__Solutions:__

1)You do not need the handbook for this type of question. It’s __one of the 3 main types of questions__ we discuss in the book.__ __It’s simply testing your basic engineering skills…only involving basic engineering formulas that you already have memorized and don’t have to waste time trying to look up the formula.

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Don’t run off searching for formulas for every problem, think like an engineer first!

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What is the problem really asking me? What is it testing me on that I should know?

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Pressure = Force / Area (!!!!)

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Therefore:

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1500 psi = Force / [π (1.5)^{2}/4]

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1500 psi (1.76 in^{2}) = Force

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Force = 2640 lb (acting on the piston)

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Therefore:

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Pressure = Force / Area

20,000 psi = 2640 lb / Area

Area = 2640 lb / 20,000 psi

Area = .132 in^{2}

**The answer is (A) .132 in ^{2}.**

(2) Does this sound complicated to you? If you do your research, you will find that most FE Exam questions can be solved by using only one formula! You can pass the exam by just becoming familiar with the basic equations in each topic area. It's that simple. This is what our book is all about. Forget theory and over complications used by our professors. It's time to transition to real life. All of the equations are found in the FE Handbook. The secret to passing is Problem Recognition (literally).

Doesn't this problem sound like conduction? This is a basic heat transfer problem. What famous equation uses conduction? The essence of our book describes the importance of knowing the layout of the FE Handbook (since the FE Exam is an open book exam!). If you haven't already, download the FE Handbook from NCEES.org. You have to create an account first. Now, use the search function (Control -F) on the pdf to look up "conduction".

What equation do you see? Yes, it's that easy!

__Solution:__ In the FE Handbook - you will find the equation:

The question is asking for the Temperature Difference, so let's rearrange the equation to find delta T.

The rest is just plug and chug. Be careful though, you need to convert all the mm to m.

ΔT=kw2t⋅P= 0.003m⋅ 92W

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175 W/m∙K (.009 m)^2

= 19.47 K

3) *Refer to page 231 (235 of 498) of the FE Handbook:*

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Straight-line depreciation.

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__This is the second type of problem__ that can show up on the exam. Equations you don’t need to necessary memorize but should know how to quickly find them using the search function on the FE Handbook.

The equation is:

__D___{j} = __C - S___{n }

n

__Givens:__

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Cost (C) = $9,000

n = 10 year life

S_{n }= Salvage value = $200

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Therefore: Just simply plug and chug.

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D_{ = }($9000 - $200) / 10

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D =$880 per year

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**The answer is (D) $880/ year**

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Please Note:

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This equation gives you the depreciation value __per year __which the question asked. But if the question instead wanted to know the “Book Value”, then simply the formula on the same page, BV = initial cost – D_{j})

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4) __Refer to page 107 (111 of 498) of the FE Handbook:__

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Resolution of a Force: Middle of Page.

__Type 1 Problem__. You need to be able to solve statics problems by heart.

Simply solve for the component forces in both the x and y directions, square them, and take the square root to find the resultant force.

Pay extra attention to the direction of the forces, some may be positive or negative, we’ll see this below.

__Sometimes, the forces are given in angles and other times as 2 sides of a triangle. This problem has sides, so you must solve for the hypotenuse for each triangle:__

__For the 211 lb force: H = √5__^{2} + 12^{2} = 13

For the 398 lb force: H = √4^{2} + 3^{2} = 5

__Let’s sum forces in the Y-direction:__

ΣF_{y }= 0: R_{y} = 12/13(211) + 3/5(398) - 75 = 0

(Note how the components are positive except the 75 lb force because it is pointing down)

R_{y} = 194.7+ 238.8 - 75 = 0

R_{y} = 358 lb

__In the X-direction:__

ΣF_{x }= 0: R_{x} = -5/13(211) + 4/5(398) = 0

(Notice how there is a negative sign in front of the 211 lb force because it is pointing to the left, Also the 75 lb does not have a x-component)

R_{y} = -81+ 318 =

R_{y} = 237 lb

Therefore:

__R = √Ry__^{2} +Rx^{2}

R = √ (358)^{2} + (237)^{2}

R = √ (358)^{2} + (237)^{2}

R = 429 lb

**The answer is (C) 429 lb.**

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5) * Refer to page 175 (179 of 498) of the FE Handbook*:

Psychometric Chart:

__Type 2 Problem.__

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The easiest way to do this type of problem is to simply look it up on the chart. Trust me.

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Find the actual air temperature (dry-bulb) of 25 ºC on the bottom of the chart and go up until you hit the wet-bulb temperature 19 º C diagonal lines. At this point, read the relative humidly line, which are the curved lines.

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Note: This chart looks extremely intimidating; however, most problems can be solved simply by understanding how to read it. Get familiar with the chart. These types of problems are easy to rack up points.

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**The answer is (A) 58%.**

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6) __Refer to page 114 (118 of 498) of the FE Handbook:__

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Particle Kinematics:

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__Type 1 Problem:__

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They are testing your understanding of calculus and dynamics in one question.

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Realize that velocity is just a derivative of distance and that the acceleration is just a derivative of velocity. Also, the term “jerk” is a derivative of acceleration.

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Therefore, to find the velocity we need to take the derivative of the distance equation:

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D = 10t^{4} – t^{3}

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V = (4)10t^{3} – (3)t^{2}

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Please Note: When taking the derivative, the exponent drops down out in front, and subtract 1 from the exponent to get the new exponent. Remember that, right?

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We just solved for the velocity equation,

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Now, when t = 3:

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V = 40t^{3} – 3t^{2}

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V = 40(3)^{3} – 3(3)^{2 } (substitute “3” in for time, T)

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V = 40(27) – 27

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V = 1,053^{}

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**The answer is (A) 1053.**

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Note: If the question wanted to know the acceleration after a certain time, just simply take another derivative of the velocity equation to find the acceleration.

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