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!
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! :)
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!
5:01 ...And yes, if you were wondering about Q1, when finding the last two links, EF and GH, instead of considering the entire force system (from top to bottom), you can focus on the spreader beam BD, which has a mass of 50 kg. Multiply this mass by the vertical acceleration and equate it to the forces in links BA (-BA) and DC (-CD), along with the weight of the spreader, plus the vertical components of links EF and GH. This approach will still yield the same magnitudes for the forces in the links. The main point is that it's possible to use other vertical components in the system to determine the forces in the links. Just don’t forget to adjust your masses according to the specific component(s) you are working with in the system. Sharp
@@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!
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?
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.
Thank you for such an informative video! My question is regarding the moment equation when used for a different point. Why do we use clockwise as positive? In dynamics so far I have been told only consider counter clockwise as positive. Regards!
You're very welcome. In 2D problems, you can use clockwise as positive or counterclockwise as positive. You will get the same answer regardless of which you pick. It's totally up to you. In 3D problems, you will need to use the right hand rule to figure out the positive vector directions. Try it out :)
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?
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.
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 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)
@@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.
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.
Please take a look at this video: ua-cam.com/video/NrL5d-2CabQ/v-deo.html It's not long, and I go through how to break forces into components using sine and cosine, when to use them, etc. :)
So if you hold something in the palm of your hand and suddenly, you move your hand to the right, where does the object move? It moves to the left right? Which means friction would be to the right, opposing that movement.
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.
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?
@@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.
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?
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!
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 UA-cam 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.
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.
@@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?
@@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.
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.
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 :)
@@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.
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??
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?
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.
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.
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.
@@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.
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.
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. 👍
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 👍
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.
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 :)
@@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?
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 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
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!
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.
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.
@@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??
@@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??
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. 👍
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: ua-cam.com/users/shortsP029mqnp4XY
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?
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.
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?
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. 😅
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?
@@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.
@@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.
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.
@@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 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
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?
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 👍
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 👍
@@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?
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 🌝
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.
@@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 :)
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.
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.
@@QuestionSolutions Thanks a lot
Nah your lectures are literally amazing.... they are so simple and straight forward 🐐
Thank you very much! :)
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!
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! :)
This man is single handedly getting me through this subject
I am glad to hear these videos are helpful to you. I wish you the best with your studies!
got a dynamics final on monday and this video is going to be the only reason ill come close to passing ty so much
You're very welcome. I wish you the best on your exam Monday. You got this!
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!
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!
@@QuestionSolutions Whoha so fast..
I shall 💪 Thank you!
@@QuestionSolutions got the highest grade on my exam..thank u so much!🥰🥰🥰
5:01
...And yes, if you were wondering about Q1, when finding the last two links, EF and GH, instead of considering the entire force system (from top to bottom), you can focus on the spreader beam BD, which has a mass of 50 kg. Multiply this mass by the vertical acceleration and equate it to the forces in links BA (-BA) and DC (-CD), along with the weight of the spreader, plus the vertical components of links EF and GH. This approach will still yield the same magnitudes for the forces in the links.
The main point is that it's possible to use other vertical components in the system to determine the forces in the links. Just don’t forget to adjust your masses according to the specific component(s) you are working with in the system.
Sharp
That's right! Many ways to get to the same answer.
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?
Can you give me a time stamp or let me know which question you are referring to? Thanks!
@@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
@@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!
thanks for such a great explanation in less time
You're very welcome!
These videos are so helpful thank you !
You're very welcome!
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?
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.
Thank you for such an informative video!
My question is regarding the moment equation when used for a different point. Why do we use clockwise as positive? In dynamics so far I have been told only consider counter clockwise as positive. Regards!
You're very welcome.
In 2D problems, you can use clockwise as positive or counterclockwise as positive. You will get the same answer regardless of which you pick. It's totally up to you. In 3D problems, you will need to use the right hand rule to figure out the positive vector directions. Try it out :)
The best! About to get a 100 on my final thx to u
Glad to hear! I wish you the best on your final.
