Sum of 4 Vectors: Magnitude & Angle

AI Thread Summary
To find the sum of the four vectors A, B, C, and D, first convert vectors B and D from polar to rectangular coordinates using trigonometric functions. The components for B are calculated as 4cos(65)i + 4sin(65)j, and for D as 5cos(-235)i + 5sin(-235)j. Next, sum the x-components and y-components separately to obtain the resultant vector. The magnitude is determined using the Pythagorean theorem, while the angle can be found using the arctangent of the ratio of the y-component to the x-component. This method ensures clarity in handling two-dimensional vectors without introducing a third dimension.
Kp0684
Messages
14
Reaction score
0
What is the sum of the following four vectors in (a) unit- vector notation, and as (b) a magnitude and (c) an angle?... A=(2.00m)i + (3.00m)j...B: 4.00m, at +65.0 degrees...C= (-4.00m)i - (6.00m)j...D: 5.00m, at -235 degrees...i understand how to get the magnitude and the angle but how would i set this up... would i start with A and C and find their sum...i believe what's confusing me is B and D...otherwise i know how to set it up...need help...
 
Physics news on Phys.org
HINT:Write all 4 vectors in component form.

Then simply add the 4 numbers for the "x" component and the 4 numbers for the "y" component.

Daniel.
 
Kp0684 said:
What is the sum of the following four vectors in (a) unit- vector notation, and as (b) a magnitude and (c) an angle?... A=(2.00m)i + (3.00m)j...B: 4.00m, at +65.0 degrees...C= (-4.00m)i - (6.00m)j...D: 5.00m, at -235 degrees...i understand how to get the magnitude and the angle but how would i set this up... would i start with A and C and find their sum...i believe what's confusing me is B and D...otherwise i know how to set it up...need help...


Here are the formula's you will need to apply.

Given two vectors with components A = (i,j,k) and B = (a,b,c)

magnitude \sqrt{a^2 + b^2 + c^2}
scalar product: A.B = ia + b j + ck
scalar product: A.B = magnitude of A * magnitude of B * cos(t) where t is the angle between the two vectors A and B

sum A+B = (i+a,j+b,k+c) (this is a new vector, the scalar product yields a number)
marlon
 
okay, i get -2.00i - 3.00j - 1.17k when i sum it up...iam still lost on this one...need help again...
 
i, j and k represent the 3 directions of the space, for example high, wide and length. The object will be placed this 3 distances in the space from the point you consider as reference
 
I'm very confused! Where did "k" come from? You original post said
"... A=(2.00m)i + (3.00m)j...B: 4.00m, at +65.0 degrees...C= (-4.00m)i - (6.00m)j...D: 5.00m, at -235 degrees..." with only two dimensions.

I presume that the angles are measured relative to the positive x axis.
B would be 4cos(65)i+ 4sin(65)j and D would be 5cos(-235)i+ 5sin(-235)j

Adding those four vectors does not give you any "k" component.

Of course, once you have found the sum, you find the length by the Pythagorean theorem and the angle is arctan((j component)/(x component)).
 
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