Rigid Bodies and Equations of Motion Translation (Learn to solve any question)

Sdílet
Vložit
  • čas přidán 6. 08. 2024
  • Learn about solving dynamics rigid bodies and their equations of motion and translation of rigid bodies with animated examples. Step by step solved solutions to cover kinetic diagrams, kinetic moments, and f=ma.
    🔹Rigid Bodies and Rotation About a fixed Axis: • Rigid Bodies: Rotation...
    Intro (00:00)
    Kinetic Diagrams (01:50)
    The 4-Mg uniform canister contains nuclear waste material encased in concrete. (03:38)
    A force of P = 300 N is applied to the 60-kg cart. (05:10)
    The dragster has a mass of 1500 kg and a center of mass at G (08:04)
    The 100-kg uniform crate C rests on the elevator floor (11:16)
    Find more at www.questionsolutions.com
    Book used: R. C. Hibbeler and K. B. Yap, Mechanics for engineers - dynamics. Singapore: Pearson Education, 2014.

Komentáře • 234

  • @atefrod680
    @atefrod680 Před 3 lety +18

    Could you cover more problems please? Thank you so much for your great videos! I find it difficult to see a pattern, so please try to emphesize on repeating routines! I shared your videos with my classmates!

    • @QuestionSolutions
      @QuestionSolutions  Před 3 lety +52

      You're very welcome. I agree that it will be hard to find patterns from 3 examples, mainly because the 3 or 4 types of problems I pick are actually a bit "different" from each other. The reason I do that is so that you can find similar questions in your textbook and you can do them, and if you get stuck, you can look back at these examples and hopefully get a hint out of it. You will see that in your textbook, there will be a few questions that is very similar to question 1, then there will be a few questions similar to question 2, and so forth. If you solve one or 2 from your textbook, along with the ones I've covered, it should be a very good foundation for you to do well on your exams. Also, try to not look for patterns or routines, its better to understand how the equations were used, rather then look for certain types of patterns, if that makes sense. I say this because most of the time, on your exams, you won't have a question where you can solve with just a single chapter, most questions are written in a way where you'd have to use multiple concepts, like momentum, conservation of energy, relative motion, etc, to solve a problem. So its almost always better to understand how the equation was used. I will keep what you said in mind, and over time, I will add more solved examples for each chapter as well. I truly appreciate the share, and your feedback. Many thanks! :)

  • @menooaavrand8896
    @menooaavrand8896 Před rokem +5

    Nah your lectures are literally amazing.... they are so simple and straight forward 🐐

  • @GenknownTutorial
    @GenknownTutorial Před 2 lety +1

    Your videos are really helpful, Thank you!

  • @carolinedowling953
    @carolinedowling953 Před 3 lety +2

    These videos are so helpful thank you !

  • @sohomraymandal535
    @sohomraymandal535 Před rokem

    thanks for such a great explanation in less time

  • @fariaalju4649
    @fariaalju4649 Před rokem +2

    How is it that you reply to everyone's comment!! I'm simply in awe!
    Also, I have a final exam tomorrow and I'm passing solely thanks to you. I wish there was a donation option or sth in my country..alas there isn't. Again, thank you, tons!

    • @QuestionSolutions
      @QuestionSolutions  Před rokem +1

      I try my best to reply because I want to clear up any questions they have. I wish you the best with your exam tomorrow. Do your best!

    • @fariaalju4649
      @fariaalju4649 Před rokem

      @@QuestionSolutions Whoha so fast..
      I shall 💪 Thank you!

    • @fariaalju4649
      @fariaalju4649 Před rokem +3

      @@QuestionSolutions got the highest grade on my exam..thank u so much!🥰🥰🥰

  • @zohaibahmed1331
    @zohaibahmed1331 Před 2 měsíci +1

    The man's a Legend. I have final exam in 10 hours, and this channel is the last hope I have now. Thanks for the great content.

    • @QuestionSolutions
      @QuestionSolutions  Před 2 měsíci +1

      You probably did your exam by now, and I hope it went well for you. Keep up the great work and best wishes with your future endeavors.

    • @zohaibahmed1331
      @zohaibahmed1331 Před 2 měsíci

      @@QuestionSolutions Thanks a lot

  • @stigza
    @stigza Před 2 měsíci

    This man is single handedly getting me through this subject

    • @QuestionSolutions
      @QuestionSolutions  Před 2 měsíci

      I am glad to hear these videos are helpful to you. I wish you the best with your studies!

  • @total_chaos4u620
    @total_chaos4u620 Před 2 měsíci

    got a dynamics final on monday and this video is going to be the only reason ill come close to passing ty so much

    • @QuestionSolutions
      @QuestionSolutions  Před 2 měsíci

      You're very welcome. I wish you the best on your exam Monday. You got this!

  • @jerickmarzan3483
    @jerickmarzan3483 Před 3 lety

    You're the best!

  • @abdallahamouda6633
    @abdallahamouda6633 Před 3 lety

    from today on you are mister fantastic

  • @ezradese477
    @ezradese477 Před rokem

    So for question 3, lets say for instance Ff max was greater then the friction required, would the slipping condition be true then? Or does it need to equal the friction calculated?

    • @QuestionSolutions
      @QuestionSolutions  Před rokem +1

      So Ffmax tells us the minimum force required to make NA = 0, in other words, the minimum needed to lift up the wheel.

  • @hexrobel1071
    @hexrobel1071 Před rokem

    Thank you very much for doing this!
    In last question taking friction backwards is very confusing. Shouldnt we apply formula: total forces - Max static friction= 0 since friction should is always negative to the force. Or am i missing something?

