The problem I was given was to initially add F1 and F2
F1 = 200 g at 30deg
F2 = 200 g at 120 deg
To add them I: added the cos for both, added the sin for both, square rooted the addition of both of the squared answers, and came up with a magnitude of 282.84.
My problem is subtracting them. When I was doing the same method but with subtraction, I got the same answer as I did when I was adding them. Can someone tell me how to subtract these two vectors?
F1 = 200 g at 30deg
F2 = 200 g at 120 deg
To add them I: added the cos for both, added the sin for both, square rooted the addition of both of the squared answers, and came up with a magnitude of 282.84.
My problem is subtracting them. When I was doing the same method but with subtraction, I got the same answer as I did when I was adding them. Can someone tell me how to subtract these two vectors?
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Add the vectors
Horizontal components:
200cos30 + 200cos120 = 73
Vertical components:
200sin30 + 200sin120 = 273
Resultant = 283 at 75 degrees
Subtract the Vectors:
Horizontal components:
200cos30 - 200cos120 = 273
Vertical components:
200sin30 - 200sin120 = -73
Resultant = 283 at -15 degrees
So the magnitudes are the same but the angle of the resultant is different by 90 degrees
What is happening is F2 is at 120 summing and at 120-180 = -60 subtracting.
The vectors have 90 degrees between them so the resultant whether adding or subtracting is the hypotenuse of a right triangle. In both cases, the triangle has the same side lengths which is why the magnitudes of the resultants are the same.
Horizontal components:
200cos30 + 200cos120 = 73
Vertical components:
200sin30 + 200sin120 = 273
Resultant = 283 at 75 degrees
Subtract the Vectors:
Horizontal components:
200cos30 - 200cos120 = 273
Vertical components:
200sin30 - 200sin120 = -73
Resultant = 283 at -15 degrees
So the magnitudes are the same but the angle of the resultant is different by 90 degrees
What is happening is F2 is at 120 summing and at 120-180 = -60 subtracting.
The vectors have 90 degrees between them so the resultant whether adding or subtracting is the hypotenuse of a right triangle. In both cases, the triangle has the same side lengths which is why the magnitudes of the resultants are the same.
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In such problem , we have nothing to do with the addition or subtraction but with their RESULTANT. And if two vectors are in same direction they are added to find the resultant and if in different direction, they are subtracted. Further process is as follows:
find 1)horizontal component,X
2)vertical component,Y
3) and resultant,R
Horizontal component, X =200cos30+200cos120=73
Vertical component,Y=200sin30+200sin 120=273
Resultant,R=sqrt(X^2+Y^2)
=sqrt(73^2+273^2)
=283
Direction=tan^-1(Y/X)
=+75 deg.
find 1)horizontal component,X
2)vertical component,Y
3) and resultant,R
Horizontal component, X =200cos30+200cos120=73
Vertical component,Y=200sin30+200sin 120=273
Resultant,R=sqrt(X^2+Y^2)
=sqrt(73^2+273^2)
=283
Direction=tan^-1(Y/X)
=+75 deg.
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Draw a diagram.
For addition, draw F1 first and then at the end of F1, draw F2, the resulting position is the sum of the two vectors, and F1 + F2 = F2 + F1.
For subtraction, draw F1 as before starting from (0,0) but then also draw F2 starting from (0,0). The difference between them is the vector from the end of F1 to the end of F2 and the sign of that difference depends on F1 - F2 or F2 - F1.
Alternatively, proceed as for addition and draw F1 from (0,0) but now draw -F2 from the end of F1. That is, reverse the direction of F2.
For addition, draw F1 first and then at the end of F1, draw F2, the resulting position is the sum of the two vectors, and F1 + F2 = F2 + F1.
For subtraction, draw F1 as before starting from (0,0) but then also draw F2 starting from (0,0). The difference between them is the vector from the end of F1 to the end of F2 and the sign of that difference depends on F1 - F2 or F2 - F1.
Alternatively, proceed as for addition and draw F1 from (0,0) but now draw -F2 from the end of F1. That is, reverse the direction of F2.