# 11.4 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.

find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
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Abhi
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can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
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Abhi
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Abhi
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Abhi
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salma
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Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
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Kristine 2*2*2=8
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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Cied
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Porter
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Porter
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Yasmin
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Cesar
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Uday
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Stotaw
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Azam
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Prasenjit
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Azam
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Prasenjit
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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