This is incredibly helpful. Thank you for taking the time to choose great examples, show everything in well designed animations, and discuss the topic in a complete yet succinct manner.
Thank you for taking the time to write a comment like this. It's always a pleasure to read and I really appreciate it. Best wishes with your studies :)
these help so much!! mostly because unlike the fundamental or preliminary problems, the answers in the book don't include the steps so if you have a small mistkake it takes ages sometimes to find it because there are so many steps.. thank you sir and bless you!! appreciation all the way from South Africa💛
question on 9:45 , I was trying to get to the same answer that you got on your pully. I first did aA=-3aB then i plugged it in the equation below it to get T=640.5aB and when i plugged it in the last formula I got a huge number. I was wondering if you can explain what have you done?
It's hard for me to say exactly what went wrong without seeing your steps, but the answers shown on screen is correct. Here it is solved: www.symbolab.com/solver/step-by-step/a%2B3b%3D0%2C%20t-%5Cleft(50%5Ccdot9.81%5Cright)%3D-50a%2C%203t-%5Cleft(15%5Ccdot9.81%5Cright)%3D-15b?or=input
It's mostly extending lines and seeing the angles the formed. But to make it easier, you can draw this on a big piece of paper with a ruler and see how the components make up the 30 degree angle. :)
For the last problem at 12:00, why we cannot substitute s=0.2m at that step and calculate acceleration? I'm aware that this produces the wrong answer but I don't understand why.
Another question, what's a consistent way to assign signs to unknowns? I noticed that in the pulley problem (9:29), the unknown accelerations of both block A and B were assigned to be negative, implying that they both go downwards. But this doesn't make sense to me since block B is supposedly being pulled upwards when block A falls. However again in the spring problem (12:00), the unknown acceleration was assigned negative due to the spring being pushed downwards, which makes sense. However in my attempt I assigned it to be positive, thinking that in the end arithmetic would fix the sign convention for me, only to got stuck at solving the square root (12:49), where the radicand is negative my calculator couldn't give an real number.
@@kwtr So to answer your first question, you can't plug in 0.2 m because notice how F_S is not a constant, it's based on the distance the spring stretches or compresses which is also connected to the acceleration of the cylinder. This is why we need to integrate and go through the whole process of figuring out the velocity. This is sort of like applying a force to a box that's based on the displacement, something like F = 500s where "s" is the displacement. And so, to find the work, you have to integrate it, same thing. To answer your other question, when it comes to pulleys, it's a whole different way to assuming things. Once you draw out your datum, and we draw our position coordinates, like S_B and S_A, you must assume your acceleration to be in the same direction as your position coordinates. This is how pulley problems are solved, it's not really an assumption for each block, but following through what we did. If you assume the position coordinates to be down, then acceleration is also down, and vice versa. At the end, a negative answer will mean it's opposite to your assumption. I cover this in this video, I encourage you to take a look, it's not long: ua-cam.com/video/IudPPGIV5QM/v-deo.html Every pulley problem is solved in this manner. With the spring question, it's more like an educated assumption. When you see the diagram, you know which way things will move, and you can guess which side is positive and negative. Over time, you will instinctively gain the ability to see how the question will progress so you know which sides to pick to be positive. In fact, being unable to solve the square root and getting an imaginary value is pretty much your que to assign a different direction for the acceleration. 😅
In case anyone else had doubts about this like I did, I'd like to clarify that the reason why in the last two problems he used f=m(-a) instead of f=ma is because he chose the positive direction for force to be opposite the positive direction for acceleration, if instead he had defined the positive direction for force to be downwards to match the acceleration, then the equations would have been f=ma but the final answers would be the same.
Hello, at 9:28 when you solve newtons second law on each block, you took both Aa and Ab to be negative. How did you know that both blocks are moving in the upwards direction? Why isnt block A moving down and block B moving up since mass of A is heavier than B?
