Why do we solve i and j components of a vector using trig?

AI Thread Summary
Vector components are solved using trigonometry because they represent the horizontal and vertical legs of a right triangle. In the example of a 5 N force vector, the components are 4 N horizontally and 3 N vertically, which correctly relate through the Pythagorean theorem, where 5^2 equals 4^2 plus 3^2. The misunderstanding arises from thinking that the magnitudes of the components should add linearly; instead, they combine vectorially, resulting in a single resultant vector. Thus, the total force is not the sum of the magnitudes of the components but rather the magnitude of the resultant vector formed by them. Understanding this relationship clarifies how the components relate to the original force vector.
ALRedEye
Messages
2
Reaction score
0

Homework Statement


I'm having trouble understanding why we solve vector components (i and j, or the horizontal and vertical legs) like a right triangle?
An example would be a 5-4-3 triangle. If 5 N was the force vector I am solving for then I would end up with 4 N in the horizontal direction and 3 N in the vertical direction. The part I don't understand is how the sum of the legs can equal more than the original force? To me it seems like i^2 + j^2 > (the original force vector)^2 and i + j = (the original force vector) makes more sense.
When I try and relate this to the real world I'm thinking maybe that 5 N force is me pushing on a box at a -53 degree angle. So that means the box is moving at 4 N along the floor and 3 N into the floor, but I'm misunderstanding how these components relate to each other since the total of the two components is 7 N and I'm only pushing at 5 N.
I'd really appreciate some help wrapping my head around this!
(Sorry if I posted to the wrong category)

Homework Equations

The Attempt at a Solution

 
Physics news on Phys.org
ALRedEye said:

Homework Statement


I'm having trouble understanding why we solve vector components (i and j, or the horizontal and vertical legs) like a right triangle?
An example would be a 5-4-3 triangle. If 5 N was the force vector I am solving for then I would end up with 4 N in the horizontal direction and 3 N in the vertical direction. The part I don't understand is how the sum of the legs can equal more than the original force?
What, exactly do you mean by "the sum of the legs"? The sum of the horizontal and vertical force vectors are equal to the total force- that's the whole point.

To me it seems like i^2 + j^2 > (the original force vector)^2
Assuming you mean the dot product here, (3i).(3i)= 9 and (4i).(4i)= 16. 9+ 16= 25= 5^2. Where did you get the idea that it was larger than
(the original force vector)^2

and i + j = (the original force vector) makes more sense.
Did you not mean to write 3i+ 4j= (the original force vector)?

When I try and relate this to the real world I'm thinking maybe that 5 N force is me pushing on a box at a -53 degree angle. So that means the box is moving at 4 N along the floor and 3 N into the floor
You mean there would be a 4N force pushing the box and a 3N force pressing it into the floor, right?

, but I'm misunderstanding how these components relate to each other since the total of the two components is 7 N and I'm only pushing at 5 N.
I'd really appreciate some help wrapping my head around this!
Your mistake is thinking that "the total of the two components is 7 N". You do not add the magnitudes of two vectors- you add the vectors themselves. These two vectors, of magnitude 3 and 4, add to a vector of magnitude 5. That is because the magnitude of a vector of the form ai+ bj is \sqrt{a^2+ b^2}.

(Sorry if I posted to the wrong category)

Homework Equations

The Attempt at a Solution

 
Do you understand that if you walk 3 miles north and 4 miles west that you can get back where you started by walking 5 miles diagonally to the southeast?
 
  • Like
Likes Chestermiller and Nathanael
That makes more sense. Thanks
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top