4:55 why don't you consider the forces in the y direction of the two tension forces we found earlier?
When we consider the object as a whole, we don't consider forces inside it.
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?
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.
Your videos are really helpful, Thank you!
You're very welcome!
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
At 6:38, notice that it is a moment equation for equilibrium, hence the reason why it's equal to zero.
@@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)
@@aashishkumarjha3993 Correct. Keep an eye out for the right side of the equal sign. 👍
@@QuestionSolutions can you give me some tips, when to apply kinetic moment or just moment equation
@@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.
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.
Please take a look at this video: ua-cam.com/video/NrL5d-2CabQ/v-deo.html It's not long, and I go through how to break forces into components using sine and cosine, when to use them, etc. :)
@@QuestionSolutions Thank you that makes sense now!
@@alsix7334 Glad to hear!
why did you choose the frictional force to be to the right at 11:50 ?
So if you hold something in the palm of your hand and suddenly, you move your hand to the right, where does the object move? It moves to the left right? Which means friction would be to the right, opposing that movement.
Hey, you forgot to upload 16.8 Relative motion analysis using rotating axes. Please upload. Thanks!
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 👍
@@QuestionSolutions Thank u. What subjects are u doing now?
@@programmingprograms726 Currently, working on thermodynamics :)
question 2 at 5:15, why aren’t we considering friction between wheel and surface?
It's not given in the question, so you can safely ignore it.
from today on you are mister fantastic
Alright!! 😅
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
Also in this question why did we ignore the friction
@@mariaalhijazeen4974 We didn't ignore the friction, it was used to find Fmax, at 10:55.
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
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.
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?
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!
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?
@@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.
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?
So Ffmax tells us the minimum force required to make NA = 0, in other words, the minimum needed to lift up the wheel.
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!
substitution into on of the two equations
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 UA-cam 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.
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?
Whatever velocity the link moves with, the box will too. So the full length is not necessary.
Hello! Thank for the the video. In the dragster problem, why did we take the moment about B and not G?
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.
@@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?
@@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.
@@QuestionSolutions thank you very much for answering!
Q2, why friction direction to the right? How to know direction rear wheel slipped
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.
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 :)
@@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.
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??
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.
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?
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.
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.
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.
for the dragster problem, why didnt we consider moment due to friction while solving for M(b)=(Mk)b
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.
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
Please give me a timestamp so I know where you're referring to. Thanks!
can the dragster problem be solved by using the moment equation about the center mass instead of point B?
You can, but you'd have to solve more equations because you're not eliminating any of the unknowns. It's too time consuming.
For 1st example, why are we not adding the mass of the spreader to find m in the equation?
Can you give me a timestamp? I'm not sure which "m" you're referring to.
@@QuestionSolutions Thanks for replying. I was talking about m at 4:28 on the F_y equation.
@@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.
Can you explain the difference between torque and moment, please? I see their equations are the same. Thanks for the great tutorials!
For a normal statics course, you can think of them as the same. 👍
@@QuestionSolutions Thanks a lot Mr. Greatest tutor of all time!
@@SG-dw8jh 😅You're very welcome!
Torque is spinning, rotation about the same axis, whereas moment is rotation about different axis.
when do we draw a kinetic diagram in solving problems and when do we don't?
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.
Why we don't consider the Tab and Tcd when calculating for the Tef and Tgh (Q1)
They become part of the whole object when we consider forces T_ef and T_gh.
Hello sir I hope you are doing well, do you use specific programs to simulate questions? It will really help me.
Thanks in advance
I don't use any programs to simulate questions. I animate them by hand the best I can :)
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)?
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. 👍
@@QuestionSolutions you're amazing sir haha
@@DrDerivative Thanks! Keep up the good work, best wishes with your studies :)
For the drag car problem wouldnt it be 2NB because the problem says rear wheels?
