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A bridge span is 5m high and 20m long. Find the verticle force and the horizontal thrust on each of the piers P in terms of weight suspended from the center of the span.

I really hope you can understand the geometry of the problem without the drawing(there was a drawing in the book). Imagine the two piers as stumps that are 20m apart. From the upper corner of each stump, on the side facing the other pier, exits a rod in the direction of the other pier and up. The two rods meet at a height of 5m right between the two piers, so the cathete that forms an angle with the rod(hypothenuse) is 10 m long.

I will name the weight F.

Heres how I tried solving:

I reasoned that the rod would transfer the force of the weight to the pier by means of a force which is parallel to the rod.

I found that force by first finding the angle between the 10m length and the rob using tan w = 5 / 10.

With the angle I find the componant parallel to the rod:

sin w = F / F(parallel to rod) --> F(parallel to rod) = F / sin w

I reason that the horizontal componant that Im searching for is the x componant of F(parallel to rod).

I find it by cos w = F(horizontal) / F(parallel to rod)

=H(horizontal) * sin w / F

which becomes: F(horizontal) = (cos w / sin w) F

= 2 F ; because cos / sin = 1 / tan, and tan w = 0.5

I know for certain that my answer is wrong.

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# Homework Help: Particle statics - A bridge span

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