1. The problem statement, all variables and given/known data 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. 2. Relevant equations [tex]\sum Moments_on_A[/tex] = 0 [tex]\sum Moments_on_D[/tex] = 0 [tex]\sum Fy[/tex] = 0 [tex]\sum Fx[/tex] = 0 3. 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 [Broken] My calculations: (AK, AL, AB and so on represent the distance between point A to point K, L, B respectively.) 1) [tex]\sum M on A[/tex] = 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) [tex]\sum M on D[/tex] = 0 = - R_A_y*AD + W_1*DK - W_2*DL - Psin(60)*DB Solved for R_A_y = -155 N 3) [tex]\sum Fx[/tex] = 0 = Pcos(60) + R_A_x + R_D_x = 0 R_A_x + R_D_x = 500N 4) [tex]\sum Fy[/tex] = 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 ?