First Condition of Equilibrium

We may say that an objectat restis in equilibrium or in static equilibrium. An object at rest is described by Newton's First Law of Motion. An object in static equilibrium has zero net force acting upon it.

TheFirst Condition of Equilibriumis that thevector sumof all the forces acting on a body vanishes. This can be written as

F = F_{1}+ F_{2}+ F_{3}+ F_{4}+. . . = 0

where , the Greek letter sigma, again means the summation of whatever follows -- the summation of the forces, in this case.

That's all there is!

However, remember the followingEnsure that you have included all the forces. This means carefully draw a free body diagram. Include gravity (the weight) and all contact forces.

Remember that forces are vectors.

That means that the first condition of equilibrium,F = 0

really means

F_{x}= 0andF_{y}= 0

Example: Consider the 98 newton weight (or 10 kg mass) supported by a rope.

The tension in the rope attached to the 98 newton weight is just 98 newtons.

But this rope is now tied together with two other ropes as shown here.

What forces are exerted by the other two ropes?

To answer this, look at the forces exerted on the knot where the three ropes are joined. The knot at rest so the sum of the forces acting on it must be zero.

Draw a free-body diagram showing just those forces.

T is the magnitude of the force on the knot from the rope attached directly to the weight. T_{L}is the magnitude of the force on the knot from the rope on the left that makes an angle of 45° with the horizontal. And T_{R}is the magnitude of the force on the knot from the rope on the right that makes an angle of 30° with the horizontal. These three forces must add to zero. Graphically, they can be added as shown here; the force vectors form a closed triangle.

Now we apply the first condition of equilibrium,F = 0

which really means

F_{x}= 0F_{y}= 0First, resolve all the forces into their x- and y-components.

F0_{x}=

FT_{x}= -_{L}cos 45^{o}+ T_{R}cos 30^{o}= 0- 0.707 T

_{L}+ 0.866 T_{R}= 00.866 T

_{R}= 0.707 T_{L}F0_{y}=

FT_{y}=_{L}sin 45^{o}+ T_{R}sin 30^{o}- 98 N = 00.707 T

_{L}+ 0.5 T_{R}- 98 N = 00.707 T

_{L}+ 0.5 T_{R}= 98 NNow we can substitute, 0.707 T _{L}+ 0.5 T_{R}= 98 N0.866 T

_{R}+ 0.5 T_{R}= 98 N( 0.866 + 0.5 ) T

_{R}= 98 N1.366 T

_{R}= 98 NT

_{R}= 98 N / 1.366T

_{R}= 71.7 N0.866 T _{R}= 0.707 T_{L}T

_{L}= (0.866 / 0.707) T_{R}T

_{L}= 1.22 T_{R}T

_{L}= 1.22 ( 71.7 N )T

_{L}= 87.8 N

Rotational EquilibriumSecond Condition of EquilibriumReturn ToC, Rotational Dynamics(c) 2005, Doug Davis; all rights reserved