Angle Between Two Forces: Understanding Vector Addition and Magnitude

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To find the angle between two equal magnitude forces when their resultant equals the magnitude of either force, one vector can be placed along the x-axis while the other is at an angle θ. The resultant vector is calculated by adding the x and y components of both vectors, leading to the equation Rx = F1 + F2 cos θ. The magnitude of the resultant must equal the length of the individual forces, allowing for the determination of θ through vector addition principles. Properly drawing the vectors and correcting any misrepresentations in the diagram is crucial for solving the problem accurately. Understanding vector components and their addition is essential for finding the correct angle.
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Homework Statement


Find the angle between two forces of equal magnitude when the magnitude of their resultant is also equal to the magnitude of either of the forces

2. The attempt at a solution

x-component of the resultant:

Rx = F1 cos 0\circ + F2 cos θ
Rx = F1 + F2 cos θ

It's an Example in my book and I want to know why F1 cos 0\circ has an angle of 0. Also, I don't how they have done it either.
 
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Well let's think about it like this. It doesn't matter where we put the vectors, all that matters is the angle between them. So for convenience, let us place one vector along the x-axis with a length A. Then the other vector will be pointing up at some angle \theta
It may help to draw the vectors out. Also draw the resultant vector. So we have two vectors of length A. We call them A1 and A2. A1 is along the x axis. We have

\vec A_1 = A \hat i

\vec A_2 = A cos \theta \hat i + A sin \theta \hat j

The resultant vector is

\vec R = \vec A_1 + \vec A_2 = (A + A cos \theta) \hat i + (A sin \theta) \hat j

The magnitude of the resultant must also be A. If you set the magnitude (the square root of the sum of the squares of the components) of the resultant vector equal to A, you should be able to solve for theta.

Give this a try and see what you get
 
Oh yes... you may not have seen this notation. When I write

\vec F = (f_x) \hat i + (f_y) \hat j

this means that F is a vector with two components. It has an x component and a y component. i hat and j hat are unit vectors. A unit vector is just a vector of length "1". We denote them by writing little hats over the lowercase letters. By convention, we use i hat for a unit vector in the x direction and j hat for a vector in the y direction.

So the point of this is that when you see a vector written in this way, the f_x is just the x component and the f_y is just the y component.

My apologies if you already know this.
 
Yep, I know about unit vectors, but Thnx for the info. :)

Well, It's been solved in the book in Example, but I can't understand it...

I've drawn the following diagram to solve it. Am I doing it right? Also, 0° is for the parallel vectors ?

angle.png
 
The idea is this. One vector is lying on the x axis, so it only has one component, (an x component).
The other vector has both an x component and a y component.
You are to add these two vectors to find a resultant vector.
To do this, you use the definition of vector addition. Vector addition is defined as simply adding the components of two vectors together to find the resultant vector.
Then the magnitude of a vector is the square root of the sum of the squares of its components.
If you understand these things, you should be able to solve the problem.
Your diagram is a bit wrong i think.
First of all, if you set F1 on the x axis, then
F_1 sin 0 = 0
So F1 sin 0 should not be on the diagram. It is just zero.
Also...
F_2 sin \theta
is a y-component... however you have drawn it at an angle. It should be drawn vertically upwards. If you make these corrections, and then add the vectors again, you will arrive at the correct answer.
 
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