How Are Reaction Forces Directed at Points A and D in a Supported Beam?

[ana111790]

Homework Statement

The uniform horizontal beam is hinged to the ground at A and supported by a frictionless roller at D. Distance A to B is 4 m, A to D is 3 m. A force P with a magnitude of 1000N makes an angle 60 degrees with the horizontal at point B. The total Weight of beam is 400 N.

By noting that 3/4 of the beam lie between A and D and 1/4 between D and B calculate the x and y components of the reaction forces on the beam at A and D.

Homework Equations

\sum Moments_on_A = 0
\sum Moments_on_D = 0
\sum Fy = 0
\sum Fx = 0

The Attempt at a Solution

I drew the given picture and then drew the Free Body diagram.
The direction of the Reactive forces on A and D is unknown so I chose a random direction (up and to the right) to portray them.

[PLAIN]http://img825.imageshack.us/img825/8133/drawing3.jpg

My calculations:
(AK, AL, AB and so on represent the distance between point A to point K, L, B respectively.)
1) \sum M on A = 0 = -W_1*AK - W_2*AL - Psin(60)*AB + R_D_y*AD
Solved for R_D_y = 1420 N (rounded to 3 sig figs)

2) \sum M on D = 0 = - R_A_y*AD + W_1*DK - W_2*DL - Psin(60)*DB
Solved for R_A_y = -155 N

3) \sum Fx = 0 = Pcos(60) + R_A_x + R_D_x = 0

R_A_x + R_D_x = 500N

4) \sum Fy = 0 = -W_1 - W_2 - Psin(60) + R_D_y + R_A_y = 0
R_D_y + R_A_y = 1270 N Answer in the book is given as: (not rounded)
R_D = 1421 N, R_A_x = -500N, R_A_y = - 155 N

So these answers are giving R_D to be going straight up, not at an angle, so R_D given in the book is equal to my R_D_y, which makes R_D_x = 0.

My question is why is that/ how would I know to set R_D_x = 0 ?

 

[mplayer]

The reaction at point D has only a y-component because it is on a roller and free to move along the x-axis. Point D can only resist forces in the direction perpendicular to the beam. Point A is hinged and will have both x and y reaction components.