    • @QuestionSolutions
      @QuestionSolutions  Před rokem

      You're very welcome.
      About your question, I am sorry, but I don't really understand it. For what part did you want to subtract static force from total forces? To find which value? What is the max static friction and what do you mean by total forces? Please kindly explain your thought process, maybe I can explain it better then.

  • @alsix7334
    @alsix7334 Před 3 lety +1

    For the first problem, how did you know to use 2Tcos(30) rather than using 2Tsin(30)? I always struggle with understanding when to use cosine and when to use sine. Any clarification would help out a lot. Maybe you could make a video about it. That would be a really good video to watch. Thank you.

    • @QuestionSolutions
      @QuestionSolutions  Před 3 lety +1

      Please take a look at this video: czcams.com/video/NrL5d-2CabQ/video.html It's not long, and I go through how to break forces into components using sine and cosine, when to use them, etc. :)

    • @alsix7334
      @alsix7334 Před 3 lety

      @@QuestionSolutions Thank you that makes sense now!

    • @QuestionSolutions
      @QuestionSolutions  Před 3 lety +1

      @@alsix7334 Glad to hear!

  • @jp9523
    @jp9523 Před 2 lety +1

    In the last problem, why was normal acceleration not pointing in the inside direction of the curvilinear path you had shown? The same with the tangential acc, why is it not tangent with the curved path?

    • @QuestionSolutions
      @QuestionSolutions  Před 2 lety

      So the coordinate system was drawn not with respect to a curve, but rather using the center of gravity for the box. In that case, straight down was the normal axis and horizontal was the tangential axis.

  • @johnlerrycarillo4755
    @johnlerrycarillo4755 Před 2 měsíci

    In the first problem 5:10, based on the solution, isn't the computed force (27615N) for EF and GH just the y-component of the tensional force? since this force is on a 30°, should 27615N be further divided to cos(30) to get the final value instead?

    • @QuestionSolutions
      @QuestionSolutions  Před 2 měsíci

      No, and I don't think I can explain it in a comment. This goes back to statics, when we write equations of equilibrium. In a very simple sense, T is the variable assigned to the EF and GH links. Through the process of writing a f=ma equation for the vertical forces only, you can directly solve for T. Nothing else needs to be done, it gives a value for the force in the members EF and GH. If you want the y-component, you will need to use cosine 30 to get it and if you want the x-components, you would need to use sine30 to get it. I doubt this comment is helpful because your question requires a lot of previous fundamental knowledge to answers. If you have time, please watch some early statics videos about equilibrium of forces. It should give you a good foundation for your question. Or, it might be to good to speak with your professor or TA and ask them to cover this question during their office hours.

  • @robertlaines8648
    @robertlaines8648 Před 2 lety

    on the crate problem where we the angular acceleration is found I don't see how the frictional static force towards the right. I thought it would be in the opposite direction of the motion.

    • @QuestionSolutions
      @QuestionSolutions  Před 2 lety +4

      Sometimes, these questions can be hard to visualize when we don't have a lot of practical experience. I think an easy way to visualize this is to take a small object that's slippery and place it on top of a cardboard or even on your phone screen. Then quickly move the cardboard/phone to the right. You will notice that the object that's on top will slide to the left, and so friction would be opposing the movement, in this case, the friction will be pointing to the right.

  • @christopherkabalan1294

    The best! About to get a 100 on my final thx to u

  • @EurekaQuest
    @EurekaQuest Před rokem

    You are the best

  • @QCJF4G
    @QCJF4G Před 3 lety

    Hello thank you so much for the videos they are super helpful. When I tried to do the cart problem at 5:10 first without knowing your solution I assumed that there would also be friction between the wheels and the ground but you didn't have it. Am I misunderstanding something?

    • @QuestionSolutions
      @QuestionSolutions  Před 3 lety +5

      I am going to give you a tip, but you can't use this in the "real world" but you can in your school courses. If the question does not give you any sort of coefficient of friction, doesn't mention friction, and there is simply no way for you to figure out the friction, then they are talking about a perfect problem where friction can be ignored. In this question, we aren't given any information that can give us a frictional value, so you can simply ignore it and solve it. 😅

    • @QCJF4G
      @QCJF4G Před 3 lety

      @@QuestionSolutions Oh okay I see, thank you very much. You have the best youtube channel for mechanics!

    • @QuestionSolutions
      @QuestionSolutions  Před 3 lety

      @@QCJF4G Thank you so much, really appreciate your comment :)

  • @AhmedMahmoud-gu4un
    @AhmedMahmoud-gu4un Před 3 lety +2

    Can you please explain why did we use the acceleration of fixed axis instead of relative acceleration? Also what is the fixed axis in this case?

    • @QuestionSolutions
      @QuestionSolutions  Před 3 lety

      Can you give me a time stamp or let me know which question you are referring to? Thanks!

    • @AhmedMahmoud-gu4un
      @AhmedMahmoud-gu4un Před 3 lety

      @@QuestionSolutions Sorry I forgot to add the time stamp 11:24 I was just confused how w =0 and at what instance is it 0 and also why is "r" DE not DG

    • @QuestionSolutions
      @QuestionSolutions  Před 3 lety +4

      @@AhmedMahmoud-gu4un Okay, thanks for time stamp. ω = 0 because at 90 degrees, everything starts from rest. So at 90 degrees, nothing is moving (as stated in the question), so angular velocity has to be zero. We used DE because the whole system moves together. When rod DE moves, the box moves. So we just need to figure out that acceleration. I hope that helps!

