Statics: Equilibrium in 2 Dimensions, Part II

 Here, we only have 2 equations in equilibrium, and we set in equilibrium conditions and use facts in order to find the forces that we don't know.  The only way the beam can be simpler is if the force is loaded right on the center. The bean pushes down on force and then we get some reaction forces. 

Remember that the beams are inelastic, meaning that they do not deform at all. 

The moment must be identified. In 2 dimensions, the moment is only in the Z direction anyways, and a reference point is moments caused by force with respect to a particular point. 

The more unknown forces going through one of the points, the easier the mathematics is if you choose that point. The clockwise moment has to be equivalent to the counterclockwise one. If not, then there will be angular acceleration. That's simple, all that happens now is that the problems get more and more involved. 

Now let's set a support at a slight angle of 30 degrees. We need to have just as many unknowns as the number of equations. Here there are 3 unknowns, Ax, Ay, and the magnitude of the direction after the 30 degree angle. 

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Now, let's go to the next question. Looks like only the Y component is pushing up here. Just remember that the sum of the forces is equivalent to 0 and the sum of the moments equal 0. 

Correction, answer is 1.4 because the angle is perpendicular to the plane, which should be 60 degrees as a result. 

30 + 90 = 120 degrees. 180 degrees - 120 degrees = 60 degrees. 

Here's another setup, this time we have a wall into which we embedded a beam of some kind; however, partway in there's a partition or something with a few forces, midway with 200lbs, with 100lbs at 30 degree, and by whatever means a pulley or a motor at the end/fan belt, an applied moment of 300 ft lbs. 

PLOT TWIST: the arrow is pointing at the other direction, but everything else is the same. 

First thing we need, is a free body diagram again. 

We bore things to the wall so that we don't want something particular to turn. 

We have 200 down and less than 100 up so as a result, we're going to need some more up force. 

We have a bit of force pulling to the right, so we know the wall has got to pull this thing back to the left as a result. 

Ax drops out because its line of action goes right through the point of M. 200 doesn't drop out. We want to see which point's "line of action" goes through the point of the moment.

 

Forces are sliding vectors, and moments are free vectors, so you can't make it go away by picking a different location, just happens to be at a certain location. 

Remember, the equation is that anything counterclockwise would be equivalent to anything clockwise. You put the moment SOLELY on what the line of action is. 100 lb cos 30 has a line of action on the little jut, everything else has a line of action on the longer line. 

Getting a negative number means we just drew things backwards in the terms of the predicted moment. We want to get the direction of the arrow (With Respect To) the line of action. We get moment = 73 ft lb as a result. 

Let's go for another one. We have a 6kN force at some angle alpha and we want to find a reaction of A, a magnitude, and direction/its 2 orthogonal components based on the alpha defined. 

up to you. 

This is a problem applied very often in the real world, going over in the next article.