Coulomb's Law, Three Charges, X Y Components, Angle

[Stenn]

Homework Statement

This is finding net force on Q3.

I was taught by my Physics teacher to always get angle from the Positive side of x-axis (+X-Axis).
For example in the above situation, if the angle is -30deg from the +x-axis (assuming the Q3 is the origin),

Why is F31 x-component = 140 N cos(360-30) NOT the case?

I mean I considered that in getting angles it should ALWAYS come from the +X-Axis

I think now, if the above x-y components are correct. then the cartesian plane (Q3 Origin) is turned upside down

Now, what's not clear to me is that. When does a Cartesian plane mirrors/turns upside down?

OR is the above situation false?
2. Homework Equations

The Attempt at a Solution

 

[berkeman]

Welcome to the PF.

Why is F31 x-component = 140 N cos(360-30) NOT the case?

Since the cos() function is even, cos(-θ) = -cos(θ) cos(θ), so it doesn't really matter whether you take that angle as positive or negative, as long as the resultant is pointing in the correct direction.

Probably a better way to think of is is to just use vector dot products to keep the directions straight. Are you familiar with the vector dot product?

EDIT -- fixed typo in my cos() equation.

 

[Stenn]

THANKS! for the warm welcome! :D

berkeman, you misunderstood me

the above is cos(360-30) = cos(330)

Also, yep I know vector dot product.
Isn't the dot product supposed to get the AREA OF TWO VECTORS? What the unknown here is the sum of two vectors, simply adding the two: Combining two forces and determining the magnitude and direction.

 

[haruspex]

the above is cos(360-30) = cos(330)

They are all the same. cos(x)=cos(-x)=cos(360-x). Flipping the image about the x-axis does not change the x components of the vectors.

the dot product supposed to get the AREA OF TWO VECTORS?

No, the cross product magnitude is the area of the parallelogram defined by the two vectors. Dot products can be used for finding the component of one vector in the direction of another, but you have to divide by the magnitude of the second vector.

 

[Stenn]

I want to change the question. How do I plot the origin? in order for me to determine the angle.

 

[Stenn]

Homework Statement

View attachment 212074
When does a Cartesian plane mirrors/turns upside down?

I want to change the question to. How do I plot the origin?, for me to determine the angle I need for breaking down x-y components of a vector

 

[haruspex]

I want to change the question to. How do I plot the origin?, for me to determine the angle I need for breaking down x-y components of a vector

I don't understand your question. To find the components you only need the angle the vector makes to the directions of the x and y axes. The origin can be anywhere.

 

[Stenn]

The origin can be anywhere.

But that can't happen, I won't know if the x or y is negative

 

[Stenn]

The origin can be anywhere.

Does the origin sits on the vertex of the vector?

 

[haruspex]

Does the origin sits on the vertex of the vector?

For the purpose of resolving into components, you can take the origin as being in the line of the force.
A force is actually defined by two vectors. One gives you the direction and magnitude of the force, while the other specifies the point of application. For resolving into components, you only need the first; you can take its point of application as being at the origin.
The point of application matters when finding the moment about an axis.

 

[Stenn]

For the purpose of resolving into components, you can take the origin as being in the line of the force.
A force is actually defined by two vectors. One gives you the direction and magnitude of the force, while the other specifies the point of application. For resolving into components, you only need the first; you can take its point of application as being at the origin.
The point of application matters when finding the moment about an axis.

I can't grasp it. What I mean is...
when I resolve for x-y component of a given vector which is the product of Coulomb's law, where do I plot an origin in order to find its angle that is to be used later in resolving x-y components.

I'm sorry, please tell me if you know that you've already answered my question, but I'm not getting your point, so I know when to stop. perhaps...

 

[haruspex]

I can't grasp it. What I mean is...
when I resolve for x-y component of a given vector which is the product of Coulomb's law, where do I plot an origin in order to find its angle that is to be used later in resolving x-y components.

I'm sorry, please tell me if you know that you've already answered my question, but I'm not getting your point, so I know when to stop. perhaps...

Likewise, I cannot understand your difficulty.
In the example in this thread, you can see that the force acts at angle -30°. That is all you need for finding the x and y components. Why do you need an origin?
Please show your working up to the point where you think you need it.

 

[Stenn]

Why do you need an origin?
Please show your working up to the point where you think you need it.

You see the above. There is given -30deg
We all know in a cartesian plane there would be 360 degrees all in all around.
Now, -30 deg from x-axis would be 30 degrees BACKWARDS
Now, I was taught by my professor to ALWAYS START FROM X-AXIS forwards (I think this part is where I get problems)
Now, I think is that I would use cos(340) to get the angle and
resolve x-y components with the angle 340 from x-axis

So, I need origin to imagin cartesian plane on an reevolve around and get the angle of a vector.

PLEASE, If you're getting confused please tell me to stop, well I know I must do in some point from now.