So I picked up to be positive for the equations and assumed both blocks to be going down, making their accelerations negative. This goes to a previous section on solving pulley problems and it's all to do with the position coordinates I picked at the beginning of the question. Both SA and SB are downwards, so that's my positive direction, I follow through with that assumption until the very end. If I get a negative answer, then I know the direction is opposite to my assumption. If you have a few minutes, please watch this video: ua-cam.com/video/IudPPGIV5QM/v-deo.html It's the foundation for pretty much all the pulley problems you face. 👍
3:22 It should be noted, although probably obvious that the sum of forces in the y is still using f=ma, it's just given that there is no upwards acceleration, therefore m(0) = 0
Usually, dynamics is a course taken after statics, which deals with all stationary objects. So when you write equations of equilibrium, all you are saying is that all forces added together must equal 0 for that object to stay stationary. Even when you write summation of moments equation, you are not considering mass at all, but the sum of all moments added together is equal to 0. So writing m(0) = 0 is not done when it comes to equations to equilibrium and most students should know this because, again, this is a course taken right after statics, in fact, it's just a carried over course using the same textbook.
@@QuestionSolutionsI see, it was confusing to me initially so I wanted to leave a comment to clarify. Thanks for the response and for your excellent videos.
So W is perpendicular to the base line (the brown line), so the angle it makes must be 30 degrees. If it's not intuitive, that's okay, you just need to consider alternate interior angles and co-interior angles.
It depends on what you're trying to find. If it's similar to the problem with the pulley, then you can use the same methods shown in the video. If not, this video might help you out: ua-cam.com/video/IudPPGIV5QM/v-deo.html
So it's 3 equations with 3 unknowns. Isolate for one variable, and then plug that value into the second equation. Then isolate for the same variable in the 2nd equation and plug that into the 3rd equation. Then you can solve. Or you can use elimination or a matrix, up to you. Here is a video someone did with 2 equations, but the process for 3 equations is pretty much the same: ua-cam.com/video/ZSJ32Bq9sbQ/v-deo.html 👍
This goes back to how you would solve pully problems, but if you look, you can see that we have a_B acceleration drawn downwards. We do that because we assume everything is going down since our position coordinates are also downwards. At the end, if you get a negative value, then we know our acceleration is opposite to our assumption. So in essence, you solve pully problems by assuming a single direction, even if we know one goes up and one goes down. See: ua-cam.com/video/IudPPGIV5QM/v-deo.html
Hi, I have one questions regarding the video: 1) In 12:32, may I know why isn't the lower bound for the displacement is 0.327 because the original displacement for the spring is 0.327 due to the weight of the cylinder and afterwards it will become s when the force is applied. btw, excellent video, I really appreciate you took the time for making such a great video and explaining everything patiently. Thank you so much!!!
Good question! So in our equation, we already accounted for the 0.327 since we wrote (s+0.327). Then when we write our integral, we just go from 0 to "s" since the 0.327 doesn't matter now, it's been accounted for. In other words, our equation took care of it already. And thank you very much, I really appreciate your comment.
So here, we are solving the 3 equations with the 3 unknowns. You can use any method you're familiar with to do it, like substitution. Please see this link if you want to see a step by step solution: www.symbolab.com/solver/step-by-step/a%2B3b%3D0%2C%20t-%5Cleft(50%5Cleft(9.81%5Cright)%5Cright)%3D50%5Cleft(-a%5Cright)%2C%203t-%5Cleft(15%5Cleft(9.81%5Cright)%5Cright)%20%3D15%5Cleft(-b%5Cright)?or=input
For the spring question why did we have to calculate the displacement of the spring without the 60N force? could we not just do one calculation with the total displacement as one variable. Like Fspring = 120(s)
@@JordanPillay-g8g So we first need to figure out the displacement of the spring with just the weight of the cylinder and we need the displacement as a variable as a force is applied. This is the only way we can get an equation that can be integrated to get the acceleration.
