Let vector B = 8.8 m, 75 degrees counterclockwise from the vertical.

a) Find the x- and y-components of vector B in the normal coordinate system (where x is a horizontal line and y is a vertical line...both intersect each other).

b) Find the x- and y-components of vector B in the coordinate system where x and y are tilted such that x is 30 degrees above a 180 degree plane and y intersects x at 90 degrees.

Sorry, it's hard to explain.
Thanks

Thanks :)

No problem! I can help you with that. Let's break it down step by step.

a) To find the x and y-components of vector B in the normal coordinate system, we can use trigonometry.

The magnitude of vector B is given as 8.8 m, and it is at an angle of 75 degrees counterclockwise from the vertical.

The vertical component, B_y, can be found using sine:
B_y = 8.8 m * sin(75 degrees)
B_y ≈ 8.587 m

The horizontal component, B_x, can be found using cosine:
B_x = 8.8 m * cos(75 degrees)
B_x ≈ 2.302 m

Therefore, in the normal coordinate system, the x-component (B_x) of vector B is approximately 2.302 m, and the y-component (B_y) is approximately 8.587 m.

b) To find the x and y-components of vector B in the tilted coordinate system, we need to consider the new angle between the x-axis and the direction of vector B.

The angle between the normal x-axis and the new x-axis is 30 degrees above a 180 degree plane. This means the new x-axis is inclined at an angle of -30 degrees from the normal x-axis.

To find the new x-component (B_x') and y-component (B_y') in the tilted coordinate system, we can use trigonometric functions:

B_x' = B_x * cos(-30 degrees) - B_y * sin(-30 degrees)
B_y' = B_x * sin(-30 degrees) + B_y * cos(-30 degrees)

Now, let's substitute the values we found in part (a):

B_x' = 2.302 m * cos(-30 degrees) - 8.587 m * sin(-30 degrees)
B_x' ≈ 1.991 m

B_y' = 2.302 m * sin(-30 degrees) + 8.587 m * cos(-30 degrees)
B_y' ≈ -8.481 m

Therefore, in the tilted coordinate system, the x-component (B_x') of vector B is approximately 1.991 m, and the y-component (B_y') is approximately -8.481 m.

No problem, I understand your question. Let's start by finding the x- and y-components of vector B in the normal coordinate system (where x is horizontal and y is vertical).

a) In the normal coordinate system, the x-component, Bx, represents the horizontal displacement and the y-component, By, represents the vertical displacement.

To find the x-component, Bx, we can use the formula:
Bx = B * cos(θ)

where B is the magnitude of vector B (8.8 m) and θ is the angle counterclockwise from the vertical (75 degrees).

Bx = 8.8 * cos(75)
Bx = 8.8 * cos(75 * π/180) [converting degrees to radians]
Bx ≈ 8.8 * 0.259
Bx ≈ 2.279 m

To find the y-component, By, we can use the formula:
By = B * sin(θ)

By = 8.8 * sin(75)
By = 8.8 * sin(75 * π/180) [converting degrees to radians]
By ≈ 8.8 * 0.966
By ≈ 8.519 m

Therefore, in the normal coordinate system, the x-component of vector B is approximately 2.279 m and the y-component is approximately 8.519 m.

Now let's move on to finding the x- and y-components of vector B in the coordinate system where x and y are tilted.

b) In this coordinate system, we have a rotation of 30 degrees counterclockwise from the normal coordinate system.

To find the x-component, Bx', we can use a similar formula and adjust for the rotation:
Bx' = B * cos(θ - α)

where α is the angle of rotation (30 degrees).

Bx' = 8.8 * cos(75 - 30)
Bx' = 8.8 * cos(45) [calculating 75 - 30]
Bx' ≈ 8.8 * 0.707
Bx' ≈ 6.223 m

To find the y-component, By', we can use a similar formula and adjust for the rotation:
By' = B * sin(θ - α)

By' = 8.8 * sin(75 - 30)
By' = 8.8 * sin(45) [calculating 75 - 30]
By' ≈ 8.8 * 0.707
By' ≈ 6.223 m

Therefore, in the coordinate system where x and y are tilted, the x-component of vector B is approximately 6.223 m and the y-component is approximately 6.223 m.

a. A = 90 + 75 = 165 deg.,CCW.

B = (8.8m,165deg.).
X = hor. = 8.8cos165 = -8.50m.
Y = ver. = 8.8sin165 = 2.28m.

b. X = 8.8cos(180-30) = -7.62m.
Y = 8.8sin(90-30) = 7.62m.