  • @bigr3578
    @bigr3578 Před 3 lety

    THANK YOU

  • @Ahmad-qv4cu
    @Ahmad-qv4cu Před 3 dny

    Hello sir I hope you are doing well, do you use specific programs to simulate questions? It will really help me.
    Thanks in advance

    • @QuestionSolutions
      @QuestionSolutions  Před 3 dny

      I don't use any programs to simulate questions. I animate them by hand the best I can :)

  • @mohamedebrahim3979
    @mohamedebrahim3979 Před 3 lety

    You are great

  • @robertlaines8648
    @robertlaines8648 Před 2 lety

    can the dragster problem be solved by using the moment equation about the center mass instead of point B?

    • @QuestionSolutions
      @QuestionSolutions  Před 2 lety

      You can, but you'd have to solve more equations because you're not eliminating any of the unknowns. It's too time consuming.

  • @clearflow7925
    @clearflow7925 Před 9 měsíci

    4:55 why don't you consider the forces in the y direction of the two tension forces we found earlier?

    • @QuestionSolutions
      @QuestionSolutions  Před 9 měsíci +1

      When we consider the object as a whole, we don't consider forces inside it.

  • @tengwenxiang1692
    @tengwenxiang1692 Před rokem

    may I know how do you set the direction of Ialpha in the kinetic diagram

    • @QuestionSolutions
      @QuestionSolutions  Před rokem

      You can make an assumption. If it's wrong, you will get a negative value, so you know it's opposite to your assumption. Generally speaking though, you can make an educated guess based on how an object would move.

  • @uvadroid
    @uvadroid Před 2 měsíci

    For the dragster question, shouldn't the maximum friction force be considered Nb * 0,6? I solved it this way but the results were similar anyways

    • @QuestionSolutions
      @QuestionSolutions  Před 2 měsíci

      Please give me a timestamp so I know where you're referring to. Thanks!

  • @seijieneng9585
    @seijieneng9585 Před 3 lety

    Why we don't consider the Tab and Tcd when calculating for the Tef and Tgh (Q1)

    • @QuestionSolutions
      @QuestionSolutions  Před 3 lety +1

      They become part of the whole object when we consider forces T_ef and T_gh.

  • @SG-dw8jh
    @SG-dw8jh Před rokem

    Can you explain the difference between torque and moment, please? I see their equations are the same. Thanks for the great tutorials!

    • @QuestionSolutions
      @QuestionSolutions  Před rokem +1

      For a normal statics course, you can think of them as the same. 👍

    • @SG-dw8jh
      @SG-dw8jh Před rokem +1

      @@QuestionSolutions Thanks a lot Mr. Greatest tutor of all time!

    • @QuestionSolutions
      @QuestionSolutions  Před rokem +1

      @@SG-dw8jh 😅You're very welcome!

    • @HashemAljifri515
      @HashemAljifri515 Před 7 měsíci

      Torque is spinning, rotation about the same axis, whereas moment is rotation about different axis.

  • @edgarheng943
    @edgarheng943 Před 4 měsíci

    How do i know when the moment i shoud be taking CW or CCW as positive?

    • @QuestionSolutions
      @QuestionSolutions  Před 4 měsíci

      It makes no difference, you can pick whatever direction you want it to be positive in 2D problems. Usually, it's better to pick whatever gives you the most positives to be your positive direction. For example, if 10 forces are creating clockwise moments and 1 is creating a counter-clockwise moment, it's better to pick clockwise to be positive, it makes the math easier. Regardless, you will get the same answer. If you get a negative value, then the magnitude of that value is correct, but the assumed direction is wrong. So if you picked clockwise to be positive and you got -10, then your answer is 10 counter-clockwise. Please see: czcams.com/users/shortsP029mqnp4XY

  • @jembosalingay1628
    @jembosalingay1628 Před 2 měsíci

    Hi! Just want to ask why you use cos in problem number 1 when the condition you use is y component, im confused why it's not sine😅

    • @QuestionSolutions
      @QuestionSolutions  Před 2 měsíci +1

      Please see this video, it's less than 60 seconds, and you will understand it much better: czcams.com/users/shortsvynnKlJD_Jo?feature=share

  • @nirupankarki9284
    @nirupankarki9284 Před 10 měsíci

    Hello! Thank for the the video. In the dragster problem, why did we take the moment about B and not G?

    • @QuestionSolutions
      @QuestionSolutions  Před 10 měsíci

      So if we take the moment about G, then we have to deal with NB and NA. If we take it about B, we eliminate NB, which is one less unknown we need to worry about.

    • @nirupankarki9284
      @nirupankarki9284 Před 10 měsíci

      @@QuestionSolutions But that wouldn't matter because in our case we are assuming NA to be zero because we assume that it lifts? Is it possible to solve the question by taking moment about G?

    • @QuestionSolutions
      @QuestionSolutions  Před 10 měsíci

      @@nirupankarki9284 NA is 0, but NB is not. So if you write it about G, you will have NB as an unknown. 😅 And yes, you can solve it by taking the moment about G, it's just more steps.

    • @nirupankarki9284
      @nirupankarki9284 Před 10 měsíci

      @@QuestionSolutions thank you very much for answering!

  • @maryabegailmirafuentes0325

    Hi, i have a question for problem #2 8:02 how did you get the two unknown normal forces from the two equations? Can you elaborate the process of solving it. Thanks!

