are the vectors directed outward from the origin? I assume so.
So here is a general solution: Measure angles from 000, true N, or on x-y, the +Y axis. I will use i,j axis in the solution. You can use x,y, or N,E, machts nicht.
A=i*MagA*sinThetaA+jmagA*cosThetaA
B=i*MagB*sinThetaB+jmagA*cosThetaB
C=A-B
C=i(5*sin35 -2*sin(180-33))+j(5*cos35-2cos(180-33)
note: I assumed A was in QuadrantI, and B is in quadrant IV, the figure is not there to look. If not, change the angles, measure from N, orOOO, or y axis).
do that calculation, then you have the vector in i,j format, and you can calculate magnitude and angle (arctan y/x) easily. Remember where your angles are measured from.
when you have a number of vectors to add,subtract, this is a sure fire way to figure it.
Then in the end, draw a graphical sketch of the vectors, to see if the result is close to what you calculated.
Can someone please help me? I am having a really hard time with physics. This is not a test question. It is a practice problem from a review website I am supposed to use, but I am having a really hard time. Can someone at least help me to begin setting up the problem? I would really appreciate it.
Two vectors A and B are shown in the x-y plane. Vector A has a magnitude of 5 m and makes an angle of 55° with the positive x-axis; vector B has a magnitude of 2 m and makes an angle of 33° with the negative x-axis. (See the figure above.) Vector C (not shown in the diagram) is the difference of A and B (C = A - B).
If you enter an expression, use degrees as arguments for sin and cos, rather than radians.
*****I do not know how to attach the image but I think the problem can be set up without an image.************
2 answers
Oh okay. Thank you very much bobpursley. I was forgetting so use sin and cos. I was trying to solve the problem just using numbers. Thank you for the help and helping me to see my mistake.