What are the Components of a Vector Given its Magnitude and Direction?

  • Thread starter Thread starter sp33dk1lls
  • Start date Start date
  • Tags Tags
    Vector
AI Thread Summary
The discussion revolves around calculating the components of force vector B given the total force vector C and its direction. The total force C is defined as the sum of vectors A and B, with a magnitude of 80 N and an angle of -110˚. Participants share calculations for the x and y components of vectors A and C, noting the importance of signs and correct trigonometric functions. To find vector B, they emphasize setting up equations based on the components and solving for Bx and By to ultimately determine the magnitude and angle of vector B. The conversation highlights the need for careful attention to detail in vector calculations and the use of diagrams for clarity.
sp33dk1lls
Messages
4
Reaction score
0

Homework Statement



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.
Capture.jpg

Homework Equations


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


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.
 
Physics news on Phys.org
sp33dk1lls said:
Cx=80sin(-110)=-75.2
Cy=80cos(-110)=27.4
You mixed up the components here. And be careful with signs.
 
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.
 
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.
 
i meant third quarter
 
the components of A and C are Ax=-75sqrt2 Ay=75. Cx=-40 and Cy=-40sqrt2
 
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.
 
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
 
Bx=102 and By=-112. Do we agree so far?
 
  • #10
the way you did it yes. the way i was doing it no.
 
  • #11
sp33dk1lls said:
now i could use a little hint about what to do next. I'm sure how to get the single magnitude for B.
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.
 
Back
Top