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Physics Question. Please Help!?
Two dimensional vectors, A and B, have the same magnitude. The angle of orientation for vectors A and B are respectively θA = 42.0° and θB = 70.0°. The vector C is related to vectors A and B by the relationship C = A − 2B. The magnitude and direction for the vector C are |C| = 5.45 and θC = 272.8°. Determine the magnitude of the vectors A and B. (Assume all angles are measured counterclockwise from the positive x axis.)
VECTORS
2 Answers
- NCSLv 76 years agoFavorite Answer
So vector A = |A|*cos42.0º i + |A|*sin42.0º j = 0.743*|A| i + 0.669*|A| j
and vector B = |B|*cos70.0º i + |B|*sin70.0º j = 0.342*|B| i + 0.940*|B| j
and vector C = 5.45*cos272.8º i + 5.45*sin272.8º j = 0.266 i - 5.44 j
x-direction: Ax - 2*Bx = Cx
0.743*|A| - 2*0.342*|B| = 0.266
y-direction: Ay - 2*By = Cy
0.669*|A| - 2*0.940*|B| = -5.44
Two equations, two unknowns. I use wolfram, which yields
A ≈ 4.49 and
B ≈ 4.49
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- xyzzyLv 76 years ago
A = a cos 42, a sin 42
B = b cos 70, b sin 70
C = a cos 42 - 2b cos 70, a sin 42 - 2b sin 70
law of cos says
||C||^2 = a^2 + 4b^2 - 4ab cos (70-42) = (5.45)^2
and
tan 272.8 = (a sin 42 - 2b sin 70)/(a cos 42 - 2b cos 70)
there is some number crunching to do, but the bottom equation will simplify
a = K b
where k is a constant
and then you will make that substitution into the equation from the law of cos and you will get everything in terms of b^2 and solving for b will be easy enough after that.