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 👍
may I know how do you set the direction of Ialpha in the kinetic diagram
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.
Why is friction acting in that direction for the last problem? ( 11:49 )
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 :)
@@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?
@@QuestionSolutions Thank you, much appreciated.
@@arjungovender3248 That is correct! 👍
You are the best
Thank you :)
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.
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 👍
@@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
Nevermind, I got it now =)
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?
@@DrDerivative glad to hear :)
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!😃
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!
@@QuestionSolutions Sorry I realized my answer right after posting the question. Thank you for replying so promptly!
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.
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.
@@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??
@@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??
@@aliemad6823 Well normal acceleration is dependent on the radius of curvature, so would they both have the same radius of curvature? :)
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.
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. 👍
How do i know when the moment i shoud be taking CW or CCW as positive?
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: ua-cam.com/users/shortsP029mqnp4XY
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?
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.
Nice analogy! Subscribed!
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?
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. 😅
@@QuestionSolutions Oh okay I see, thank you very much. You have the best youtube channel for mechanics!
@@QCJF4G Thank you so much, really appreciate your comment :)
THANK YOU
You're very welcome!
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?
Could you provide a timestamp to the location you're referring to? Thanks!
@@QuestionSolutions 6:29, you chose a moment equation about G.
@@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.
@@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.
@@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!
Why is frictional force taken forward, should it not be backward??
I don't know where you're referring to. Please use timestamps. Thanks.
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😅
Please see this video, it's less than 60 seconds, and you will understand it much better: ua-cam.com/users/shortsvynnKlJD_Jo?feature=share
You're the best!
😊
12.57 why did you get anormal zero ? can you explain in detaily
At 12:21, we already establish that angular velocity is 0, which means normal acceleration is also zero.
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
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.
@@QuestionSolutions Thanks that was a quick response and a great one too!!
@@melekmnif1933 You're very welcome! Best of luck with your studies.
@@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?
@@arinzeanthony7447 In cases that's exactly like this, yes, that is how the friction will act.
eyvallah
Not sure what you said, but I will think it's something positive 👍
It means Thank You in Turkish
@@muhammetfurkan6818 You are very welcome!
for the dragster problem, for the kinetic moment about B why dont you include Ig*alpha?
Are you referring to angular acceleration?
@@QuestionSolutions yes! when you write the kinetic moment of B
@@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
@@sack260 What is the angular acceleration of the dragster? Moreover, what is the I_g value for a dragster? You're overthinking :)
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?
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 👍
@@QuestionSolutions thank you very very much you are the best
@@abdallahamouda6633 You are very welcome!
You already drawn fbd for cannister then why you took weight of cannister in fbd of beam?
Please give me a timestamp so I know where to look, thanks!
@@QuestionSolutions 4:51
@@saiprasadsatya3677 oh, because the beam carries the weight of the cannister as well.
@@QuestionSolutions oh thanku
In the second problem you said acceleration upwards is negligible. So you put 60(0) why cant we take a=-9.81?
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 👍
Nice
Thank you so much! :)
may i know why for the last problem the r= 1.5 rather than r=1.5+0.6
I assume you're referring to 12:33? If so, it's because the box will have the same acceleration value as rod DE. 👍
@@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?
@@theheroyouneed2270 Yes, but you need a position vector from D to the center of the box, so you'd need to calculate that distance.
god bless you
Thank you very much!
You are great
Thank you so much
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 🌝
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.
@@QuestionSolutions So this answer is for both wheels
How did you 4-Mg as 4000 kg?
1 Mg = 1000 kg. en.wiktionary.org/wiki/megagram 👍
@@QuestionSolutions and why make the left positive when left is negative. That is in the second problem.
@@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 :)
@@QuestionSolutions okay. And think I noticed something. Think Positive in this case means the court is moving. If negative it would slow down.
@@darrylcarter3691 You can assume if the cart moves to the left, it's positive. 👍
Thank you. God bless you. Jesus loves you.
You're welcome!