Adding Vectors: Understanding Direction and Length

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Adding vectors involves determining their components and using them to find the resultant vector and its angle. The placement of x and y components can be done in any order, as long as they are connected head-to-tail, resulting in the same resultant vector regardless of the configuration. While the angles of the triangles formed may differ, the resultant vector's direction and magnitude remain consistent. The angle of the resultant with respect to the axes is what matters, not the specific triangle formed by the components. Understanding this principle is key to correctly adding vectors and interpreting their results.
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Homework Statement


Hi Guys, I am a bit unclear regarding adding vectors and hoping someone can clear up the confusion for me.

In the image below, we are adding two vectors and we used the vector components to find find x and y. Finally we then used x and y to get the resultant and the angle.

There is a diagram after "Now place the resultant components head-to-tail to form a right angled triangle", where we begin with x going east, then y going north and touching the tip of x.

My question is how do we determine to place the x and y in this way? Is there any criteria or methodology that we use because I could also place x starting at the tip of resultant vector and going east and then from the head of x vector going north to form y vector. If I do this the angle differs as adjacent and opposite sides of the angle change. Could anyone please help me understand this.

I hope my questions is clear.Problem: Determine the length and direction of a + b if a is 4 m [N30°E] and b is 2 m [S40°W].

Homework Equations


N/A

The Attempt at a Solution


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aatari said:
Is there any criteria or methodology that we use because I could also place x starting at the tip of resultant vector and going east and then from the head of x vector going north to form y vector.
It doesn't matter whether you place the start of the y component vector at the tip of the x component vector (not the resultant vector!), or the reverse. You get the same answer either way.

aatari said:
If I do this the angle differs as adjacent and opposite sides of the angle change.
Try again. If you get different resultant vectors, you're doing something wrong. (Note that the triangle you get might be different, but the resultant vector will be the same.)
 
aatari said:
My question is how do we determine to place the x and y in this way? Is there any criteria or methodology that we use because I could also place x starting at the tip of resultant vector and going east and then from the head of x vector going north to form y vector. If I do this the angle differs as adjacent and opposite sides of the angle change. Could anyone please help me understand this.
I'm not sure I understand your question, but the key here is to combine the x and y vectors. You have a y component that is positive 1.93, towards the North. You have a positive 0.71 x component, towards the East. So whether you put the x arrow down and then place the tail of the y with the head of the x, or you put the y arrow down first and then place the x at its end, you still end up in the same place. That same place is 1.93 to the North and 0.71 to the East of where ever you started from. Also, in either case, once you draw the resultant arrow, you will have a right triangle and you will be able to use their formula to compute the length of that arrow (2).

However, if you use the different triangle, you'll be looking at a different angle. Instead of computing 20 degrees east of north (as shown in the example) you could end up with 70 degrees north of east. But that would just be an alternate way of specifying the same bearing.
 
Doc Al said:
It doesn't matter whether you place the start of the y component vector at the tip of the x component vector (not the resultant vector!), or the reverse. You get the same answer either way.Try again. If you get different resultant vectors, you're doing something wrong. (Note that the triangle you get might be different, but the resultant vector will be the same.)

Your are right, resultant does not change. However, my angle changes.
 
.Scott said:
However, if you use the different triangle, you'll be looking at a different angle. Instead of computing 20 degrees east of north (as shown in the example) you could end up with 70 degrees north of east. But that would just be an alternate way of specifying the same bearing.

So my answer will still be correct?
 
aatari said:
However, my angle changes.
The angle of your triangle changes. But the angle that the resultant makes with the horizontal and the vertical does not. You may have to convert one to the other.
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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