Question

The two vectors A and B in the figure have equal magnitudes of 13.5 m and the angles are θ1 = 33° and θ2 = 110°. How to find (a) the x component and (b) the y component of their vector sum R , (c) the magnitude of R, and (d) the angle R ?

Answer 1

Here's what I got.

Explanation:

I don't wave a good way of drawing you a diagram, so I'll try to walk you through the steps as they come along.

So, the idea here is that you can find the `x`-component and the `y`-component of the vector sum, `R`, by adding the `x`-components and `y`-components, respectively, of the `vec(a)` and `vec(b)` vectors.

For vector `vec(a)`, things are pretty straighforward. The `x`-component will be the projection of the vector on the `x`-axis, which is equal to

`a_x = a * cos(theta_1)`

Likewise, the `y`-component will be the projection of the vector on the `y`-axis

`a_y = a * sin(theta_1)`

For vector `vec(b)`, things are a little more complicated. More specifically, finding the corresponding angles will be a little tricky.

The angle between `vec(a)` and `vec(b)` is

`theta_3 = 180^@ - theta_2 = 180^@ - 110^@ = 70^@`

Draw a parallel line to the `x`-axis that intersects the point where the tail of `vec(b)` and head of `vec(a)` meet.

In your case, line `m` will be the `x`-axis and line `a` the parallel line you draw.

In this drawing, `angle6` is `theta_1`. You know that `angle6` is equal to `angle3`, `angle2`, and `angle7`.

The angle between `vec(b)` and the `x`-axis will be equal to

`180^@ - (theta_1 + theta_2) = 180^@ - 143^@ = 37^@`

This means that the `x`-component of vector `vec(b)` will be

`b_x = b * cos(37^@)`

Now, because the angle between the `x`-component and the `y`-component of a vector is equal to `90^@`, it follows that the angle for the `y`-component of `vec(b)` will be

`90^@ - 37^@ = 53^@`

The `y`-component will thus be

`b_y = b * sin(37^@)`

Now, keep in mind that the `x`-component of `vec(b)` is oriented in the opposite direction of the `x`-component of `vec(a)`. This means that the `x`-component of `vec(R)` will be

`R_x = a_x + b_x`

`R_x = 13.5 * cos(33^@) - 13.5 * cos(37^@)`

`R_x = 13.5 * 0.04 = color(green)("0.54 m")`

The `y`-components are oriented in the same direction, so you have

`R_y = a_y + b_y`

`R_y = 13.5 * [sin(110^@) + sin(37^@)]`

`R_y = 13.5 * 1.542 = color(green)("20.82 m")`

The magnitude of `vec(R)` will be

`R^2 = R_x^2 + R_y^2`

`R = sqrt(0.54""^2 + 20.82""^2)" m" = color(green)("20.83 m")`

To get the angle of `vec(R)`, simply use

`tan(theta_R) = R_y/R_x implies theta_R = arctan(R_y/R_x)`

`theta_R = arctan((20.82color(red)(cancel(color(black)("m"))))/(0.54color(red)(cancel(color(black)("m"))))) = color(green)(88.6""^@)`