(a) Vector E has magnitude 50.0 and is directed 40.0° counterclockwise from the +x axis. Express it in unit-vector notation.
(b) Vector F has magnitude 50.0 and is directed 40.0° counterclockwise from the +y axis. Express it in unit-vector notation.
(c) Vector G has magnitude 50.0 and is directed 40.0° counterclockwise from the -y axis. Express it in unit-vector notation.
I will be happy to critique your thinking.
This is what I went about doing but the signs don't make sense. Thanks for any help!
a.) 50*sin40=y y= 37.25j
50*cos40=x x= -33.35i
b.) 50*sin40=x x= 37.25i
50*cos40=y y= -33.35j
c.) 50*sin40=y y= 37.25j
50*cos40=x y= -33.35i
To express a vector in unit-vector notation, we use the i and j unit vectors, which are the unit vectors in the x and y directions, respectively.
(a) Vector E is directed 40.0° counterclockwise from the +x-axis. We can break it down into its x-component and y-component using trigonometry.
The x-component of Vector E is given by E_x = E * cos(theta), where E is the magnitude of the vector and theta is the angle it makes with the positive x-axis. Substituting the given values, E_x = 50.0 * cos(40.0°).
The y-component of Vector E is given by E_y = E * sin(theta), where E is the magnitude of the vector and theta is the angle it makes with the positive x-axis. Substituting the given values, E_y = 50.0 * sin(40.0°).
So, Vector E in unit-vector notation is E = E_x * i + E_y * j.
(b) Vector F is directed 40.0° counterclockwise from the +y-axis. Similarly, we can find its x-component and y-component using trigonometry.
The x-component of Vector F is given by F_x = F * cos(theta), where F is the magnitude of the vector and theta is the angle it makes with the positive x-axis. Substituting the given values, F_x = 50.0 * cos(40.0° - 90°).
The y-component of Vector F is given by F_y = F * sin(theta), where F is the magnitude of the vector and theta is the angle it makes with the positive x-axis. Substituting the given values, F_y = 50.0 * sin(40.0° - 90°).
So, Vector F in unit-vector notation is F = F_x * i + F_y * j.
(c) Vector G is directed 40.0° counterclockwise from the -y-axis. Similar to above, we can calculate its x-component and y-component.
The x-component of Vector G is given by G_x = G * cos(theta), where G is the magnitude of the vector and theta is the angle it makes with the positive x-axis. Substituting the given values, G_x = 50.0 * cos(40.0° + 90°).
The y-component of Vector G is given by G_y = G * sin(theta), where G is the magnitude of the vector and theta is the angle it makes with the positive x-axis. Substituting the given values, G_y = 50.0 * sin(40.0° + 90°).
So, Vector G in unit-vector notation is G = G_x * i + G_y * j.