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.************
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.
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.
Of course, I'd be happy to help you with your physics problem! Let's work through it step by step.
To begin, we have two vectors A and B with their respective magnitudes and angles. 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.
We are asked to find vector C, which is the difference of A and B (C = A - B). To find vector C, we need to break down vectors A and B into their x and y components, and then subtract their corresponding components.
To find the x and y components of vector A:
- The x component (Ax) is given by: Ax = magnitude of A * cos(angle of A with positive x-axis)
Therefore, Ax = 5 m * cos(55°)
- The y component (Ay) is given by: Ay = magnitude of A * sin(angle of A with positive x-axis)
Therefore, Ay = 5 m * sin(55°)
Similarly, we can find the x and y components of vector B:
- The x component (Bx) is given by: Bx = magnitude of B * cos(angle of B with negative x-axis)
Therefore, Bx = 2 m * cos(33°)
- The y component (By) is given by: By = magnitude of B * sin(angle of B with negative x-axis)
Therefore, By = 2 m * sin(33°)
Now, we can calculate the x and y components of vector C (C = A - B):
- The x component of C (Cx) is given by: Cx = Ax - Bx
- The y component of C (Cy) is given by: Cy = Ay - By
Finally, we can find the magnitude of vector C using the Pythagorean theorem:
- The magnitude of C is given by: magnitude of C = square root of (Cx^2 + Cy^2)
I hope this explanation helps you begin setting up the problem! If you have any further questions or need assistance with the calculations, feel free to ask.