Vector C has a magnitude 23.4 m and is in the direction of the negative y axis.

Question

Vector C has a magnitude 23.4 m and is in the direction of the negative y axis. Vectors A and B are at angles alpha=43.4 degrees and beta=27.7 degrees up from the x-axis respectively. If the vector sum A+B+C=0, what are the magnitudes of A and B? Okay, I found out that the magnitude of A is 22 by using the Law of Sines. Now, how do I figure out the magnitude of B? I keep trying but I can't seem to get the correct answer.

Explanation / Answer

First start with defining vector A : This is at an angle of 43.3 deg to x-axis so it can be written as A = p cos(43.4) i + p sin(43.4) j p being the magnitude of A Similarly for B, B = q cos(27.7) i + q sin (27.7)j C is given as 23.4j also we have, A+B+C = 0; so [p cos(43.4) + q cos(27.7)] i + [p sin(43.4) + q sin(27.7)+23.4] j =0 i.e. p cos(43.4) + q cos(27.7) = 0 ---> p/q = cos(27.7) /cos(43.4) = 1.22 and p sin(43.4) + q sin(27.7)+23.4 = 0 ---> p*sin(43.4) + q* sin(27.7) +23.4 = 0 but from first equation, p = 1.22*q so, 1.22*q * sin(43.4) + q* sin(27.7) + 23.4 = 0 --> q =23.4/(sin(43.4)*1.22+sin(27.7)) = 23.4/1.303 = 17.65 so magnitude of A = p = 1.22*17.65 = 21.533 magnitude of B = q = 17.65

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