seegongsik
statics explained

rigid body equilibrium conditions

Both translation and rotation zero, three planar equations ΣFx, ΣFy, ΣM = 0, reactions, determinacy

A particle is a point, so it stayed still whenever the sum of the forces was 0. A body with size, a rigid body such as a bridge, a crane, or a desk, is a different story, because even balanced forces can twist as a couple and spin the body around. For a rigid body to stay truly at rest, two things are needed at once: it must not be pushed sideways or up and down, and it must not turn either way. As equations, the sum of the forces is 0, and the sum of the moments is 0 too. In a plane these two conditions resolve into exactly three equations, and those three are the key to computing invisible reactions such as the force a bridge support pushes up with. Start with the case where the forces balance and yet the body still turns.

A rectangular rigid body feels two forces of equal size and opposite direction. The sum of the forces is exactly 0, which would be equilibrium for a particle, but press the button to set the two forces off the same line and the resultant is still 0 while the body begins to spin. A couple has appeared and the moment is no longer 0, so the forces are balanced while the rotation is not. Place the two forces back on the same line of action and the moment becomes 0 too, so the body stops completely. This is why a rigid body carries one more condition: not only the sum of forces but also the sum of moments must be 0 before it is in equilibrium.

So the condition for rigid-body equilibrium is two lines. First, the vector sum of all forces is 0, which means the body does not translate in any direction. Second, the sum of all moments is 0, which means the body does not turn about any point you choose. A central load P is held up by two supports, and you adjust the two reactions R1 and R2 so the top force gauge and the bottom moment gauge reach 0 together. Only one of them at 0 is not enough: if just the force is 0 the body tips in place, and if just the moment is 0 it slides as a whole. When both are 0, and only then, the rigid body truly comes to rest.

The two conditions written as vectors resolve, in a plane, into three scalar equations by components. The sum of forces being 0 splits into horizontal and vertical as ΣFx = 0 and ΣFy = 0, and the sum of moments being 0 is ΣM = 0, so together there are exactly three equations. Drag the angle of the inclined load and the three reactions from the pin and the roller move so that all three equations stay satisfied. Three equations means you can solve for exactly three unknowns, which matters a great deal. A particle had two equations and two unknowns, but a rigid body gains one more equation for rotation and can handle three unknowns. The reference point of the moment equation may be anywhere, though a point that many unknown forces pass through makes the equation simpler.

Now see the real power of the three equations. A simply supported beam rests on two supports, with a load P placed off to one side rather than in the middle, and the reactions R1 and R2 that the two supports push up with come straight from the equilibrium equations. Take moments about the left support and set that sum to 0: R1, which passes through the point, drops out, R2 is solved in one step, and then the sum of vertical forces at 0 gives R1. Drag the load position and the closer the load comes to a support, the larger that support's reaction grows. It is the same logic as a seesaw: pile a heavy load to one side and that leg carries more.

Finally, note the limit that three equations can solve only three unknowns. You compare the number of unknown reactions the supports create against the number of equations, and the buttons that change the support arrangement show the difference. A simply supported beam on a pin and a roller has three unknowns, so the three equations solve it cleanly, and that is called statically determinate. A pin at each end makes four unknowns, one too many, and that is statically indeterminate: equilibrium alone cannot solve it, and the deformation of the material has to be counted too. The other way, too few supports and two or fewer unknowns, and the body moves, an unstable structure, a mechanism. Matching these counts is the starting point of sound structural design.

In PracticeTo sum up, a rigid body is in equilibrium only when two things are 0 at once, the sum of forces 0 and the sum of moments 0. In a plane that is ΣFx = 0, ΣFy = 0, and ΣM = 0, three equations that solve up to three unknowns such as support reactions. Choose the moment reference freely, but a point an unknown force passes through drops that force from the equation and makes the work easier. Three unknowns is statically determinate and equilibrium alone is enough, four or more is indeterminate and needs deformation, and fewer than three is a moving mechanism. That closes the equilibrium of forces and moments. The next lesson is the diagram every one of these solutions starts from, the free-body diagram that draws every force on a body.
← statics explained
Was this helpful? Support seegongsik