Resolved resultant force along a specific axis

  • #1

Homework Statement



http://img687.imageshack.us/img687/8767/meprob27.jpg [Broken]

If the resultant force acts along the positive u axis, determine the resultant force and angle theta (red angle)

Homework Equations



law of sines

pythagorean theorem, etc

trig

parallelogram, head to tail methods, something probably missing, etc

The Attempt at a Solution



The resultant force comes from the 3 kN and 2 kN forces.

I try

-30 deg = tan-1(Fy / Fx)

Fy = (2 kN)(sin θ)

Fx = (3 kN) + (2 kN)(cos θ)

and I get stuck here, trying to go by just what I thought was usual trig / force vectors

If I have to draw another triangle to help solve, then I am not sure how.



Thank you.
 
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Answers and Replies

  • #2
741
27
Try writing expressions for the two components of u nd see if that helps....
 
  • #3
I'm not sure how to go about that but do you mean something like:


F(u) = 3000(cos 30) + 2000(cos -(30 - theta)) ?


I'm really not sure what to do, if it means that the resultant force has to go along the u axis or not.

ie Is it saying that ALL the resultant force has to act along the u axis?

I left it out of my original post that it was supposed to be to determine the MAGNITUDE of the resultant force (if the resultant force acts along the positive u axis).

It seems kind of hard when you aren't given that other angle either... any tips?
 
Last edited:
  • #4
741
27
You have successfully written down the two components of the resultant force u. That is
Fy = (2 kN)(sin θ)

Fx = (3 kN) + (2 kN)(cos θ)
So you have two equations and three unknowns. You need another equation.
Suppose Fx and Fy were real numbers like say 4 kN and 5 kN. How would you combine them?
 
  • #5
How about

-30 deg = tan-1[ (2 sinθ)/[3 + 2(cos θ)] ]


Factor out cosine from bottom:

-30 deg = tan-1[ (2 sinθ) / cos θ[3/(cos θ) + 2]


-30 deg = tan-1[ (2 sinθ) / cos θ[3/(cos θ) + 2] make sin / cos = tan:

-30 deg = tan-1[ tan (2)/[3/(cos θ) + 2]

-30 deg = (2)/[3/(cos θ) + 2]


-30 deg = 2 / [3/(cos θ) + 2]

(not sure if I can flip everything over like this and if the other algebra is correct below) :


1 / (-30 deg) = [3/(cos θ) + 2] / 2

(2)/ (-30 deg) = [3/(cos θ) + 2]

Divide everything by 3 (kN, so the kN units cancel out now for sure):

(2)(-10 deg) = 1/(cos θ) + 2/3


(2)(-10 deg) - 2/3 = 1/(cos θ)


Flip everything again:

[(2)(-10 deg) - 2/3]^-1 = cos θ


cos^-1[ [(2)(-10 deg) - 2/3 ]^-1 ] = θ



and i get an error doing this

If I do Pythagorean theorem with Fy and Fx algebraically it just causes R to come up and theta stays so that doesn't work.
 
  • #6
741
27
Colour of Cyan: This line of yours:
-30 deg = (2)/[3/(cos θ) + 2]

should be Tan(-30 deg)=(2tanθ)/[3/(cos θ) + 2]
and you know that Tan(-30) is -1/sqrt 3
hence theta should be attainable. However, if you look at this problem graphically, you will see that there are two possible solutions for theta. Have you tried this approach? First draw a horizontal line 3 units long. From the right hand end draw a line of indefinite length at an angle of 150. With your compass on the left end of the 3, draw an arc 2 units long to intersect the sloping line.
 
  • #7
I didn't think about moving tan^-1 to the other side

so

tan(-30 deg)=(2tanθ)/[3/(cos θ) + 2]

-0.577 = (2tanθ)/[3/(cos θ) + 2]

-1 / 0.577 = [3/(cos θ) + 2]/(2tanθ)

-2tan θ/0.577 = (3/cos θ) + 2

Divide everything by 3, change cos

-2tan θ/-1.732 = sec θ + (2/3)

Divide everything by sec θ:

-2tanθ / -1.732(sec θ) = 1 + (2/3secθ)

change sec back to 1/cos, change tank to sin/cos, factor out:

-2sinθ/-1.732 = 1 + 2cosθ

I'm not sure how to solve this from here, algebraically


This is the right approach to this problem though, right? No shorter way? When I do get this right I hope most of my other homework problems aren't this long, heh.
 

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