Question

Two forces if magnitudes 10N and 6N acting at an angle 60° with each other. The ratio of the magnitudes of the resultant of the two vectors to the difference of two vectors is

Answer 1

Answer:

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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 unit.

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.

Answer 2

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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 unit.

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.

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@MissWorst