How to Solve a Three-Force Vector Problem Using the Component Method?

[Lifeinsteps]

Homework Statement

R is the resultant of the three forces A, B, and C, that is R = A + B + C. If
A = (A) N, 210°,
B = 200 N, Θ1,
C = 200 N, 65°,
and R = 250 N, 125°.

Use the component method to determine A and Θ1. (There may be two solutions to this one.) Check your result by adding directly using your calculator.

Homework Equations

N/A

The Attempt at a Solution

I don't even know. I've tried breaking this down into its components and such, but I just don't understand. Having both ungiven magnitude and direction makes this seem like the only way I can solve it is by plugging numbers. Is there something I'm missing? I'm stressing out over this one problem.

 

[cmmcnamara]

How familiar are you with vector algebra? You only have two unknowns to solve for and you have two coordinate types available to you for this problem. This means you should be able to generate a system of two equations and two variables and achieve a solution. You can create these equations by using the proper vector algebra quite easily which is why I ask.

 

[Lifeinsteps]

How familiar are you with vector algebra? You only have two unknowns to solve for and you have two coordinate types available to you for this problem. This means you should be able to generate a system of two equations and two variables and achieve a solution. You can create these equations by using the proper vector algebra quite easily which is why I ask.

I thought I was familiar enough to handle these problems, but apparently, I'm not!

Is there some kind of equations or something I should know that can help me out here? I am totally stumped! I calculated all the components to the two that it's possible for, but I just have no idea what to do to use the component method to solve this thing.

The last 3 hours have been nothing but gain-less stress for me, so thanks for the response...

[Edit: For the record, I did some equation systems in high-school, but I can't really remember them and I would have no idea how to use what I can remember on this problem...]

 

[cmmcnamara]

Note that each vector decomposes into x and y components correct? The resultant, R, is the sum of the other three vectors. You can create your two equations by decomposing all the vectors into components and note that Rx=Ax+Bx+Cx and Ry=Ay+By+Cy. This will allow you to solve for your two unknowns. Does this help you see what to do now?

 

[Lifeinsteps]

Note that each vector decomposes into x and y components correct? The resultant, R, is the sum of the other three vectors. You can create your two equations by decomposing all the vectors into components and note that Rx=Ax+Bx+Cx and Ry=Ay+By+Cy. This will allow you to solve for your two unknowns. Does this help you see what to do now?

Well, yes, I got this far:

-143 = Ax + Bx + 84.5
205 = Ay + By + 181

Before, but I wasn't sure what to do. I didn't think of creating a system of equations! I still can't really remember how to do it though... Is there some kind of instructional or something you can recommend? My professor didn't lecture this, so I guess he just expected us to know it.

 

[cmmcnamara]

Good so far but you can find the components of A and B in terms of the variables you are looking for. Recall for any vector V, the components are:
Vx=Vcos(a)
Vy=Vsin(a)

Where V is the vector magnitude and a is the angle the vector makes with the positive x-axis. Can you rewrite the A and B components in terms of your variables now?

 

[Lifeinsteps]

Good so far but you can find the components of A and B in terms of the variables you are looking for. Recall for any vector V, the components are:
Vx=Vcos(a)
Vy=Vsin(a)

Where V is the vector magnitude and a is the angle the vector makes with the positive x-axis. Can you rewrite the A and B components in terms of your variables now?

Um,

Ax = Acos(210°)
Ay = Asin(210°)

Bx = 200cos(Θ1)
By = 200sin(Θ1)

That's what you mean, right?

 

[cmmcnamara]

Yes exactly, put those into your equations and now you can solve for those two variables by finding the solution to that system!

 

[Lifeinsteps]

Yes exactly, put those into your equations and now you can solve for those two variables by finding the solution to that system!

-143 = Acos(210°) + 200cos(Θ1) + 84.5
205 = Asin(210°) + 200sin(Θ1) + 181

Like that? But I'm still not really sure how to solve them... I remember something looking like this in high school Pre-Calc but I really don't remember how to do it off the of the top of my head.

 

[cmmcnamara]

It's a little bit different because of the trig functions. In this case, I would solve both equations for A. This will give you two functions for A in terms of your angle. You can set these two equal because A must be equivalent in both equations. This sets up a simple equation to solve for the angle which can then be used recursively to find A.

 

[Lifeinsteps]

It's a little bit different because of the trig functions. In this case, I would solve both equations for A. This will give you two functions for A in terms of your angle. You can set these two equal because A must be equivalent in both equations. This sets up a simple equation to solve for the angle which can then be used recursively to find A.

Uh, I can give it a shot, is it something like this?

1)
-143 = Acos(210°) + 200cos(Θ1) + 84.5
-84.5
-227.5 = Acos(210°) + 200cos(Θ1)
Simplify
-227.5 = A(-.87) + 200cos(Θ1)
/(-.85)
261.5 = A + 200cos(Θ1)
-200cos(Θ1)
261.5 - 200cos(Θ1) = A

2)
205 = Asin(210°) + 200sin(Θ1) + 181
-181
24 = Asin(210°) + 200sin(Θ1)
simplify
24 = A(-.5) + 200sin(Θ1)
/-.5
-48 = A + 200sin(Θ1)
Move over <-

-48 - 200sin(Θ1) = A

So...

