A vector with a magnitude of 17.6 m is directed at angle θ = 43o counterclockwise from the +x axis. What are the components (a) ax and (b) ay of the vector? A second coordinate system is inclined by angle θ' = 20o with respect to the first. What are the components (c) a'x and (d) a'y in this primed coordinate system?
17.6m[43o]
a. X = 17.6*cos43
b. Y = 17.6*sin43
To find the components of a vector, we can use trigonometry. The magnitude of the vector represents the hypotenuse of a right triangle, and the components represent the lengths of the adjacent and opposite sides.
(a) To find the x-component (ax) of the vector, we can use cosine since it represents the adjacent side:
ax = magnitude * cos(θ)
ax = 17.6 m * cos(43°)
(b) To find the y-component (ay) of the vector, we can use sine since it represents the opposite side:
ay = magnitude * sin(θ)
ay = 17.6 m * sin(43°)
(c) To find the x-component (a'x) of the vector in the primed coordinate system, we need to consider the angle θ' and use cosine:
a'x = magnitude * cos(θ - θ')
a'x = 17.6 m * cos(43° - 20°)
(d) To find the y-component (a'y) of the vector in the primed coordinate system, we need to consider the angle θ' and use sine:
a'y = magnitude * sin(θ - θ')
a'y = 17.6 m * sin(43° - 20°)
To calculate the values, plug in the given angle values and evaluate the trigonometric functions using a scientific calculator or software.