=-16.39m$

b)

$mFm=Fm*0.2\\\\mFm=-30V_3x0.2\\\\mF_m=-6v_3\\\\mF_v=-6Nm\\\\Thereforem=16.39Nm(cw)$

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