2d static body; three unknown forces

[togo]

Homework Statement

Member ABD is supported by a pin at A and a pin in a smooth slot at B. Member BC serves as a support post. A 100 N force is applied at D. Determine the reactions at A and B necessary for equilibrium of ABD.

The Attempt at a Solution

Ma = 0 = (.5)(Fb) - (1)(83.74)
Fb = 167.48

but the answer (for B) is around 190 Newtons. I know I didn't break this question into x and y factors. Could someone explain how to do that, thanks.

edit: the angle between the normal of the lever and the 100 N force is 33.13 degrees, resulting in mostly a Y force operating on the end but a small X force as well.

 

[PhanthomJay]

You have the right idea, but moment is force times perpendicular distance, or , alternatively, F(d)sin theta, where theta is the angle in between the force and position vector (the angle between the 100 N force and the rod, in this example). It may be easier to use this latter approach rather than try to break up F into its components parallel and perpendicular to the rod, where the parallel component does not contribute to the moment. In either case, you need to do some geometry to calculate the angles or force componentsts.

 

[togo]

I did that and got: Ax = 59.1 N and Ay = 80.6 N but failed to get the answer of 191.4 N as force at B

 

[PhanthomJay]

Calculate using geometry the angle between the applied force and the bar. Call it theta.
The force B must be perpendicular to the bar, since the slot does not allow for any forces to exist there in the direction parallel to the bar.

Now sum moments about A to solve for B, noting that the moment of the applied force , F, is F(r)sintheta, where F =100, r =1, and theta you calculate from the geometry of the setup, and the moment of B is force times perpendicular distance. When you solve for the Force B, the direction of the force is perpendicular to the bar. The Bx and By force components you can get from the trig.