Find the direction of the resultant of the forces in the diagram.

Answer Details

Reading the diagram, the three forces make these angles, measured anticlockwise from the positive \(x\)-axis:

• \(8\sqrt2\ \text{N}\): \(45^\circ\) from the \(y\)-axis in the first quadrant, i.e. at \(45^\circ\)
• \(10\ \text{N}\): \(30^\circ\) from the \(y\)-axis in the second quadrant, i.e. at \(120^\circ\)
• \(20\ \text{N}\): \(60^\circ\) below the negative \(x\)-axis in the third quadrant, i.e. at \(240^\circ\)

Resolve horizontally:

\[\sum F_x=8\sqrt2\cos45^\circ+10\cos120^\circ+20\cos240^\circ.\]

\[\sum F_x=8\sqrt2\cdot\tfrac{\sqrt2}{2}+10(-0.5)+20(-0.5)=8-5-10=-7\ \text{N}.\]

Resolve vertically:

\[\sum F_y=8\sqrt2\sin45^\circ+10\sin120^\circ+20\sin240^\circ.\]

\[\sum F_y=8+8.660+(-17.321)=-0.660\ \text{N}.\]

Both components are negative, so the resultant lies in the third quadrant, just below the negative \(x\)-axis. The acute angle it makes with the negative \(x\)-axis is

\[\alpha=\tan^{-1}\!\left(\frac{|\sum F_y|}{|\sum F_x|}\right)=\tan^{-1}\!\left(\frac{0.660}{7}\right)=\tan^{-1}(0.0943)\approx 5.4^\circ.\]

Measured anticlockwise from the positive \(x\)-axis, the direction of the resultant is

\[\theta=180^\circ+5.4^\circ\approx \mathbf{185.4^\circ}.\]

That is, the resultant points along the negative \(x\)-direction, inclined about \(5.4^\circ\) below it. (For reference its magnitude is \(\sqrt{(-7)^2+(-0.66)^2}\approx 7.03\ \text{N}\).)