FE Dynamics Review Session 2022

FE Exam Review Session: Dynamics
Problem sheets are posted below. Take a look at the problems and see if you can solve them ahead of time. See the video for solutions. Feel free to ask questions!
See for problem sheets: sites.google.c...
This is a question based review with questions that will cover topics that address each topic in the NCEES FE Civil exam specification.
These topics include:
A. Kinematics (e.g., particles, rigid bodies)
B. Mass moments of inertia
C. Force acceleration (e.g., particles, rigid bodies)
D. Work, energy, and power (e.g., particles, rigid bodies)
For the series see the playlist at • FE Civil Review 2022

[@itspoolyy]

CONVERSIONS on TI-36XPro 2nd, 8(Convert), 3, km/h to m/s

[@oliviabrady6673]

Hi Mark! I watched a bunch of your videos and passed the FE first try last month. Thanks so much for making these. I will definitely be recommending your videos to my peers.

[@MarkMattsonPE]

Congrats! Thanks!

[@BELQISYAQUBI-o3t]

The question is same the Mark Chanel or same questions is in this channel?

[@engraghad8663]

Hi Mark! Finally I paaaaaaseeeed the FE civil exam from the second try 😍😍 I am soooo happy and want to thank you for all your videos I watched it several times Also I want to recommend the book of “ Islam “ which contains 800 questions it was also amazing and let me understand many additional things that not included in the yellow book So it was great knowledge from your videos, the yellow book, and Islam book after completing it I passed the exam 😍😍😍

[@MarkMattsonPE]

For more information (above and beyond what is needed for the FE) on the dilemma zone, feel free to check out this article. www.sciencedirect.com/science/article/pii/S2095756415302804

[@Noneofyourbusiness2000]

Why do the civil guys have the best KZbin lectures? It's almost impossible to find a dynamics review for Mechanical where English is the speakers native language. Unfortunately, this is too sparse for me.

[@isaac10231]

Just took the FE! Did pretty good on the 1st section, only flagged like 12 questions which is definitely passing territory. 2nd half though I did really bad, didn't study those later topics much, and I was guessing so much I just felt sad at the end ☹️ But if i do have to do it again, at least I know exactly what to do and how to study for it!

[@isaac10231]

And the 25 minute break is awful. Just pretend you have like, 10 minutes instead.

[@eugeniojimenez7268]

When did you take it? I have it tomorrow!!!!

[@isaac10231]

@@eugeniojimenez7268 Today!

[@eugeniojimenez7268]

@@isaac10231 how did it go brother

[@Visueart]

did you pass Isaac?

[@mohamedfergany5611]

Great Video Matt I have a quick question if I may. In problem no.2 shouldn't the answer be 16 (10 + 2 + 4) since it's asking for the time from (time 0)? thanks

[@MorselOfBread]

16:51 the "2" in the denominator is not just a transportation thing. It comes directly from the Dynamic equations for a constant "a". It is the "1/2" in the s(t) equation. If you eliminate the time variable by finding it from the velocity equation t = - v0/a, then plug that into distance. You will get: s = - v0^2/(2a). There is more to that story, but I'm sure you will be able to figure it out.

[@pb5212]

In the last question, why isn't the potential energy equal to Torque*Angular distance (i.e. rotational potential energy)? So instead of mgh, shouldn't it be mgh*(pi/2)? Where pi/2 is the angular distance travelled, i.e. 90 degrees. If we're saying that the potential energy is due to gravity and that's why we are using gravitation PE, then yes, PE=mgh, but then, shouldn't this h be equal to height of the centroid of the object above a given datum, and not distance from the pivot point?

[@carultch]

A difference in potential energy only depends on the initial and final states. The fact that it rotates, is not relevant to the difference in potential energy. What ultimately matters is the difference in elevation between the center of mass in the two states. In this problem, the center of mass is 100 cm from the pivot. At the horizontal position, it is at the same elevation as the pivot. At the vertical position, it is 100 cm below the pivot. So h would equal 1 meter in GPE=m*g*h. It would be the same change in GPE, if we A) disconnect it from the pivot, B) drop it down 1 meter, C) slide it horizontally until the center of mass is below the pivot, and then D) rotate it about the center of mass to reconnect the pivot. Of these actions, only Action B involves work done by or against gravity. The torque due to gravity also isn't constant on a pendulum. If you wanted to use W= tau*theta, you'd have to promote this multiplication problem into an integral, which would be W=integral tau dtheta. Tau as a function of theta would be m*g*dcm*sin(theta), and we'd integrate it from theta = -pi/2 to 0. The only term that changes with theta is sin(theta), and integrating sin(theta) from -pi/2 to 0 gets us 1. So as we expect, m*g*dcm is the work done by gravity as the pendulum drops, which equals the decrease in GPE.

