Vector problem: flagpole with wires at different angle

[prime-factor]

Homework Statement

Two supporting wires keep a flagpole in its
vertical position. The horizontal components of the tensions in
the supporting wires must be equal and opposite,
otherwise the pole will move laterally. The
vertical components may be different.
The longer wire forms a 30 degree angle with the ground, and
the shorter wire forms a 45 degree angle with the ground.
a Calculate the ratio of the tension in the
shorter wire to the tension in the longer wire.
b If the tension in the shorter wire is 600 N, calculate the magnitude of the tension
in the longer wire.

(Refer to picture) in the word document.

Homework Equations

basic trigonometry:

sine, cosine, tan,

possibly cosine/sine rules

The Attempt at a Solution

I am not sure, what to do with this problem.
I have attempted it, but haven't a clue
how to do it correctly. Please help.
I've just started this topic, and haven't been
taught this yet.

 

[Carid]

Do you know how to resolve a vector into two component vectors? For example, if you are driving North East at speed V you have a North vector and an equal East vector. These two vectors would each be of size V*cos(45°). If you now turn a bit to the left so you are driving at 30° East of North then the North vector will now be V*cos(30°) and the East vector will be V*cos(60°). Does this help?

 

[prime-factor]

I am having trouble understanding.

So with my question, V would be the magnitude of the vector, and I would have:

r.cos(30) , and r.cos(45) ?

I don't really get it. Because I don't have a magnitude.

 

[Carid]

You need tensions; just call them T1 and T2. They won't be equal. The horizontal components are equal and the vertical ones are different.

 

[prime-factor]

Ahhh :)

Okay.

So let horizontal components equal x, and vertical components equal T1, T2

=>cos (theta₁) = x/T1
T1 = x / cos (theta₁)

=>cos(theta₂) = x/T2
T2 = x / cos (theta₁)

Then let's say I choose an arbitrary value...2

2/ cos(30) = 2.30

2/ cos(45) = 2.82

=> 2.82 : 2.3 = 1: 1.2(approx) , so that's my ratio?

Or I could have just done cos(30) : cos(45)

 

[Carid]

You answer is incorrect I fear.
That wasn't quite what I had in mind.

Longer cable has tension T1.
The horizontal component will then be T1*cos(60°)
The vertical component will then be T1*cos(30°)
Repeat this exercise for the shorter rope with tension T2 and equate the horizontal components.
From here you can get the ratio of tensions.

 

[prime-factor]

I did that and got:

cos(30)(T1) = cos(45)(T2)

and get the same: 1.224:1

which is the same as what I get doing it the way you feared was incorrect.

It ends up being the same, because I was equating the horizontal components

 

[Carid]

My horizontal component is T1*cos(60°); yours is cos(30)(T1). Not the same!

 

[Carid]

prime-factor

I broke the golden rule and didn't draw a diagram.
Your solution was correct.

 

[prime-factor]

Hey. No problems :). I appreciate your help nonetheless. I needed to resolve to resolve the components and you reminded me too, so Thankyou.