Calculating Support Forces in Statics: Solving Equations for A and B

[aaronfue]

I've been trying to figure this out for 4 days now. I'm completely stumped. I've attached a picture of the diagram: http://postimage.org/image/t3d5i0l1v/

I found the force in F:
F = 25.04 lb ---> 16.92i + 18.46j
Now I need to find the support forces at A and B.

I'm not sure if this is right but I came up with the following equations:

I used the angle at the supports as 42°.

ƩFx=0= -Bx*cos42 - Ax*cos42 +16.92
ƩFy=0= - By*sin42 - Ay*sin42 +18.46

Then I thought about taking the moment about point A:

ƩMa=0=Fx*2 + Fy*9 - Bx*cos42 *(9) - Bysin42 *(8)

Then the moment about the other points F and B. A the end I got the following equations.
ƩFx (Eq. 1): (0.743)Ax + (0.743)Bx = 16.9
ƩFy (Eq. 2): (0.670)Ay + (0.743)By = 18.5
ƩMa (Eq. 3): (6.7)Bx + (5.63)By = -200
ƩMb (Eq. 4): (6.7)Ax - (5.63)Ay = 100

Now when I try to solve these equations my unknowns cancel out! What’s the deal!??

 

[SteamKing]

It's not clear why F = 25.04 lbs. What was the original problem statement?

 

[aaronfue]

I'm sorry I didn't clarify the force, F.

The given information was: "The force F creates a 200 lb moment about point A and a 100 lb moment about point B. Calculate the force F." I created the equations to find the force using the geometry and points given in the diagram.

The question did not ask about the support forces but I want to figure them out just for knowledge.

 

[aralbrec]

I used the angle at the supports as 42°.

ƩFx=0= -Bx*cos42 - Ax*cos42 +16.92
ƩFy=0= - By*sin42 - Ay*sin42 +18.46

I'm not sure why you are multiplying Bx, By, Ax, Ay by angles. The component of B in the x direction is Bx. The component of B in the y direction is By.

If B was your vector and B was the magnitude of the vector B then I could see Bx = B cos 42, By = B sin 42, etc. But you notation is shorthand for B = Bx i + By j

 

[PhanthomJay]

Please confirm coordinates of the point of application of F. The sketch shows (4,3) but it scales as (3,4).

Also, Bx and By are not trig related as stated, since member AB is not a 2 force member.

 

[aaronfue]

The coordinates for F are (4,3).

 

[aaronfue]

You're right aralbrec...I usually use them as a reference for which axis I'm working with. I completely forgot about that! Thanks for the reminder!

 

[PhanthomJay]

I've been trying to figure this out for 4 days now. I'm completely stumped. I've attached a picture of the diagram: http://postimage.org/image/t3d5i0l1v/

I found the force in F:
F = 25.04 lb ---> 16.92i + 18.46j
Now I need to find the support forces at A and B.

I'm not sure if this is right but I came up with the following equations:

I used the angle at the supports as 42°.

from the vertical

ƩFx=0= -Bx*cos42 - Ax*cos42 +16.92
ƩFy=0= - By*sin42 - Ay*sin42 +18.46

your error here has already been addressed

Then I thought about taking the moment about point A:

ƩMa=0=Fx*2 + Fy*9 - Bx*cos42 *(9) - Bysin42 *(8)

Since the support at B is a roller, it can only support forces perpendicular to AB; so in your above equation, take note of this

Then the moment about the other points F and B. A the end I got the following equations.
ƩFx (Eq. 1): (0.743)Ax + (0.743)Bx = 16.9
ƩFy (Eq. 2): (0.670)Ay + (0.743)By = 18.5
ƩMa (Eq. 3): (6.7)Bx + (5.63)By = -200
ƩMb (Eq. 4): (6.7)Ax - (5.63)Ay = 100

Now when I try to solve these equations my unknowns cancel out! What’s the deal!??

make corrections...

 

[aaronfue]

OMG...I can't believe I didn't see that. Thank you friend. I'm done now!