Two forces of magnitude 6n and 10n are inclined at an angle of 60 degrees with each other. What is the magnitude of resultant and the angle made by resultant with 6n force?

 Two forces of magnitude 6n and 10n are inclined at an angle of 60 degrees with each other. What is the magnitude of resultant and the angle made by resultant with 6n force?

Solution

Sketch a triangle starting with a line (Call it b) of 6 units, turn through an angle of 60° (relative to the direction you were heading) and draw line (c) a length of 10 units. Now connect the beginning of line (b) to the end of line (c). Call this line (a). You should now have a triangle with an angle of 120° opposite the side (a). Mark this angle (A). Mark the angles (B), opposite side b, and (C), likewise. If you haven’t got the triangle described, try again.

If you drew this accurately and to scale you could directly measure the length of (a) to obtain its magnitude and the angle (C) to obtain the required angle, but we can determine both the length (a) and the angle (C) using the Cosine and Sine rules.

We first use the Cosine rule to determine a → a.a = b.b+c.c-2b.c.CosA → a = sqrt(b.b+c.c-2b.c.CosA)

a = sqrt(36+100–120.Cos120) = 14N

Now we’ll use the Sine rule to find angle C: SinA/a = SinC/c → C = arcsin(c.sinA/a) → C = arcsin(10x(-0.5)/14) = -20.9° (the negative is merely the direction of the turn.

Answer to the question: the resultant vector is a force of 14N at an angle of 20.9° to the 6N vector.