Vector Question of two workers

[robbondo]

Homework Statement

Two workers pull horizontally on a heavy box, but one pulls twice as hard as the other. The larger pull is directed at 25.0 degrees west of north, and the resultant of these two pulls is 350.0 N directly northward.

Use vector components to find the magnitude of each of these pulls and the direction of the smaller pull.

Homework Equations

Vector A + VectorB = VectorR
Magnitude of R = 350

The Attempt at a Solution

Well, Since Rx = 0 = Ax + Bx then Ax=-Bx

and since Ay+By=350 then Ay=350-By and By=350-Ay

I know that Ay=A*cos(25)
and that Ax=A*sin(25)
Also that By=B*sin (x)
and Bx=B*cos(x) Where x is the angle from the x-axis to the vector B.

A=2B since A is twice that length of B

So, the problem here is trying to get these things in terms of each other so that i can solve for only one variable.

I've figured out that since By=350-Ay then By=350-A*cos(25) which is equal to By=350-2B*cos(25).

So that's pretty much where I'm stuck. I know that there has to be some way to relate the fact that Ax=-Bx and A=2B and that the angle of 25 degrees for the C vector, that I have sufficient information to figure this problem out. It seems that I can never get it down to one variable. Maybe there's an equation that I'm forgetting. I've spent more than a few hours on this problem and I'm getting frustrated, it's the last problem in the assignment. I appreciate any help.

 

[cristo]

Well, you know that A_x=B_x (note no - sign, since the small pull must be to the easterly side of the resultant). The above gives, in your notation, Asin(25)=Bcos(x), but then, since A=2B, this becomes, 2Bsin(25)=Bcos(x), which you can solve for the angle.

Does this help?

 

[robbondo]

Well I thought since the pulls were in opposite directions then one would be going West (-) and the other East (+) and because the resultant is on the y-axis my understanding is that since Rx=(Ax+Bx) and Rx equals zero then Ax=-Bx. This is new to me... and for 2Bsin(25)=Bcos(x) I don't understand how to solve if there is two variables B and X.?

thanks for the reply

 

[cristo]

Well I thought since the pulls were in opposite directions then one would be going West (-) and the other East (+) and because the resultant is on the y-axis my understanding is that since Rx=(Ax+Bx) and Rx equals zero then Ax=-Bx.

Rx=Ax-Bx=0 (since, as you said, the two components are in opposite directions.) The easier way to see this is if you simply resolve the vectors into horizontal components then you see that they must equal one another, otherwise there would be some resultant force in the x direction.

and for 2Bsin(25)=Bcos(x) I don't understand how to solve if there is two variables B and X.?

Divide both sides by B (we can do this, since clearly B is non-zero). The you will obtain 2*sin(25)=cos(x). Can you solve this?

 

[robbondo]

Sweet ass... Thanks a lot man, that really helped a lot. I didn't even think to divide out a variable like B that was on both sides. I kept getting two variables and thinking that I was stuck. I should go back and take algebra again. Thanks again cristo. What I got was that x=32.30 and the mag of A=298.28 and B=149.12.

 

[mazrit]

What I got was that x=32.30 and the mag of A=298.28 and B=149.12.

arccos (2.sin 25) = 32.30, but how did you get A and B?

 

[davidorviet]

What if it asks to find south east instead of north east? How would you solve it then?