261.5 - 200cos(Θ1) = -48 - 200sin(Θ1)
+48
309.5 -200cos(Θ1) = -200sin(Θ1)

I'm not sure how to get any further than that because of those trig functions. Did I mess up somewhere?

 

[cmmcnamara]

No you are fine. So you aren't very familiar in solving trigonometric equations? To get you started, you can use a trig identity to substitute for either cos or sin giving you and equation of only one trig function. You can then solve for the angle. I believe the solution here is the inverse cosine of a quadratic function which may or may not give you two different angle values. Only one should make sense or it may not lie in the domain of the arccos function.

 

[Lifeinsteps]

No you are fine. So you aren't very familiar in solving trigonometric equations? To get you started, you can use a trig identity to substitute for either cos or sin giving you and equation of only one trig function. You can then solve for the angle. I believe the solution here is the inverse cosine of a quadratic function which may or may not give you two different angle values. Only one should make sense or it may not lie in the domain of the arccos function.

Okay, so I have to use either arcsin or arccos?

I'm sorry; I'm not really sure how to do that right here. Won't that end up kind of encompassing the whole other side?

Thanks again, by the way, you've been a huge help.

 

[cmmcnamara]

Yea you'll have to use one or the other depending on which trig function you chose to isolate. You can use the calculator function to do that. No problem, anything to help

 

[Lifeinsteps]

Yea you'll have to use one or the other depending on which trig function you chose to isolate. You can use the calculator function to do that. No problem, anything to help

I'm still not really sure, exactly, because of the unknown variable...

Won't it end up looking like this?

309.5 -200cos(Θ1) = -200sin(Θ1)

arccos{309.5 -200cos(Θ1)} = arccos{-200sin(Θ1)}

Because when I try to put it either of those I just get a domain error. I can only assume that I'm not typing it right?

Edit: To clarify, I'm just not really sure what to type in, or where I'm supposed to use the arccos.

Oh wait, you said to isolate one. Like this?

309.5 - 200cos(Θ1) = sin(Θ1)
-200

But I'm still not sure how to use arcsin because I'll get the domain error if I try to encompass the thing on the left in arcsin.

 

[Lifeinsteps]

Bump!

Sorry, I hope that's allowed, I have to get this problem figured out by 1:00 PM, and it's 8:59 AM now.

Can anyone help me with this last little step? If I can get this angle then the rest of the problem is pie.

 

[SammyS]

I'm still not really sure, exactly, because of the unknown variable...

Won't it end up looking like this?

309.5 -200cos(Θ1) = -200sin(Θ1)

arccos{309.5 -200cos(Θ1)} = arccos{-200sin(Θ1)}

Because when I try to put it either of those I just get a domain error. I can only assume that I'm not typing it right?

Edit: To clarify, I'm just not really sure what to type in, or where I'm supposed to use the arccos.

Oh wait, you said to isolate one. Like this?

309.5 - 200cos(Θ1) = sin(Θ1)
-200

But I'm still not sure how to use arcsin because I'll get the domain error if I try to encompass the thing on the left in arcsin.

Technically, you should not bump until after 24 hours, but you've worked quite a bit on this so you might get away with it.

Assuming the following is correct to to enough decimal places:

309.5 -200cos(θ₁) = -200sin(θ₁)​

First divide by -200 →

-1.5475 + cos(θ₁) = sin(θ₁)​

Subtract cos(θ₁) from both sides. Then square both sides being sure to use FOIL on the right. Of course, sin²(θ₁) + cos²(θ₁) = 1

Then you will have a 2sin(θ₁)cos(θ₁) which is sin(2θ₁).

Then your almost there.

 

[Lifeinsteps]

Technically, you should not bump until after 24 hours, but you've worked quite a bit on this so you might get away with it.

Assuming the following is correct to to enough decimal places:

309.5 -200cos(θ₁) = -200sin(θ₁)​

First divide by -200 →

-1.5475 + cos(θ₁) = sin(θ₁)​

Subtract cos(θ₁) from both sides. Then square both sides being sure to use FOIL on the right. Of course, sin²(θ₁) + cos²(θ₁) = 1

Then you will have a 2sin(θ₁)cos(θ₁) which is sin(2θ₁).

Then your almost there.

Thanks so much, I figured out the problem based on your last comment, and just in time.

I ended up being the only one in the class who was able to complete the assignment, which made me feel almost bad considering how much help I had to get for it.

The professor liked it though!

Thanks again, you two. I know where to go with my physics questions from now on. I was actually taught something instead of just given the answer, and I'm sure the knowledge I gained will come in handy in the future!

 

[SammyS]

...

Thanks again, you two. I know where to go with my physics questions from now on. I was actually taught something instead of just given the answer, and I'm sure the knowledge I gained will come in handy in the future!

You're welcome.