[@bassamalshaha4972]

Hi Mark, I just passed the FE using your videos and some more questions. Thank you so much you I’m not joking when I say you saved my life my job and my time spending with my toddler daughter. May Allah bless with his mercy

[@MarkMattsonPE]

Congrats! Pass it on!

[@scottsimon8771]

Question 8: Granted, we did have the masses for the individual parts from a previous problem, but is it possible to solve for the individual masses just using dimensions and the Mass Moment of Inertia for the rigid body? I can't seem to. Eventually, I arrive at a place where 0.25(m1) + 1.21(m2) = 15.1445. Two unknowns, one equation.

[@MarkMattsonPE]

Thanks, @Scott Simon! I added the masses back into a revised problem on the problem set... you need those values, and I totally took them from the previous problem.

[@carultch]

@@MarkMattsonPE It turns out, it is possible to solve for the masses of the individual components. It seems like an unrealistic task to expect a student to do on an exam, but it is possible. It was hard enough to set it up in EES, where the software does the heavy lifting equation solving for you. You'd also have to assume equal densities, which wasn't given. By my calculation, I got that the rod is 2.4 kg and the end cylinder is 12 kg, with a density of 12000 kg/m^3.

[@rodrigodeleon8352]

Go Al! lol Taking my exam in two weeks. These videos highlight the important topics very well.

[@wajeehaansari-u1t]

we can use unit method here as well like we have 12 rad/sec at 2 sec 9 rad/sec at 4 sec that makes 6 rad/sec at 6 3 rad/sec at 3 and 0 rad/sec at t=10 sec

[@bloody3red]

Hi Mark, For question 3. I solved for time using the Vt=-gt+v0sintheta and got 0.61sec for time. It takes 0.61 seconds to get to the peak, yes? So should the time to reach the final position should be multiplied by 2? ( 0.61seconds*2=1.22seconds ) My question is, will this always be true?

[@MarkMattsonPE]

Hi Daniel, time up should equal time down unless other factors are added in, like for example, if you have different start and end elevations.

[@bloody3red]

Thanks for the clarification.

[@tundratales449]

I took the derivative of the “y” equation and set it equal to zero to find the middle point. Then multiplied by two.

[@carultch]

@@tundratales449 That's a creative way to do it as well.

[@omerpervez4679]

hey mark with question no 2 I just literally did that in my head as at 2 seconds the angular velocity to be 12 , at 4 = 9 so at 6 seconds to be 6 and at 8 seconds to be 3 so at 10=0. is common sense not gonna work or am I missing something ?

[@carultch]

For that problem, it's a simple matter of constructing a line between the two given points, and finding the intercept on the time axis. The reason this works is that constant angular acceleration means the slope of the angular velocity vs time graph, is constant, and is thus a straight line. Had they asked about total number of rotations to stop, or given a situation without constant acceleration, you'd have a more complicated calculation, since the graph you'd be following to its time-intercept wouldn't be a straight line. You'd also have to translate units to rotations.

[@keannadakwa1210]

Hi Mark, I have a question, why on question 8 do we omit the first 1/2(mv^2) out of the kinetic energy equation? I know we were given v anyway but I am curius to why it is cut out

[@twigsfloat2773]

You're the best Mark Mattson!

[@krishnaakkaram]

For question 7: When calculating sum of forces, is there a there you didn't consider F of A in y-direction. You seem to add up the forces in x-direction(for block A) & y-direction (Block B)

[@carultch]

For block A, you need to consider forces in both the x-direction and y-direction, and it helps to rotate the coordinate system, where y' is perpendicular, and x' is parallel to the incline. If it were a frictionless incline, you don't really care about force balance in the y'-direction, because it doesn't affect acceleration, and there's no need to find the normal force. However, due to the fact that it is an incline with friction, you need to find the normal force, since it directly determines the kinetic friction. If you have enough experience solving these problems, you instinctively know that normal force equals m*g*cos(theta), without setting up the details.

[@RicardoLebaquin]

Hi Mark, firstly, I would like to thank you for making these great FE Civil review videos. The concepts are easy to understand because of your simple but logical explanations. However, I want to clarify something about the last question, can you explain again why we only need to use Ic(w^2)/2 and exclude mv^2/2 to calculate Kinetic Energy? Thank you.

