Given that r = (10 N , 200º) and n = (16 N , 020º), find (3r – 2n).
- (62 N , 240º)
- (62 N , 200º)
- (62 N , 280º)
-
(62 N , 020º)
The correct answer is: D
Explanation
r = (10 N, 200º) and n = (16 N, 020º)
In rectangular form:
r = 10cos 200ºi + 10sin 200ºj = -9.397i - 3.420j
n = 16cos 20ºi + 16sin 20ºj = 15.035i + 5.472j
3r = -28.191i - 10.260j
2n = 30.070i + 10.945j
3r - 2n = (-28.191i - 10.260j) - (30.070i + 10.945j)
3r - 2n = -58.261i - 21.205j
|3r - 2n| = √((-58.261)\(^2\) + (-21.205)\(^2\)) = 62 N
\(tan θ =\frac{-21.205}{-58.261} = 0.3640\)
\(θ = tan^{-1} (0.3640) = 20^o\)
∴ (62 N , 020º)