Coplanar force 4N, 8N, 6N, 4N and 5N act at a point as shown in the diagram. If the 6N force acts in the direction 090°, calculate the : (a) magnitude of th...

Answer Details

Reading the diagram, the 6N force acts due East (direction 090°). Measuring anticlockwise from this 6N line, the other forces sit at these angles:

• 6N along East: \(0^\circ\)
• 8N: \(55^\circ\) above the 6N line, so \(55^\circ\)
• 4N (upper-left): a further \(100^\circ\) beyond the 8N, so \(55^\circ+100^\circ=155^\circ\)
• 4N (lower-right): \(25^\circ\) below East, so \(-25^\circ\) i.e. \(335^\circ\)
• 5N (lower-left): \(130^\circ\) round from the lower 4N, i.e. \(-25^\circ-130^\circ=-155^\circ\) i.e. \(205^\circ\)

(a) Resolve each force into components along the 6N direction (\(x\)) and perpendicular to it (\(y\)).

\[\sum F_x = 6\cos0^\circ+8\cos55^\circ+4\cos155^\circ+5\cos205^\circ+4\cos335^\circ\]

\[\sum F_x = 6+4.589-3.625-4.532+3.625 = 6.057\ \text{N}\]

\[\sum F_y = 6\sin0^\circ+8\sin55^\circ+4\sin155^\circ+5\sin205^\circ+4\sin335^\circ\]

\[\sum F_y = 0+6.553+1.690-2.113-1.690 = 4.440\ \text{N}\]

Magnitude of the resultant:

\[R=\sqrt{(\sum F_x)^2+(\sum F_y)^2}=\sqrt{6.057^2+4.440^2}=\sqrt{36.69+19.71}=\sqrt{56.40}\]

\[R\approx 7.51\ \text{N}\]

(b) Direction of the resultant. Both components are positive, so the resultant lies above the 6N (East) line.

\[\theta=\tan^{-1}\!\left(\frac{\sum F_y}{\sum F_x}\right)=\tan^{-1}\!\left(\frac{4.440}{6.057}\right)=\tan^{-1}(0.7332)\approx 36.2^\circ\]

The resultant is about 7.51 N acting at \(36.2^\circ\) above the 6N direction. Since the 6N points along a bearing of \(090^\circ\), this direction corresponds to a bearing of \(090^\circ-36.2^\circ\approx \mathbf{054^\circ}\).