# 9.4 Applications of statics, including problem-solving strategies

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• Discuss the applications of Statics in real life.
• State and discuss various problem-solving strategies in Statics.

Statics can be applied to a variety of situations, ranging from raising a drawbridge to bad posture and back strain. We begin with a discussion of problem-solving strategies specifically used for statics. Since statics is a special case of Newton’s laws, both the general problem-solving strategies and the special strategies for Newton’s laws, discussed in Problem-Solving Strategies , still apply.

## Problem-solving strategy: static equilibrium situations

1. The first step is to determine whether or not the system is in static equilibrium    . This condition is always the case when the acceleration of the system is zero and accelerated rotation does not occur .
2. It is particularly important to draw a free body diagram for the system of interest . Carefully label all forces, and note their relative magnitudes, directions, and points of application whenever these are known.
3. Solve the problem by applying either or both of the conditions for equilibrium (represented by the equations $\text{net}\phantom{\rule{0.25em}{0ex}}F=0$ and $\text{net}\phantom{\rule{0.25em}{0ex}}\tau =0$ , depending on the list of known and unknown factors. If the second condition is involved, choose the pivot point to simplify the solution . Any pivot point can be chosen, but the most useful ones cause torques by unknown forces to be zero. (Torque is zero if the force is applied at the pivot (then $r=0$ ), or along a line through the pivot point (then $\theta =0$ )). Always choose a convenient coordinate system for projecting forces.
4. Check the solution to see if it is reasonable by examining the magnitude, direction, and units of the answer. The importance of this last step never diminishes, although in unfamiliar applications, it is usually more difficult to judge reasonableness. These judgments become progressively easier with experience.

Now let us apply this problem-solving strategy for the pole vaulter shown in the three figures below. The pole is uniform and has a mass of 5.00 kg. In [link] , the pole’s cg lies halfway between the vaulter’s hands. It seems reasonable that the force exerted by each hand is equal to half the weight of the pole, or 24.5 N. This obviously satisfies the first condition for equilibrium $\text{(net}\phantom{\rule{0.25em}{0ex}}F=0\right)$ . The second condition $\text{(net}\phantom{\rule{0.25em}{0ex}}{\tau }_{}=\text{0)}$ is also satisfied, as we can see by choosing the cg to be the pivot point. The weight exerts no torque about a pivot point located at the cg, since it is applied at that point and its lever arm is zero. The equal forces exerted by the hands are equidistant from the chosen pivot, and so they exert equal and opposite torques. Similar arguments hold for other systems where supporting forces are exerted symmetrically about the cg. For example, the four legs of a uniform table each support one-fourth of its weight.

In [link] , a pole vaulter holding a pole with its cg halfway between his hands is shown. Each hand exerts a force equal to half the weight of the pole, ${F}_{R}={F}_{L}=w/2$ . (b) The pole vaulter moves the pole to his left, and the forces that the hands exert are no longer equal. See [link] . If the pole is held with its cg to the left of the person, then he must push down with his right hand and up with his left. The forces he exerts are larger here because they are in opposite directions and the cg is at a long distance from either hand.

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