Engineering QuestionOctober 18, 2009 10:03 PM   Subscribe

Any engineers or physicists who can help with a vector mechanics problem?

http://i36.tinypic.com/35a2fc8.jpg That is the diagram. The problem is as follows: A 0.5 X .8 m gate AB is located at the bottom of a tank filled with water. The gate is hinged along its top edge A and rests on a frictionless stop at B. Determine the reactions at A and B when cable BCD is slack.

I've been trying all weekend and I really have no idea how to do this problem. Any help would be nice
posted by nickhb to Education (5 answers total)

There's not enough information there to solve the problem, because the weight of the gate isn't given.

If you neglect that and assume that the gate is weightless, you can assume that any force due to water pressure is normal to the surface of the gate.

Does that help?
posted by flabdablet at 10:30 PM on October 18, 2009

Response by poster: I got the point that the total weight is the area of the trapezoid defined by the weight of the water. It is applied normal to the gate on the line of the centroid of that area. But now I have four unknowns (Ax, Ay, Bx, By) and only three equilibrium equations.
posted by nickhb at 10:55 PM on October 18, 2009

I would change your axis. Make the X along the gate and Y perpendicular to that. Then you just have one force from the bottom into the gate and the X and Y forces at A.
posted by DJWeezy at 11:07 PM on October 18, 2009

The reaction at B is going to be normal to the gate from the way that it's holding the gate shut. This should give you another equation on how Bx and By relates to each other. You'd be better off rotating the FBD so that reaction at B and the force from the water are vertical after you determine the magnitude of pressure from the water (this essentially removes Bx after rotating).
posted by woolylambkin at 11:12 PM on October 18, 2009

Response by poster: Got it. I forgot that the support reaction of a frictionless surface is perpendicular to itself, so B is normal to the gate. I had one less unknown and I could solve it. Thanks for your help.
posted by nickhb at 11:16 PM on October 18, 2009

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