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?
To find the components of a vector, you can use trigonometry.
Given that the magnitude of the vector is 17.6 m and it is directed at an angle of 43 degrees counterclockwise from the +x axis, we can find the components in the original coordinate system.
The x-component (ax) can be found using cosine function:
ax = magnitude * cos(angle)
ax = 17.6 m * cos(43°)
Using a calculator, we find:
ax ≈ 12.51 m
The y-component (ay) can be found using sine function:
ay = magnitude * sin(angle)
ay = 17.6 m * sin(43°)
Using a calculator, we find:
ay ≈ 11.13 m
Therefore, in the original coordinate system:
(a) ax ≈ 12.51 m
(b) ay ≈ 11.13 m
Now, let's find the components in the primed coordinate system, which is inclined by an angle of 20 degrees with respect to the original coordinate system.
To transform the coordinates, we can use rotation matrices. The x-component in the primed system (a'x) can be obtained by rotating the x-component in the original system (ax) by the angle of rotation (θ'):
a'x = ax * cos(θ') - ay * sin(θ')
Substituting the values:
a'x = 12.51 m * cos(20°) - 11.13 m * sin(20°)
Using a calculator, we find:
a'x ≈ 10.93 m
Similarly, the y-component in the primed system (a'y) can be obtained by rotating the y-component in the original system (ay) by the angle of rotation (θ'):
a'y = ax * sin(θ') + ay * cos(θ')
Substituting the values:
a'y = 12.51 m * sin(20°) + 11.13 m * cos(20°)
Using a calculator, we find:
a'y ≈ 16.47 m
Therefore, in the primed coordinate system:
(c) a'x ≈ 10.93 m
(d) a'y ≈ 16.47 m