The following forces act at a point 20N inclined at 30° towards north of east, 25N towards north, 30N towards northwest 35N inclined at 40° towards southwest. Find the magnitude and directions of the resultant force.
Answers
Resolving the forces along x and y-axes, and then determine algebraic sum of horizontal components and vertical components:
\xrightarrow {+} \sum H=30 – 35 \cos 45^{\circ} – 25 \cos 40^{\circ}
+
∑H=30–35cos45
∘
–25cos40
∘
= –13.90 N (here ‘–’ sign indicate that \sum H∑H is acting toward negative x direction i.e. towards West)
+\uparrow \sum V=20 + 35 \sin 45^{\circ} – 25 \sin 40^{\circ} – 40+↑∑V=20+35sin45
∘
–25sin40
∘
–40
= –11.32 N (here ‘–’ sign indicate that \sum V∑V is acting toward negative y direction i.e. towards South)
Now we know that resultant R is calculated as,
R=\sqrt{(\sum H)^{2}+(\sum V)^{2}}R=
(∑H)
2
+(∑V)
2
\therefore R=\sqrt{(13.90)^{2}+(11.32)^{2}}∴ R=
(13.90)
2
+(11.32)
2
\therefore R = 17.92 N ∴ R=17.92N
Let \thetaθ be the angle of resultant R with x-axis, then
\tan \theta =\frac{\sum V}{\sum H} tanθ=
∑H
∑V
\therefore \tan \theta =\frac{11.32}{13.90} \therefore \theta =39.15^{\circ}∴ tanθ=
13.90
11.32
∴ θ=39.15
∘
Now angle of resultant with East = 180° + 39.15° = 219.15°