## Equilibrium

We continue with our established pattern of identifying all the forces, making a clear "free body diagram", and applying Newton's Second Law,

F= maFor bodies at rest -- we say they are in equilibrium, we know the acceleration. It is zero. If a = 0 then we know F = 0. Remember, tho', force is a vector. So saying

F = 0really means

F_{x}= 0and

F_{y}= 0

Example:Consider a lamp hanging from a chain. What is thetensionin the chain?As always, begin with a "free body diagram". Tension

Tactsupwardon the lamp while the force of gravity pullsdownwith forcew, theweightof the lamp. Thenet forceis thevector sumof these two forces. The lamp isnot acceleratingso the forceupmust equal the forcedown. In terms of magnitudes, this meansT = w

Tension:Tension is themagnitudeof the force exerted by a chain or a rope or a string. The direction of that force depends upon the rest of the situation and the object that we are concentrating on at the moment. If we concentrate on the chain shown below, the downward forceT'is the force exerted on the chain by the lamp while the upward forceT''is the force exerted on the chain by the ceiling. There is no substitute for good free-body diagrams.

Example:Consider a traffic light suspended by cords as shown in the sketch below. What is the tension in each of those cords?Tension T

_{3}is easy so we will look at the first. As we have seen in the previous two examples; this tension in a vertical cord supporting a weight is just equal to the weight. In the diagram below, we have drawn in the forces acting on the traffic light. The only forces acting on the traffic light arew, the weight acting downward, andT_{3}, the upward force due to the vertical cable. T_{3}is the tension in this cable. ClearlyT _{3}= wBut what of the tensions in the other two cables, T

_{2}and T_{1}? To find those, we must look at the junction where the three cables come together. That junction is in equilibrium soF_{net}= 0

F_{net}=F=T_{1}+T_{2}+T_{3}= 0However, we must remember that this single

vector equationis elegant, shorthand notation fortwo scalar equations,F _{net,x}= F_{x}= T_{1 x}+ T_{2 x}+ T_{3 x}= 0F

_{net,y}= F_{y}= T_{1 y}+ T_{2 y}+ T_{3 y}= 0So we must resolve all these forces into their x- and y-components,

T _{1x}= - T_{1}cos 37^{o}= - 0.8 T_{1}T

_{1y}= T_{1}sin 37^{o}= 0.6 T_{1}T

_{2x}= T_{2}cos 53^{o}= 0.6 T_{2}T

_{2y}= T_{2}sin 53^{o}= 0.8 T_{2}T

_{3x}= 0T

_{3y}= - T_{3}= - wThe

signsareimportant!Now we can go back to the component equations and solve for tension T_{1}and T_{2}.F _{net,x}= F_{x}= T_{1 x}+ T_{2 x}+ T_{3 x}= 0T

_{1 x}+ T_{2 x}+ T_{3 x}= 0- 0.8 T

_{1}+ 0.6 T_{2}+ 0 = 0T

_{1}= 0.75 T_{2}F

_{net,y}= F_{y}= T_{1 y}+ T_{2 y}+ T_{3 y}= 0T

_{1 y}+ T_{2 y}+ T_{3 y}= 00.6 T

_{1}+ 0.8 T_{2}2 - w = 00.6 T

_{1}+ 0.8 T_{2}= w0.6 (0.75 T

_{2}) + 0.8 T_{2}= w1.25 T

_{2}= wT

_{2}= 0.8 wT

_{1}= 0.75 (0.8 w)T

_{1}= 0.6 w

(c) Doug Davis, 2005; all rights reserved