Hi, thank you so much for creating this video. I have a question of the pulley problem - why is the acceleration of B downward? won't it be moving upward?
@@Hadeshinai1 So you have to be able to break forces into components, which should have been covered in statics before doing dynamics. Please see: ua-cam.com/video/NrL5d-2CabQ/v-deo.html
I used the substitution method. So isolate for N in the first equation and plug that in to the 2nd equation. You can also use cymath.com or wolframalpha to check your answers. I am thinking maybe you're confused about the Pcos30 part? Convert it to decimal form, so cos30 = 0.866. Then you get 0.866P instead of Pcos30.
Love the animation in your videos. You make engineering an art which makes it very easy on the eye. Mind sharing which programs you use in creating your content
You’ve really inspired me man. I intend on “opening” a UA-cam channel that focuses on college level engineering subjects and I had this idea of describing and explaining concepts using animation. That’s when I came across your channel. You’re a beast. Thank you for the prompt response.
Another question regarding the pulley problem. If A and B are released from rest, they would both accelerate in opposite directions. Why then do you assume both of their accelerations to be downwards in your free-body-diagram?
The position coordinates were picked to be downwards, meaning position, velocity, and acceleration are all downwards, that assumption is carried out until the end of the question. This is independent of the f=ma equation, this is to do with pulley problems. Hope that helps 👍
@@QuestionSolutions Hey once again haha, Hope youre having a great day. Would it be possible to send yyou a question and my solution to it (its the question in this video about pulleys), and have some sort of short discussion with regards to a small doubt? Prehaps by email or something?
@@sameer9839 The acceleration isn't a fixed value. It's with respect to displacement, so whenever acceleration, or any other value for that matter, is not a fixed amount, equations of motion/suvat equations cannot be used.
never mind i got it, the force from the spring is a function of distance, since one force is changing while the other two forces are constant, the acceleration in f=ma is also changing.
Fantastic video, I have a quick question on the last example. I was wondering why we don't just plug s into the equation of motion and solve for the spring force right away. I understand you had integration in mind so you didn't, but how do we know we need to integrate. Thanks so much.
Good question, I will answer with another question so I can understand your question better. Let's say we do exactly like you said, we plug in s and figure out the spring force. Using that, how will we find the velocity? Like what steps will you take? :)
@@QuestionSolutions Well I'd use the equation of motion in the y-direction to find my acceleration. Then I'd use the kinematic equation v^2 = (v0^2) + 2a(s - s0). Since it started moving from equilibrium s0 and v0 would be 0. Then v^2 = 2as would be all I have left, plug in the acceleration I found from the equation of motion and get my velocity. But, it gives the wrong value so I know that method isn't interchangeable with the integration we did. Is it because the kinematic equation I used was for constant acceleration?
@@alaukwuanozie1415 That is absolutely correct. Well done! :) The kinematic equation you used was for constant acceleration, when this is not the case with this question. Whenever we have accelerations or velocities that are not constant, don't try with the kinematics equations, it will give the wrong values. Over time, you will read a question and know "okay, this won't have constant accelerations, so I have to integrate at some point." Best of luck with your studies!
At the time around 12 min., it is mentioned that as the movement is toward negative direction, the acceleration is,also negative. It does not sound correct, does it? I think the acceleration is against the movement direction, that causes the movement to stop, and become reverse.
You are confusing positive and negative signs. In the question, we are assuming up to be positive, so any force upwards is going to positive and any force downwards is negative. The acceleration is facing down, which means it will be positive. So it is actually correct. You can tell it's correct because you get a positive answer at the end, if you get a negative answer, then the assumption is wrong. I hope that makes sense.