    • @tomasclaeys7747
      @tomasclaeys7747 Před rokem

      substitution into on of the two equations

    • @QuestionSolutions
      @QuestionSolutions  Před rokem +1

      So one of the ways of doing it is to isolate for one variable in one equation. For example, you can isolate for N_A in the first equation. Then you plug that value into the second equation and solve for N_B. After you find N_B, you can plug that value into the previous equation to get N_A. It's called solving simultaneous equations using the substitution method. If you search for that on CZcams or Google, you will find a lot websites/videos going through it step by step. It's also possible you know how to do it, but maybe overthinking it. For example, if you know how to solve "2x+y=5" and 3x+2y=10" then it's the same exact thing with these equations. If it makes it easier, convert sine and cosine values into decimal form. So like sin30 = 0.5, and cos30 =0.866. If you want to double check your answers, use cymath.com.

  • @mariaalhijazeen4974
    @mariaalhijazeen4974 Před 2 měsíci +1

    Hi ,, for question 2 can calculating the moment around wheel A be the third equation
    because I tried it but it didn’t give the same answer

    • @mariaalhijazeen4974
      @mariaalhijazeen4974 Před 2 měsíci

      Also in this question why did we ignore the friction

    • @QuestionSolutions
      @QuestionSolutions  Před 2 měsíci

      @@mariaalhijazeen4974 We didn't ignore the friction, it was used to find Fmax, at 10:55.

  • @Mac-hr9rg
    @Mac-hr9rg Před 3 lety

    For the drag car problem wouldnt it be 2NB because the problem says rear wheels?

    • @QuestionSolutions
      @QuestionSolutions  Před 3 lety

      It doesn't really matter, you can consider the set of wheels as a single entity since all we're looking for is to see if the front wheels can go up. So you consider the front wheels as a set and the back wheels as a set. Try to make questions as simple as possible when solving 👍

  • @debaratiroy2792
    @debaratiroy2792 Před 2 lety

    for the dragster problem, why didnt we consider moment due to friction while solving for M(b)=(Mk)b

    • @QuestionSolutions
      @QuestionSolutions  Před 2 lety +1

      The moment was written about point B, which means any force that has it's line of action going through that point will be 0.

  • @KontKledula
    @KontKledula Před 7 měsíci

    Shouldn't you take the wheel reaction forces as 2Na and 2Nb? I thought there were 4 wheels in total. at8:02 Thanks for tutorial btw. You are awesome

    • @QuestionSolutions
      @QuestionSolutions  Před 7 měsíci

      You can solve it that way, but it requires you to multiply your answer by 2 at the end since the question wants you to find it for both wheels. It's easier to assume them to be one unit (a set of 2) and just solve for an answer instead of thinking them as 4 wheels.

  • @e-nes4042
    @e-nes4042 Před 2 lety

    12.57 why did you get anormal zero ? can you explain in detaily

    • @QuestionSolutions
      @QuestionSolutions  Před 2 lety

      At 12:21, we already establish that angular velocity is 0, which means normal acceleration is also zero.

  • @kwtr
    @kwtr Před 2 lety

    For the last problem, when solving for normal acceleration of the box, i.e. (a_g)t, why is the distance 1.5, not 1.5 + 0.6 (distance from the elevator surface to centre of mass of the crate)? I feel confused because the notation is a_g, but the distance r used is r_de, not r_dg.

    • @QuestionSolutions
      @QuestionSolutions  Před 2 lety +1

      The simplest way to think about this is that the acceleration "felt" by the box is the same as member DE. Because they are all moving together, not independently.

    • @aliemad6670
      @aliemad6670 Před 2 lety

      @@QuestionSolutions does that mean if we had angular velocity in this problem , then the normal acceleration of aG would be same as the normal acceleration of member DE ??? or does that principle only apply to tangential acceleration and linear velocity??

    • @aliemad6823
      @aliemad6823 Před 2 lety

      @@QuestionSolutions does that mean if we can say that the normal acceleration of member DE will be same as the normal acceleration of the box?? Or does that only apply to tangential acceleration??

    • @QuestionSolutions
      @QuestionSolutions  Před 2 lety +1

      @@aliemad6823 Well normal acceleration is dependent on the radius of curvature, so would they both have the same radius of curvature? :)

  • @joshuacharlery5826
    @joshuacharlery5826 Před 3 lety +1

    Nice

  • @DrDerivative
    @DrDerivative Před 2 lety

    For question #2 about the cart, are we forced to take the moment about G in order to find the equations for reactions at A and B? I am attempting to take the moment about A or B, but I am getting the incorrect values for reactions A and B.

    • @QuestionSolutions
      @QuestionSolutions  Před 2 lety +1

      You can do it at either location, make sure you don't miss any forces when taking the moment. You will get the same answer 👍

    • @DrDerivative
      @DrDerivative Před 2 lety

      @@QuestionSolutions for moments about A, I am using, Px(0.4m) - Py(0.08m) -W(0.3m) + Nb(0.5m) = 0,
      Where Px = 300cos(30), Py = 300sin(30), and W = (60)(9.81) = 588.6

    • @DrDerivative
      @DrDerivative Před 2 lety

      Nevermind, I got it now =)

    • @DrDerivative
      @DrDerivative Před 2 lety

      When you dont sum the moments about the center of mass, them the sum of the moments equals I(alpha), not 0.
      Why is this true?

