Finding 2 of 3 Forces Given 1 and a Resultant

[cadcdac]

Homework Statement

Force A has a magnitude of 200lb and points 35° N of W. Force B points 40° E of N. Force C points 30° W of S. The resultant of the three forces has a magnitude of 260lb and points 85° S of W.
(a) What is the magnitude of Force B?
(B) What is the magnitude of Force C?

Homework Equations

F = √(Fx² + Fy²)
Fx = FcosΘ
Fy = FsinΘ

where

F = Force
Θ =Angle

The Attempt at a Solution

Having drawn these vectors, I've been trying to use components which could've been easier if 2 forces were given.

Resultant's X = R_x = 260cos85°
Resultant's Y = R_y = 260sin85°
Force A's X = A_x = 200cos35°
Force A's Y = A_y = 200sin35°
B_x = Bsin40°
B_y = Bcos40°
C_x = Csin30°
C_y = Ccos30°
​

And formed equations

R = A + B + C
R_x = Acos35° + B_x + C_x
R_y = Asin35° + B_y + C_y
​

With these, I don't really know because 2 are missing while having no other equations for them like A+B=1 or stuff like that that give hints. Seems like something is missing, but i don't know because others have solved it. Don't know what is next.

 

[BvU]

Hello cad, welcome to PF !

You have two unknows (B and C) and two equations.
You are making life difficult for yourself by dragging in B_{x, y} and C_{x, y}.

[edit]By the way, I hope you don't mean R = A + B + C but $\vec R = \vec A + \vec B + \vec C$

 

[cadcdac]

Hello cad, welcome to PF !

You have two unknows (B and C) and two equations.
You are making life difficult for yourself by dragging in B_{x, y} and C_{x, y}.

So 260 = 200 + $\vec B + \vec C$ would be the first equation and i'll have to find the other equation
By any chance, does the 2nd equation contain angles in it? Or anything?

[edit]By the way, I hope you don't mean R = A + B + C but $\vec R = \vec A + \vec B + \vec C$

Ah yes, sorry about that

 

[BvU]

260 and 200 are not vectors. I take it you do know about vector addition ?

You have two equations

R_x = Acos35° + B_x + C_x
R_y = Asin35° + B_y + C_y​

with two unknowns: the magnitudes B and C.​

Perhaps I confused you with my blunt comment

You are making life difficult for yourself by dragging in B_{x, y} and C_{x, y}.

but the intention was that you would substitue the magnitudes of B and C

 

[cadcdac]

260 and 200 are not vectors. I take it you do know about vector addition ?

You have two equations

with two unknowns: the magnitudes B and C.​

Perhaps I confused you with my blunt comment
but the intention was that you would substitue the magnitudes of B and C

I see. I came up with estimated answers B = 1780.15 and C = 2006.18. I also checked and resulted to 259.99... Am I doing this right?

 

[BvU]

Does it match with the drawing you mentioned in post #1 ?

 

[BvU]

Check your angles. Do they have the same reference ?
In math and physics, east is zero angle, north $\pi/2$

 

[BvU]

Never mind, it matches. Note that finding the magnitude of $\vec A + \vec B + \vec C$ is 260 is not enough to ensure $\vec R = \vec A + \vec B + \vec C$; the direction has to be the same too.
But from you answer I expect you did the right thing and equated x and y components. WEll done.

 

[Mystery]

Hi, I am a newbie here... can someone explain how did he got the values of B and C??

 

[Mystery]

I see. I came up with estimated answers B = 1780.15 and C = 2006.18. I also checked and resulted to 259.99... Am I doing this right?

how did you got B and C??