Find the angle of a force by means of graphic solution

[kaffekjele]

Homework Statement

A beam ABC is fastened in C and is supported by the axially loaded bar BD. There is a load F working in A.
Disregard any friction.
Use the given measurements to find the force in the axially loaded bar BD, the force in C and the angle of the force in C by means of graphic solution.

Original figure can be found here: http://i45.tinypic.com/14b2m9d.jpg

Homework Equations

The line of action of the three forces must meet in one point for equilibrium to occur.

The Attempt at a Solution

http://i46.tinypic.com/34ih0y0.jpg

What I've done is draw the line of action for the forces working through F, B and C until they meet in a single point.
I've then drawn the resulting vector triangle by starting with force F since that's the only one that is known. After that I continued the triangle by displacing the force working through C and then moved the force BD to complete the triangle. (I'm not sure if this is correct, so feel free to comment.)

I'm not sure how to find the angle of the force working through point C. I assume they mean the angle of the force relative to the beam, but I'm not sure.
Anyway, what I've tried is to use invers tangens 2,2 divided by 3,3(by means of using the measurements from the original figure) which gave an angle of 33,7 degrees, but this was not correct.

 

[haruspex]

What I've done is draw the line of action for the forces working through F, B and C until they meet in a single point.

So far so good. Call the intersection point P.

I've then drawn the resulting vector triangle by starting with force F since that's the only one that is known. After that I continued the triangle by displacing the force working through C and then moved the force BD to complete the triangle.

I don't understand how you decided where to move BD to. It doesn't look parallel to the original BD. Maybe it's just the perspective in the image.

I'm not sure how to find the angle of the force working through point C. I assume they mean the angle of the force relative to the beam, but I'm not sure.

Yes, it's unclear - it could mean angle from vertical.

Anyway, what I've tried is to use invers tangens 2,2 divided by 3,3(by means of using the measurements from the original figure) which gave an angle of 33,7 degrees, but this was not correct.

Those are the wrong lengths. You want the angle PCA; which lengths tell you the tangent?

 

[kaffekjele]

I don't understand how you decided where to move BD to. It doesn't look parallel to the original BD. Maybe it's just the perspective in the image.

It's not parallell to the orignial BD, no. I moved it because I didn't see any other way to complete the triangle without having to move some of the other forces as well. I looked at an example in my coursebook and it seemed this was an ok thing to do, but I might have gone about it the wrong way so feel free to comment.

Those are the wrong lengths. You want the angle PCA; which lengths tell you the tangent?

Uhm. I fear I'm about to feel stupid now, but I don't see any other way. I have the length of the beam (2,1+1,2= 3,3m) which I guess would be my adjacent side, and the length between fastenings C and D(2,2m) which would be my opposite side.

In addition you could use the Pythagoream theorem to calculate the length of the axially bar BD, but I don't see how that would be of any use since the forces are displaced.

 

[haruspex]

It's not parallell to the orignial BD, no.

It needs to be. You can change the line of action to create the triangle, but the lengths must relate to the magnitudes and the directions must be preserved.

I have the length of the beam (2,1+1,2= 3,3m) which I guess would be my adjacent side, and the length between fastenings C and D(2,2m) which would be my opposite side.

But using those lengths will give you the tangent of angle CAD, whereas you want angle PCA. Do you see that? So you need to find the length AP. You can get that from finding two similar triangles.

 

[Nickie]

Your approach of drawing the force vectors and forming a vector triangle is correct. To find the angle of the force in C, you can use the law of cosines. In this case, we have a triangle with sides of 3.3, 2.2, and 3.3 (since the force in C is equal to the force in BD). The angle opposite the side with length 3.3 is the angle we are looking for. So we can use the law of cosines to solve for this angle:

c^2 = a^2 + b^2 - 2abcosC

(3.3)^2 = (2.2)^2 + (3.3)^2 - 2(2.2)(3.3)cosC

10.89 = 4.84 + 10.89 - 14.52cosC

14.52cosC = 4.84

cosC = 4.84/14.52 = 0.333

C = arccos(0.333) = 70.5 degrees

So the angle of the force in C is 70.5 degrees relative to the beam. To find the forces in BD and C, you can use the law of sines. Since we know the angle at C and the lengths of two sides (3.3 and 2.2), we can solve for the force in BD:

sinB/3.3 = sinC/2.2

sinB = (3.3)(sin70.5)/2.2 = 2.63

B = arcsin(2.63) = 70.5 degrees

Therefore, the force in BD is equal to 2.63 units. To find the force in C, we can use the same equation, but with the angle at B:

sinC/3.3 = sinB/2.2

sinC = (3.3)(sin70.5)/2.2 = 2.63

C = arcsin(2.63) = 70.5 degrees

So the force in C is also equal to 2.63 units. This makes sense since the triangle is symmetrical and the forces should be equal.