    • @QuestionSolutions
      @QuestionSolutions  Před 2 lety +1

      @@DrDerivative glad to hear :)

  • @abdallahamouda6633
    @abdallahamouda6633 Před 3 lety

    I have two questions they may be stupid so sorry but i asked myself, if I have no angular velocity at 90 degrees from rest how could it be that I have acceleration because they nothing moves, is it because it really takes an instant I mean after some micro second we have movement but shouldn't it also apply to angular velocity? Second one why don't we take the distance from A to the center of mass G and this would be a hypothenuse?

    • @QuestionSolutions
      @QuestionSolutions  Před 3 lety

      For the future, please kindly use time stamps so I know where you are referring to. I think you are asking about the last question. In the last question, we don't need the distance from A to the center of mass G because rod EB will follow the movement of rod DE since they are connected. Because of this, what we are looking for is the tangential acceleration of rod DE. So we don't need a position vector from D to G, instead, all we need is one from D to E. Second, at rest, acceleration is zero, just like you said, if an object is at rest, it cannot have an acceleration since it's in equilibrium. However, I think you might have gotten confused about what we are trying to find. In a nutshell, we are going to rotate link DE very quickly, but we want to rotate it with an angular acceleration that won't make the box slip. This is the angular acceleration we are trying to find, what is the maximum angular acceleration we can give it to make the box not slip. So initially it's at rest, then we apply an acceleration and we need to figure that value out. I hope that makes sense. Also, don't be sorry for asking questions 👍

    • @abdallahamouda6633
      @abdallahamouda6633 Před 3 lety

      @@QuestionSolutions thank you very very much you are the best

    • @QuestionSolutions
      @QuestionSolutions  Před 3 lety

      @@abdallahamouda6633 You are very welcome!

  • @padamyonjan5608
    @padamyonjan5608 Před 4 měsíci

    For this question, Why is the tangential acceleration is not tangent to the path of center of mass of box and normal acceleration towards the point of rotation?

    • @QuestionSolutions
      @QuestionSolutions  Před 4 měsíci

      I am not sure which question you are referring to. Please provide me with a timestamp to the location and I will take a look. Thanks!

    • @padamyonjan5608
      @padamyonjan5608 Před 4 měsíci

      11:51 where t and n are just like x and y axis. Why not t tangent to the path and n towars the point of rotation?

    • @QuestionSolutions
      @QuestionSolutions  Před 4 měsíci

      @@padamyonjan5608 You can do it the way you described, by using a different rotation for the axes, but that creates more work. You will need to break all the forces into components, whereas having the axes in a vertical and horizontal formation means you won't have to break it into components since the forces lie on the axes.

  • @elijahmayhew2141
    @elijahmayhew2141 Před 2 lety

    In the cart problem. Why can't we take sum of the moments around point A to get reaction Nb. Then take sum of the forces in the y+ direction to determine Na. I try this and get different answers. But it should work right?

    • @QuestionSolutions
      @QuestionSolutions  Před 2 lety

      Could you provide a timestamp to the location you're referring to? Thanks!

    • @elijahmayhew2141
      @elijahmayhew2141 Před 2 lety

      @@QuestionSolutions 6:29, you chose a moment equation about G.

    • @QuestionSolutions
      @QuestionSolutions  Před 2 lety

      @@elijahmayhew2141 You can do what you said, most likely you got an error because of a numerical problem. Did you use the correct components of the 300N force? You also need to use the weight when you find the moment about point A. I find it easier to get the 2 unknowns, get 2 equations and solve for them at the same time, but you can always solve these problems in more than one way.

    • @MrEmayhew
      @MrEmayhew Před 2 lety

      @@QuestionSolutions Ahh man I made a stupid mistake and didn't account for all the moments acting in the kinetic diagram. Once I changed that I got the right answers! Thanks for your reply, I needed to know my thinking was correct and I wasn't going mad haha. You deserve more views! I'll do my best to get your name out there and share your vids. All the best mate.

    • @QuestionSolutions
      @QuestionSolutions  Před 2 lety

      @@MrEmayhew Glad to hear it worked out :) Thanks for trying to get my name across, really appreciate it. I wish you the best with your studies!

  • @YouTuberDrama101
    @YouTuberDrama101 Před měsícem

    How did you know in the last question that the Force Body Diagram would be equal to the Kinetic Diagram? Thank you very much for your vides!😃

    • @QuestionSolutions
      @QuestionSolutions  Před měsícem

      Sorry, I don't understand your question. The kinetic diagram allows us to factor in acceleration, which can be done to any object you want. So it's just another way to representing forces and it applied to all objects. If you can maybe rephrase or give me some more details about your question, I can try to help you out better. Thanks!

    • @YouTuberDrama101
      @YouTuberDrama101 Před měsícem

      @@QuestionSolutions Sorry I realized my answer right after posting the question. Thank you for replying so promptly!

  • @aashishkumarjha3993
    @aashishkumarjha3993 Před rokem

    question 2 at 5:15, why aren’t we considering friction between wheel and surface?

  • @DrDerivative
    @DrDerivative Před 2 lety

    For the last problem, we set omega to zero because it starts at rest? Then why do we not do the same for the angular acceleration (alpha)?

