(adsbygoogle = window.adsbygoogle || []).push({}); 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 ?

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# Homework Help: Direction of Reaction forces?

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