1. ## Vectors

C is the resultant of the forces, 2A and 3B, that is C=2A+3B.
If A=300N<43 deg. and C =100N<97 deg. , use the component method to determine the magnitude and direction of B.

I am not sure if this problem is done correctly. Here is my attempt:

Ax = 600 cos 43 = 438N
Ay = 600 sin 43 = 409 N

Bx = ?
By = ?

Cx=100 sin 43 = 68 N
Cy=100 cos 43 = 73 N

438 + Bx= 68
409 + By = 73

Therefore
Bx= -370N
By= -336N

B = sqrt( (-370)^2 + (-336)^2) So B = 499N

Is this correct? What about the 3B?

2. your method is basically correct, however it would be the result

If A + B = C

You have 2A + 3B = C

so rework 2Ax + 3Bx = Cx

etc

3. Originally Posted by Calculus26
your method is basically correct, however it would be the result

If A + B = C

You have 2A + 3B = C

so rework 2Ax + 3Bx = Cx

etc
Ok so I did notice one error in my attempt above..This is my second attempt. Does it look correct? Can someone please work it out and let me know

Ax = 600 cos 43 = 438N
Ay= 600 sin 43 = 409 N

Bx=?
By=?

Cx=100 sin 7 = 12N
Cy=100 cos 7 = 99N

438 + 2Bx = 12
409 + 2By= 99

Bx=-213
By=-155

B = 263 N

Theta = 216 deg. from due east(+)

4. Ok

Note Cx should be negative (draw it out)

You should go with Cx = 100cos(97) = -12

Cy = 100 sin(97) = 99

438 + 2B = -12

B= -225

which makes theta 211