    • @QuestionSolutions
      @QuestionSolutions  Před 2 lety +1

      So the question asks us to find the largest initial acceleration, which we found is not zero, so we can't really set it to zero. 😀Think of it as instantaneous acceleration, which "drives" the angular velocity. If you stop at a red light in your car, your velocity is zero. At the instant you press on the gas, your velocity is still zero, but your acceleration must be greater than zero, otherwise, the car would never move. I hope that makes sense. 👍

    • @DrDerivative
      @DrDerivative Před 2 lety

      @@QuestionSolutions you're amazing sir haha

    • @QuestionSolutions
      @QuestionSolutions  Před 2 lety +1

      @@DrDerivative Thanks! Keep up the good work, best wishes with your studies :)

  • @hfzhr
    @hfzhr Před měsícem

    Q2, why friction direction to the right? How to know direction rear wheel slipped

    • @uzemakistyle
      @uzemakistyle Před měsícem

      Funny, I needed a reminder for that too. Just used chatgpt and it had a decent answer:
      When a wheel is slipping, the frictional forces act in the direction opposite to the relative motion between the wheel and the surface it is on. To understand this in the context of a slipping wheel, consider the following scenarios:
      1. Forward Slipping (Braking)
      If a vehicle is braking and the wheels are slipping (locked wheels):
      The wheels are moving forward, but they are not rolling; instead, they are sliding.
      The frictional force opposes the forward motion of the wheels, acting backward.
      This means the frictional force acts in the direction opposite to the wheel's velocity relative to the ground.
      2. Rearward Slipping (Accelerating)
      If a vehicle is accelerating too quickly and the wheels are slipping (spinning):
      The wheels are spinning faster than the vehicle is moving forward, causing them to slip backward relative to the ground.
      The frictional force opposes this backward slipping motion, acting forward.
      This means the frictional force acts in the direction opposite to the wheel's relative motion, which is backward in this case.
      Summary
      In both scenarios, the frictional force always acts opposite to the direction of the relative slipping motion between the wheel and the ground.
      For braking (forward slipping), the frictional force points backward.
      For accelerating (rearward slipping), the frictional force points forward.
      Thus, frictional forces when a wheel is slipping are angled opposite to the direction of the wheel's relative motion to the surface.

    • @QuestionSolutions
      @QuestionSolutions  Před měsícem +1

      The reply by uzemakistyle is appropriate for your question. You can intuitively understand this easier if you cut up a piece of eraser into a circle shape. Then try rolling it on your desk and making it slip. Feel it with your fingers how the friction is opposing this slip. It will help you understand it better :)

    • @uzemakistyle
      @uzemakistyle Před měsícem

      @@QuestionSolutions Big fan of your videos man, they really help a lot.
      Also quite surprised how well chatgpt is able to help in some cases. Have had quite a few questions that had mistakes in the text. Just taking a picture or typing out the problem usually has the ai point out the inconsistencies in the question, usually at a point we're i'm having trouble wrapping my head around what I'm doing wrong.
      I wish they spend more time fixing these mistakes instead of publishing new editions. I have multiple books and quite often they copy over questions wrongly, or change minor things and forget values.

  • @chizwheeze
    @chizwheeze Před rokem

    when do we draw a kinetic diagram in solving problems and when do we don't?

    • @QuestionSolutions
      @QuestionSolutions  Před rokem

      Usually, if we have too many variables and we can't solve for what we need, we need to bring in more equations. One way of doing that is to draw a kinetic diagram, and compare the two, to get more equations. For example, questions 1 and 2 didn't need one, because we can solve what was being asked with the info given. But for question 3, we didn't have enough information, so we needed a kinetic diagram.

  • @melekmnif1933
    @melekmnif1933 Před 3 lety

    Hi can someone please explain to me why the direction of friction is to the right at 8:35? I am a bit confused in general on friction direction

    • @QuestionSolutions
      @QuestionSolutions  Před 3 lety +4

      Please see this article: www.school-for-champions.com/science/friction_rolling_starting.htm, it's explained well, scroll down a bit to the heading "Friction causes forward motion." If you have a toy car at home, place it on the table, and then spin one of the wheels clockwise while it's against the table. So you're just spinning one wheel. You will notice that friction has to be opposite to the clockwise movement for the car to go forward.

    • @melekmnif1933
      @melekmnif1933 Před 3 lety

      @@QuestionSolutions Thanks that was a quick response and a great one too!!

    • @QuestionSolutions
      @QuestionSolutions  Před 3 lety

      @@melekmnif1933 You're very welcome! Best of luck with your studies.

    • @arinzeanthony7447
      @arinzeanthony7447 Před 2 lety

      @@QuestionSolutions thank you very much,just went through the info through the link you gave.
      So,I have one last question, does it mean that in every case such as this, the direction of frictional force=direction of motion.
      Since it is opposing torque?

    • @QuestionSolutions
      @QuestionSolutions  Před 2 lety

      @@arinzeanthony7447 In cases that's exactly like this, yes, that is how the friction will act.

  • @programmingprograms726
    @programmingprograms726 Před rokem +1

    Hey, you forgot to upload 16.8 Relative motion analysis using rotating axes. Please upload. Thanks!

    • @QuestionSolutions
      @QuestionSolutions  Před rokem +2

      I didn't forget, it just wasn't done because that's usually not covered. In the future, I'll upload it after finishing some other subjects 👍

    • @programmingprograms726
      @programmingprograms726 Před rokem

      @@QuestionSolutions Thank u. What subjects are u doing now?

    • @QuestionSolutions
      @QuestionSolutions  Před rokem +2

      @@programmingprograms726 Currently, working on thermodynamics :)

  • @bas73971
    @bas73971 Před 3 lety

    At 12:40, why are you using 1.5m for r, why not the distance from D to G (which is 2.1m at theta=0)?

    • @QuestionSolutions
      @QuestionSolutions  Před 3 lety

      The rod EB will follow the movement of rod DE since they are connected. Because of this, what we are looking for is the tangential acceleration of rod DE. So we don't need a position vector from D to F, instead, all we need is one from D to E. I hope that makes sense.

