Intuition behind this question/answer? - looking to understand

[Inertigratus]

Force vectors

Nevermind, I figured this one out.

Homework Statement

There's a pole standing straight in the ground, it is kept stable by two wires so the problem in the textbook looks kind of like a 60 triangle with the pole cutting it in half. One of the wires are longer though, and there are two forces pulling the pole in each direction.

They are asking for the magnitude of the force F2 which is a vector located on the right
hand wire with a direction towards the bottom right corner.

No numbers are known, the force F1 on the other wire is said to be equal to "P". There are two angles aswell, alpha (on the left side corner) and beta (on the right side corner).
There is also a sum of forces, F, which is vertical from the top corner.

They're asking for the magnitude of F as well.

Homework Equations

F = F1 + F2, obviously.
F1 = P (a constant).
e(x) & e(y) = unit vectors.

The Attempt at a Solution

As I said, I'm new to using vectors. I tried breaking down the vectors into x and y components though. The book we're using really doesn't explain much at all, I've seen people use equations like magnitude*cos(angle) to find the x-component and magnitude*sin(angle) to find the y-component but I have yet to figure out why.

The book says that the x-component is a projection of the vector in the x-direction and you get it by doing the dot product between the vector and the unit vector. However, the same thing is said for the y- and z-components, cos for those two as well.
The only place that the sine function is in, is the cross product.

Anyway, I got F1 = Pcos(a) * e(x) + Pcos(a) * e(y) and F2 = |F2|cos(b) * e(x) + |F2|cos(b) * e(y).
I asked someone for help and they said that the x-components have to be equal, and from that you can find out |F2|, why the x-components and not the y-components?
So now that I got |F2| I have no idea how to get the magnitude of the sum of F1 and F2.
Do I add the x- and y- components of F1 and F2 and then take the absolute value?
sqrt(2*(Pcos(a) + |F2|cos(b))^2 ?
The answer is supposed to be F = Psin(a) + |F2|sin(b), why?

- Figured it out, obviously for the sum of forces to be vertical it means that the two x-components have to be equal and take each other out, leaving the y-components. Then you just use sine to get the y-components and add them together. I got mixed up with the trigonometry and blindly followed what I read, which was that you use the cosine from the dot product.

 

[anathema]

Hello,

As a fellow scientist, I completely understand your confusion with using vectors and trigonometry in this problem. Let me try to explain it in a clearer way.

First, let's define some variables:
- F1 = the force on the left wire (known as P in the problem)
- F2 = the force on the right wire (what we are trying to find)
- F = the sum of forces (what we are trying to find)

Now, let's think about the forces acting on the pole. There are two forces pulling in opposite directions - F1 and F2. These forces can be represented by vectors, which have both magnitude and direction. In order to find the magnitude of these forces, we need to break them down into their x and y components.

To do this, we use trigonometry. The x-component of a vector is equal to the magnitude of the vector multiplied by the cosine of the angle between the vector and the x-axis. Similarly, the y-component is equal to the magnitude of the vector multiplied by the sine of the angle. This is why you see equations like magnitude*cos(angle) and magnitude*sin(angle) - they are used to find the x and y components of a vector.

In this problem, we are given the angle alpha (on the left side corner) and the angle beta (on the right side corner). We also know that the x-components of F1 and F2 must be equal in order for the pole to remain stable. This is because the x-components will cancel each other out, leaving only the y-components to balance the vertical force F.

So, to find the magnitude of F2, we need to set the x-components of F1 and F2 equal to each other and solve for |F2|:
|F2|cos(beta) = F1cos(alpha)
|F2| = F1cos(alpha)/cos(beta)
Since we know that F1 = P, we can substitute that in:
|F2| = Pcos(alpha)/cos(beta)

Similarly, the y-component of F is equal to the sum of the y-components of F1 and F2:
Fsin(F) = F1sin(alpha) + F2sin(beta)
We can substitute in our values for F1 and F2:
Fsin(F) = Psin(alpha) + |F2|sin(beta)
And now we can substitute in our equation for |F2| that we