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OpenStudy (anonymous):

any one knows how to find the reaction force in the horizontal and vertical directions in a bridge made with triangles?

OpenStudy (anonymous):

put up a picture or more info

OpenStudy (anonymous):

OpenStudy (anonymous):

For the bridge shown in the diagram, find the reaction force at A: using the choices given below, enter the magnitudes of the horizontal and vertical components, in this order. E.g., enter dd. a) 2F b) 3F c) 3F/4 d) F/3 e) F/2 f) 3F/2 g) F/4 h) 2F/3 i) F j) 0

OpenStudy (anonymous):

That is the entire question.

OpenStudy (anonymous):

I think vertical is 2F and horizontal is zero because if you look at vertical lines on either side of AB there are 3 up and one down making the difference 2 and in the horizontal the lines are equal either side of F so the are balanced so I think 0. Haven't done these for a while have to go out if you didn't get another answer I get back to you later.

OpenStudy (anonymous):

Thank you :)

OpenStudy (anonymous):

Unfortunately, that was not the answer. But I see how the reasoning would have worked.

OpenStudy (anonymous):

Your reasoning helped me get the answer though. The diagonal line = 1/2. so it would be F/2.

OpenStudy (anonymous):

great

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