1. Sep 2, 2010

### sp33dk1lls

1. The problem statement, all variables and given/known data

Two force vectors are shown in the figure (not drawn to scale), and the total force vector C is the sum of A and B:C=A+B. If the magnitude of the total force is C = 80 N and its direction is specified by the angle measured from +x axis to be θC = −110˚, find B and φ for the force vector B.

2. Relevant equations
Cx= Ax+ Bx
Cy= Ay+ By
C= sqrt(Cx^2+Cy^2)?
tanθc=Cy/Cx?

3. The attempt at a solution
Ax=-150cos30=-130
Ay=150sin30=75
Cx=80sin(-110)=-75.2
Cy=80cos(-110)=27.4

Cx= Ax+ Bx
-75.2= -130 + B > B= 54.8
27.4= 75 + B > B=-47.6

Not sure if I'm doing this right.

2. Sep 2, 2010

### Staff: Mentor

You mixed up the components here. And be careful with signs.

3. Sep 2, 2010

### sp33dk1lls

ok i got that. now i could use a little hint about what to do next. i'm sure how to get the single magnitude for B.

4. Sep 2, 2010

### woodyallen1

Both components of B sould be positive. Try make a drawing and work with reduced angles. 110 is at the second quarter, so work with 30 at the second quarter.

5. Sep 2, 2010

### woodyallen1

i meant third quarter

6. Sep 2, 2010

### woodyallen1

the components of A and C are Ax=-75sqrt2 Ay=75. Cx=-40 and Cy=-40sqrt2

7. Sep 2, 2010

### woodyallen1

Bx=-40-75sqrt2 By=75-40sqrt2 and tanθB= By/Bx and it is at the second quarter. to find the angle subtract from 180, θB.

8. Sep 2, 2010

### sp33dk1lls

what was your final answer by doing your method because it seems different from mine. i got B=182 N and the angle 34.3

9. Sep 2, 2010

### woodyallen1

Bx=102 and By=-112. Do we agree so far?

10. Sep 2, 2010

### sp33dk1lls

the way you did it yes. the way i was doing it no.

11. Sep 2, 2010

### Staff: Mentor

Just set up the equations for components:

Ax + Bx = Cx
Ay + By = Cy

And solve for Bx and By. Using the components, you can compute the magnitude.

Vector Addition of Components Jan 29, 2016