Equations of Equilibrium - First year engineering

[aidan.s]

Homework Statement

As shown, a roller at point A and a pin at point B support a uniform beam that has a mass 24.0kg . The beam is subjected to the forces F₂ = 75.0N and F₂ = 58.0N . The dimensions are L₁ = 0.550m and L₂ = 1.70m . What are the magnitudes F_A and F_B of the reaction forces F_A and F_B at points A and B, respectively? The beam's height and width are negligible.picture:
http://session.masteringengineering.com/problemAsset/1180309/3/1118668_004.jpg"

Homework Equations

\sumF_X = 0
\sumF_Y = 0
\sumM_O = 0

The Attempt at a Solution

24.0 kg * 9.8 = 235.2N of force at (.55+1.7)/2 = 1.125m

trying to solve for the normal force at A, N_A **anticlockwise = positive**
(75N * 1.7m) + (235.2N * 1.125m) - (2.25m * N_A * 3/5) = 0
N_A = 270N

solving for B_X
sum of forces in x direction = 0; **positive direction to the right**
-B_X - 58.sin(15) + (N_A*4/5) = 0
B_X = 200

solving for B_Y
sum of forces in the y direction = 0l **positive direction is upwards**
-235.2 - 75 - 58.cos(15) + (N_A * 3/5) + B_Y = 0
B_Y = 204

so as the question asks, F_A = 270N ? is this correct and;
F_B = \sqrt{} 200^2 + 204^2 = 290N ? - is this the right way to find the magnitudes of the force?this is the basic method I've been trying with a few minor variants but this program says there all wrong so I'm completely lost, I'm not expecting anyone to give me a straight answers but some direction as to how I've gone wrong would be great

im pretty sure its just the fact that the roller at point A is on an angle is what's stuffing me up

thanks, aidan

 

[tvavanasd]

You may have your 3/5 and your 4/5 swapped (3 places) - check.

 

[aidan.s]

./facedesk

thanks, i think I am going to stick to just converting them triangles into actual angles and use sin n cos otherwise i keep stuffing them up :/

cheers though yeah :)