    • @bas73971
      @bas73971 Před 3 lety

      @@QuestionSolutions Oh wait, maybe I understand it now. So the tangential acceleration of point G in respect to D is the same as from E to D, because you can ignore the extra distance from E to G (which normally would result in a larger a_t, due to the bigger r), because there is a joint which prevents further tangential acceleration between point E to G?

    • @QuestionSolutions
      @QuestionSolutions  Před 3 lety +1

      @@bas73971 That is correct. It will have the same acceleration.

    • @bas73971
      @bas73971 Před 3 lety

      @@QuestionSolutions Thanks

    • @QuestionSolutions
      @QuestionSolutions  Před 3 lety

      @@bas73971 You're very welcome!

  • @marwanayman9731
    @marwanayman9731 Před 2 lety

    In the second problem you said acceleration upwards is negligible. So you put 60(0) why cant we take a=-9.81?

    • @QuestionSolutions
      @QuestionSolutions  Před 2 lety

      The cart isn't moving upwards, only in the horizontal direction. So think of it like this, when you're standing in your room, do you have an acceleration, in other words, are you floating up or going down? If you had an acceleration of -9.81, you'd be going through the floor all the way to the core of the earth. This is negated by the normal force. :) In the cart problem, we don't have an acceleration in the vertical direction, the right side of the equal sign is for movement accelerations. So we get zero for vertical movement since again, the cart doesn't move up or not, just in the horizontal direction. The weight, which is acceleration due to gravity times the mass is accounted for on the left side. I hope that helps 👍

  • @rawadhasan9111
    @rawadhasan9111 Před 2 lety

    For 1st example, why are we not adding the mass of the spreader to find m in the equation?

    • @QuestionSolutions
      @QuestionSolutions  Před 2 lety

      Can you give me a timestamp? I'm not sure which "m" you're referring to.

    • @rawadhasan9111
      @rawadhasan9111 Před 2 lety

      @@QuestionSolutions Thanks for replying. I was talking about m at 4:28 on the F_y equation.

    • @QuestionSolutions
      @QuestionSolutions  Před 2 lety

      @@rawadhasan9111 Oh, because we are only looking at the canister. So our free body diagram is only of the canister. If we drew both together and considered both, then we will have to add both masses.

  • @raniaalioueche1840
    @raniaalioueche1840 Před 8 měsíci

    god bless you

  • @AbhayJagdale-dc8jd
    @AbhayJagdale-dc8jd Před 11 měsíci

    Why is frictional force taken forward, should it not be backward??

    • @QuestionSolutions
      @QuestionSolutions  Před 11 měsíci

      I don't know where you're referring to. Please use timestamps. Thanks.

  • @ericcabrera8239
    @ericcabrera8239 Před 8 měsíci +2

    Hello, I had a question about the last example, isn't frictional force supposed to act against the other forces. Since (ag)t is going to the right shouldn't Ff go to the left?

    • @QuestionSolutions
      @QuestionSolutions  Před 8 měsíci +4

      So to understand this example, I encourage you to put something on the palm of your hand. Something without a lot of friction, so like a small box, or even a coin works. Then very suddenly, move your hand to the right. You will notice that the object actually moves or tries to move to the left at first. So the friction will face to the right.

    • @norukamo
      @norukamo Před 2 měsíci

      Nice analogy! Subscribed!

  • @theexecutioner8181
    @theexecutioner8181 Před 11 měsíci

    At 12:32 while you put r = 1.5...shouldnt r value be from D to center of gravity G? Putting r= 1.5 gives us the tangential acceleration at point E nah??

    • @QuestionSolutions
      @QuestionSolutions  Před 11 měsíci +1

      The box will have the same acceleration value as rod DE and since the question is asking for just the initial angular acceleration, it's all we need.

  • @aashishkumarjha3993
    @aashishkumarjha3993 Před rokem

    question2 at 6:38, during the calculation of moment you haven’t considered the kinetic force i.e. ma but in question 3 at 9:45, you have considered the kinetic force i.e. ma
    Please elaborate when to consider kinetic force

    • @QuestionSolutions
      @QuestionSolutions  Před rokem

      At 6:38, notice that it is a moment equation for equilibrium, hence the reason why it's equal to zero.

    • @aashishkumarjha3993
      @aashishkumarjha3993 Před rokem

      @@QuestionSolutions you mean at 6:38 we are just writing moment equation but in question 3 at 9:45 we are equating moment of free body diagram with the kinetic moment.(Dalemberts principle)

    • @QuestionSolutions
      @QuestionSolutions  Před rokem

      @@aashishkumarjha3993 Correct. Keep an eye out for the right side of the equal sign. 👍

    • @aashishkumarjha3993
      @aashishkumarjha3993 Před rokem

      @@QuestionSolutions can you give me some tips, when to apply kinetic moment or just moment equation

    • @QuestionSolutions
      @QuestionSolutions  Před rokem +1

      @@aashishkumarjha3993 SO usually, if an object is in equilibrium, or if you want to figure out a force before an object moves, then you can use a moment equation. When the object is in motion and you need to figure out a force, then you can applying a kinetic moment equation.

