Vector A⃗ points in the negative x direction. Vector B⃗ points at an angle of 30.0∘ above the positive x axis. Vector C⃗ has a magnitude of 19m and points in a direction 37.0∘ below the positive x axis. Given that A⃗ +B⃗ +C⃗ =0, find the magnitudes of A⃗ and B⃗ .
Mathematics · High School · Tue Nov 03 2020
Answered on
Given information:
Vector A points in the negative x direction.
Vector B points at an angle of 30° above the positive x-axis.
Vector C has a magnitude of 19m and points in a direction of 37° below the positive x-axis.
A + B + C = 0
We have to find the magnitude of A and B.
Vector A in the negative x direction is-a.
Vector B = b cos 30° - b sin 30°
Vector C = 19 cos 37° - 19 sin 37°
The equations are:
19 cos 37° + b cos 30° - a = 0 …eqn¹
19 in 37° - b sin 30° = 0 eqn²
With eqn²
19 in 37° = b sin 30°
19(0.6018) = b (0.5)
11.4342 = 0.5b
b = 11.4342/0.5
b = 22.8684
|b| = 22.8684 m
Put b's value in eqn¹
19 cos 37° + b cos 30° - a = 0
19(0.7986) + 22.8684(0.8660) - a = 0
15.1734 + 19.8040 = a
a = 19.8040
|a| = 19.8040 m
Therefore, the magnitudes of A and B are 34.9774 m and 22.8684 m.