[@kenlua6064]

For item 1, I dont quite get the yellow light analogy.. Adding the 2 in the denominator for the transpo equation decreases the computed time.. you get shorter time so it is less conservative than the physics computation?

[@sharil6436]

Mark, I really appreciate your question sets and videos. Do you have plan to do the same about chapter 7: materials 5-8 points. I will take the FE on 7/2. Thank you very much.

[@MarkMattsonPE]

I hope to add that session. I will post any updates on my channel. I'm glad the problems and videos help with a study plan and overall review!

[@divagirl987654321hah]

Hi Mark! For question 2, I interpolated and got the correct answer. If its constant acceleration and I'm given 2 points, will interpolation always work?

[@carultch]

Essentially, what you are doing is fitting a line to the two given points, and finding the t-intercept of that line. Since a graph of v vs t with constant acceleration is a straight line by definition, then this method is certainly valid. This is more of extrapolation than interpolation, because you are looking outside the given data, rather than looking between the given data.

[@josevela9018]

Im mechanical so we go more in debt with dynamics we actually have 10 problems in our FE

[@Structuralmechanic]

Hello Matt, in the last question, I took the datum to calculate the potential energy is the 6: 0 clock position and in that position, the PE = 2.5x9.81x(110/100)+12x9.81x(110/100), this when equated to the kinetic energy will lead to a angular velocity of 4.5 rad/s and not a 4.3 rad/ s. I did not understand the way you calculated PE. Isn't the PE simple (2.5+12)*(110/100)? as both masses are at the same height with respect to the datum? -Thanks in advance.

[@noulima4533]

Hi Mark for Q.5. Does capital M means moment of inertia which means density times L.A? i noticed you use mass for capital M

[@misterpowers9812]

Great teaching. On question 2 I would argue that the answer is t = 12 seconds. You solved for the time (t) given initial time 2 seconds, The problem statement asks for the answer from initial time 0 seconds. I'd only argue if it caused me to fail the test HAHA.

[@LetsGoBoyce]

you've done it again Mark, thanks for the great video

[@derricklundy775]

In the last question when solving for the potential energy, why is h=0.5 for the rod and h=1.1 for the cylinder instead of h=1 for the rod and h=1.2 for the cylinder?

[@scottsimon8771]

For the rod: The total length is 1 m. However, the center of mass is located in the center of the rod, 0.5 m. The same is true for the cylinder: Center of mass = 0.2m/2=0.1m, but you then must add the length of the rod, which represents the distance from the pin to the center of mass=1.1m

[@MangoJim]

1:09:00 How come for Q6 you are using the full W=mg to calculate the friction force? From the statics video, is the friction force not based on solely the component of the weight normal to the road plane? (mg*cos20 here)?

[@MarkMattsonPE]

The entire weight of the car is acting vertically down on the road (assuming no superelevation). For the statics video, I believe the block was on a 20 degree slope.

[@amanuelzemedhun8768]

Hey Matt, just going through your videos again... In question 7-step 2, you were adding up the forces and you subtracted the T and Fb. But aren't those in the conventional Y direction? How can we add them up to a Y direction that is perpendicular to the plane? And can't we just treat the free body diagram on the left to just treat the question with the smaller mass given that we have calculated T to be 196.2?

[@bouchraech-chetyouy8363]

i have the same comment !!

[@snsh-qh9ig]

I used a different approach for Q2, I used y=mx+b= y=-1.5x+15 and then just set y=0 and got 10 sec, not sure if this is right but that is what came to my mind at first when I tried to solve the question.

[@MarkMattsonPE]

That works... you recognized that the angular velocity had to be 15 to start, so to go to zero takes 10 seconds when you decrease by 1.5 rad/sec/sec. You're using your brain, which is a good thing on this exam!

[@Theejohnjohn]

You’re the best Mark! 🎉🎉🎉

[@danielj.berrios960]

Question 3 : Hey Matt calculating the time, when you factored 't'. Why is it that disappears and you just consider -gt/2 + vosin(20) ?? what happened to the t out of the parenthesis?

[@MarkMattsonPE]

The elevation will equal zero when either one of the factors equal zero. So when t = 0 or when -gt/2 + vosin(2theta) = 0. We solve for the second factor as that's what we're looking for; we already know y = 0 when t = 0.

[@carultch]

For that problem, t=0 is a trivial solution. It's immediately obvious, and not really of interest to the purpose of the problem. This is common with solving quadratic equations, where you have two solutions by the nature of the the fact that it is a quadratic equation, but only one solution is of interest to us.

[@amberhall1119]

The joke made my day. I love that so much!

[@be1062]

Thanks a lot.