Three hollow metal rods, of equal length and negligible mass, are connected by p

Question

Three hollow metal rods, of equal length and negligible mass, are connected by pins to form an triangular frame as shown. The frame stands upright, and a weight is hung from the vertex of the triangle frame.

What is the tension in the horizontal rod which forms the base of the triangular frame?
Answer in units of N.

Please provide an indepth explanation. I need to be able to rework this problem and understand it.

Explanation / Answer

Let A, B, C be the vertices of the triangular frame, with A as vertex. Force 503 N acts at A vertically downwards. The two rods AB and AC will share the force 503 N equally. Let F be the force acting on each side AB and AC, along their lengths. Balancing forces at A, F cos ( 30* ) + F cos ( 30* ) = 2 F cos ( 30* ) = 503 N (* denotes degrees) => F = 290.4072 N Now, the two forces along AB and AC are inclined at 60* to the horizontal BC. Calculating the total force on BC by body force balance, Horizontal force = 2 F cos (60*) = 290.4072 N Vertical force = 2 F sin (60*) = 503 N Now, this result was obvious considering the system as a whole, that the vertical force on the horizontal base had to be equal to the weight-force. The tension in the horizontal rod is, thus = 290.4072 N in horizontal direction pulling from both sides. OR, it is 290.4072 N/ 5.5 m = 52.8 N/m at the horizontal base. Hope it helps!

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