Statics: Force on a member of a frame

[dorado29]

Homework Statement

Determine the forces acting on member ABCD

The 100 lb force acts horizontally at point D, the large, bold circle is a wheel, and the smaller, less bold circles indicate pins connecting the members. Member BE is solid.

I forgot to label point F in the picture. Whenever I refer to point F, it will be the point at the wheel. Also, lengths AB = 12, CE = 6, EF = 6, BC = 6 and CD = 6

The Attempt at a Solution

When taking the moment around A, the x components of the B and C forces will create a moment that "cancels out" the moment from the 100 lb force (right?). However, this is probably one of the later steps. I'm pretty sure I should start by analyzing member CEF.

There are x and y components at points C and E and just a y component at F. I'm not sure where to go from here.. I think that because BE is a two force member, the force vector at E is directly towards B and B's vector directly at E.

 

[dorado29]

FBD of member CEF

It just dawned on me that if the magnitude of Cx = Ex then the magnitude of Bx is also the same, right? So from there I could find the moment about A:

(assuming clockwise is positive)

M_A = 0 = (100lbs)(24in) + (Bx)(12in) - (Cx)(18in) would then turn into

2400 + 12x - 18x = 0, x = 400 so the magnitudes of Cx, Ex and Bx all = 400Andddd then: since BE is basically a 45-45-90 triangle, Ey and By would also = 400?

 

[pongo38]

Although you are right in your answers, this is possibly more by luck of this particular geometry. You could obtain Fy by taking moments about A for the whole thing. Then deduce Ax and Ay. Graphically, the applied load and Fy meet at a point, through which the reaction at A must pass. Hence the force triangle for external forces gives you all the reactions and their components. The lesson to learn is that when taking moments, you must be careful to identify clearly the object about which you are making an equilibrium statement, and include all the forces action on it.