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Load formula charter

hey guys welcome back in this video we are doing two more examples of distributed loads so we have a constant distributed load here and also a triangular distributed load and so let's see what we can do if we can solve for both of these when we do solve for these we need two pieces of information we need the entire load that is being exerted on the object by the distributed load and we also need to find out the location of the centroid of the shape of this distributed load because that will be important for us when we're solving for the moment that it's causing on the object about some point so for the first one here for a rectangular or a constant distributed load it's pretty easy we have 20 kilonewtons per meter and this is acting on six meters and so the whole load that this distributed load is applying is 20 times 6 that's going to be 120 kilo Newtons all right the other thing that we need to do is I said we need to centroid and because this is a rectangular we know that the centroid is going to be right in the center of that rectangle located halfway along the base and so yeah let's go ahead let's start our force balance here we have the sum of forces in the Y direction maybe we should actually draw on the reactions here too so we have a Y and we have we'll have B Y and there's no horizontally applied forces so we don't need to worry about their horizontal reaction at eight all right so the sum of forces in the Y direction we have a y plus b1 is equal to or whatever minus 120 kilonewtons we can just say brain to the side right away is equal to 120 kilonewtons to get that force balance because this object is in static equilibrium and then what we want to do is we want to solve for the moments sum of moments although let's say a will define on here a little positive sense all right so the things that are going to cause moments about a is it's going to be this reaction here and that was supposed to be called B Y B Y will the moment that V Y causes will be the for the magnitude of that force times distance perpendicular so times six meters and then the magnitude and that was positive right because that would give the tendency to rotate about point a in this direction and that's what we're saying at the closet direction where you can see for the distributed load it would give the tendency for this object to rotate the other way and we're saying up the negative directions what we're going to do is we'll subtractive and in order to do this that the moment that's caused by this distributed load is the whole load itself is 120...