Question

In the sum A→+B→=C→, vector A→ has a magnitude of 13.1 m and is angled 43.6° counterclockwise from the +x direction, and vector C→ has a magnitude of 15.4 m and is angled 15.9° counterclockwise from the -x direction. What are (a) the magnitude and (b) the angle (relative to +x) of B→? State your angle as a positive number.

Answer 1

Solution:

Given,

|A| = 13.1 m , |C| = 15.4

_A = 43.6^o with +ve x-axis and _B = 15.9^o below -ve x-axis = 15.9^o + 180^o = 195.9⁰ with +ve x axis

A = 13.1[Cos_A i + Sin_A j ] = 9.49 i + 9.03 j

C = 15.4[Cos_C i + Sin_C j] = -14.81 i - 4.22 j

Given,

A + B = C

B = C - A

= [-14.81 -9.49] i + [-4.22 - 9.03] j

= [-24.3 i - 13.25 j] m

The magnitude of B,

|B| = 27.7 m

_B = Tan^-1 [-13.25/-24.3] = 180 + Tan^-1 [13.25/24.3] = 208.6^o. counter clockwise from +ve x-axis.

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