  • @arjungovender3248
    @arjungovender3248 Před 3 lety

    Why is friction acting in that direction for the last problem? ( 11:49 )

    • @QuestionSolutions
      @QuestionSolutions  Před 3 lety +4

      Good question, might be a bit hard to visualize. So for this question, we are just focusing on the initial movement, because we are looking for the initial angular acceleration. So if we suddenly move the bar to the right, where would the box try to go initially? It'll try to go to the left, which makes friction point to the right. You can try this at home. Take something slippery, so a glossy piece of cardboard and lay it flat on your palm. Then put something on top of the carboard, preferably, an object that slips easily. Then jerk your hand to the right and see where the object on top goes. You will see that when you move the cardboard to the right, the object on top goes left. Friction always opposes the direction of travel, so friction will point right :)

    • @arjungovender3248
      @arjungovender3248 Před 3 lety

      @@QuestionSolutions I think I remember this from AP physics; for example, when a car accelerates, you sort of jerk backward. It's the same concept, right?

    • @arjungovender3248
      @arjungovender3248 Před 3 lety

      @@QuestionSolutions Thank you, much appreciated.

    • @QuestionSolutions
      @QuestionSolutions  Před 3 lety

      @@arjungovender3248 That is correct! 👍

  • @sack260
    @sack260 Před 2 lety

    for the dragster problem, for the kinetic moment about B why dont you include Ig*alpha?

    • @QuestionSolutions
      @QuestionSolutions  Před 2 lety

      Are you referring to angular acceleration?

    • @sack260
      @sack260 Před 2 lety

      @@QuestionSolutions yes! when you write the kinetic moment of B

    • @sack260
      @sack260 Před 2 lety

      @@QuestionSolutions Based on the definition of the kinetic moment, I thought it would be (1500a_G)(0.25) + (I_g)(alpha) where alpha is the angular accel

    • @QuestionSolutions
      @QuestionSolutions  Před 2 lety

      @@sack260 What is the angular acceleration of the dragster? Moreover, what is the I_g value for a dragster? You're overthinking :)

  • @arinzeanthony7447
    @arinzeanthony7447 Před 2 lety

    8:32 Please I don't soo much understand the part where frictional force is in the same direction as that of motion of the vehicle.

    • @QuestionSolutions
      @QuestionSolutions  Před 2 lety

      The wheel is slipping, so you have to think in the opposite sense. I think a good way to see it is to cut out a circle out of cardboard or something, and then spin it so that it's slipping. Then visualize the direction of the friction. 👍

  • @omizay9207
    @omizay9207 Před měsícem

    at 12:55 , why is r equal to 1.5 meters. don't you need the distance from D to G. Why did you take the distance from D to E?

    • @QuestionSolutions
      @QuestionSolutions  Před měsícem

      Whatever velocity the link moves with, the box will too. So the full length is not necessary.

  • @saiprasadsatya3677
    @saiprasadsatya3677 Před 2 lety

    You already drawn fbd for cannister then why you took weight of cannister in fbd of beam?

  • @theheroyouneed2270
    @theheroyouneed2270 Před rokem

    may i know why for the last problem the r= 1.5 rather than r=1.5+0.6

    • @QuestionSolutions
      @QuestionSolutions  Před rokem

      I assume you're referring to 12:33? If so, it's because the box will have the same acceleration value as rod DE. 👍

    • @theheroyouneed2270
      @theheroyouneed2270 Před rokem

      @@QuestionSolutions yes 12.33. Understood now, thank you. Also if the rod is directly connected to centre of box then it would be r=1.5+0.6 as the box now has different acceleration?

    • @QuestionSolutions
      @QuestionSolutions  Před rokem

      @@theheroyouneed2270 Yes, but you need a position vector from D to the center of the box, so you'd need to calculate that distance.

  • @muhammetfurkan6818
    @muhammetfurkan6818 Před 3 lety +2

    eyvallah

  • @darrylcarter3691
    @darrylcarter3691 Před 3 lety

    How did you 4-Mg as 4000 kg?

    • @QuestionSolutions
      @QuestionSolutions  Před 3 lety

      1 Mg = 1000 kg. en.wiktionary.org/wiki/megagram 👍

    • @darrylcarter3691
      @darrylcarter3691 Před 3 lety

      @@QuestionSolutions and why make the left positive when left is negative. That is in the second problem.

    • @QuestionSolutions
      @QuestionSolutions  Před 3 lety

      @@darrylcarter3691 You can pick whichever side you want to be positive. It's completely up to you. If you end up with a negative answer, then your assumption was wrong, but other than that, the answer itself doesn't change. Also, I think you are under the perception that left has to be negative, which is not true. Pick whatever you like :)

    • @darrylcarter3691
      @darrylcarter3691 Před 3 lety

      @@QuestionSolutions okay. And think I noticed something. Think Positive in this case means the court is moving. If negative it would slow down.

    • @QuestionSolutions
      @QuestionSolutions  Před 3 lety

      @@darrylcarter3691 You can assume if the cart moves to the left, it's positive. 👍

  • @MrPJBarney
    @MrPJBarney Před 3 lety

    Thank you. God bless you. Jesus loves you.

  • @HashemAljifri515
    @HashemAljifri515 Před 7 měsíci

    Your answer is incorrect in 7:58, you must divide them by two because it says determine the normal reactions at both wheels at A and both wheels at B 🌝

    • @QuestionSolutions
      @QuestionSolutions  Před 7 měsíci

      No, the answer is correct. If the question requests you to solve it for an individual wheel, you should divide it by 2. When we wrote the equations, we assumed 2 wheels at the front, 2 at the back, and then solved for them. Dividing by two will only give the result of a single wheel while the question asks you to find it for both wheels.

    • @HashemAljifri515
      @HashemAljifri515 Před 7 měsíci

      @@QuestionSolutions So this answer is for both wheels