This example problem consists of a two dimensional rigid body equilibrium problem. In this case we have a plate that is supported by a pin at point A and by a roller at point B. There are also two forces acting on the object and a couple acting in the upper left-hand corner of the object. Professor Olgesby will take us through the solution to this problem.

The first step is to draw a free-body diagram showing the applied loads and the reactive forces. The pin at A will have two forces one in the horizontal and one in the vertical direction The roller at B has simply a vertical unknown. Once we have drawn the free body diagram, we apply the basic equilibrium equations and solve. We will start out using summation of moments about point in order to elminate A_x and A_y from the equation. When we write this equation we will be able to find the reaction at point B.

Excuse me, Professor Olgesby, but I see from this equation we that we have simply added in the couple that acts at the upper-left hand corner of the plate. Why can we simply add that on?

The couple is a free vector, so the two-hundred ft^.lb couple in the upper-left point acts anywhere on the plate. If we write the moment equation about any point on the plate that couple will be in the equation.

For that matter, is it really nessecary to write our moment equation about a point that is actually part of the object?

It doesn't have to be on the object, but it is easier because we can eliminate forces.

What we would do then once we have written our moment equation and found the reaction at point B, is to sum forces in the horizontal, or x, direction and sum forces in the vertical, or y, direction to find A_x and A_y. This wil l give us a complete solution to the problem.

An alternate solution, would exist in that we could use two moment equations and one force equation as long as the direction we sum forces in is not perpendicular to the line between the two moment points. In this case, we c an use moments about point A and a summation of moments about point B. The force equation which we write is in the horizontal direction. It is not a line or a direction that is orthogonal to the line AB. These equations will give us a solution to the p roblem also.

A third possiblity, would be to use three moment eqautions as long as the three points are not on a straight line. So, if we used the summation of moments about points A, B, and about point G, which does not line on the AB line, these would also give a valid solution to the problem.

Thank you, Professor Olgesby.