@@arash4232 Okay, let me try and explain. So we are applying a force from top. That means the cylinder will start to move downwards, which means acceleration has to be downwards as well. If the acceleration is upwards, the cylinder will not move down, rather, it would just go up even with our force applied. I think what's happening is, you are thinking near the end of the force, so we compressed it so much that it's now slowing down the spring and trying to go back up. This is not the case, we are looking at the initial part, where we are just applying a force. Imagine you take your finger and you press down, if the cylinder moves down, the velocity is down and so is the acceleration, it's picking up speed as we push down until we get to a certain point where the resistance of the spring is now higher than the force we apply, creating a negative acceleration. Let's look at the equation we get, we got a = 15-30s. So from this, we see that acceleration becomes negative if "s" is greater than 0.5 m. But we are looking at values when s = 0.2 m. So unless we look at the question after 0.5 m, our acceleration is still down. After we go beyond 0.5m, the acceleration is upwards. Does that now make sense?
@@arash4232 Also, you said "acceleration is against the movement direction, that causes the movement to stop, and become reverse." The acceleration is dependent on the force being applied and the resistance of the spring, not the acceleration itself. If the resistive force of the spring is higher than the force we are applying, then acceleration will be negative. Keep in mind that acceleration is not independent here.
For finding 'a' in the second example, you use a kinematic equation. However, for this equation to be valid, 'a' has to be constant. How do you know 'a' is going to be constant?
Two reasons, first, we aren't worrying about static friction, so there is nothing to "overcome" meaning it'll be linear. Also, the force P is a constant. And in most cases, if it's not a constant, the question will surely say it. :)
Why is it not possible to solve for the normal using the y-component of the weight? That is using mgsin theta instead of substituting in an equation for N.
So that would work if there is only weight affecting an object. That does not work if there are other forces effecting the object. For example, if there is a force pulling upwards, then the normal force wouldn't be "mg."
I look at all the comments 😅 Are you asking how I solved the 3 equations to get an answer? Or how we came up with Aa and Ab? If it's the first, you can use substitution or elimination. If you are tight on time, you can also just graph them equations. If it's the latter, please take a look at this video: ua-cam.com/video/IudPPGIV5QM/v-deo.html, I go through how to get velocity and acceleration when it comes to pulleys.
In the third question, when I worked it out, I go -9.0 as acceleration for A and 3.0 as acceleration for B. And 50 N for the tension. I plug exactly what you have, but for some strange reason I am not getting the same answer you have on screen.
Please see: www.wolframalpha.com/input/?i=a%2B3b%3D0%2C+t-%2850%289.81%29%29%3D%28-50%29%28a%29%2C+3t-%2815%289.81%29%29%3D15%28-b%29 Do you have the same equations as I have on screen?
@@QuestionSolutions thank you. And On the spring question. The equation for Hooke’s Law is Actually F = -kx, but I see why you skipped the negative sign. Stationary springs: F=0 and x=0. Stretching: F < 0 and x > 0, Compression: F > 0 and x < 0. Because we ignored the applied force on top of the cylinder, we can say that the cylinder is falling do to gravity and therefore cause the spring to compress. So Hooke’s Law will be positive in this case.
@@darrylcarter3691 Hooke's law is actually F = kx. Hooke's law for a spring is sometimes, but rarely, stated under the convention that Fs is the restoring force exerted by the spring on whatever is pulling its free end. In that case, the equation is F = -kx You can read more here: en.wikipedia.org/wiki/Hooke%27s_law
4. A body whose mass is 735 kg, is kept in equilibrium suspended from two ropes, as indicated in the figure, one of the ropes pulls in a horizontal direction to the east, the other makes an angle of 37° with the ceiling holding it, calculate the tension in the strings. please help
Sorry but I don't solve problems like this because then I have to solve everyone's problems 😅 Please consider using a forum to get help with problems you get stuck with.
for the last problem why do not we just substitute s =0,2m to find the acceleration and use the constant acceleration equation s=vˆ2/2a to find the velocity
@@mebawubeshet6729 So you can only do what you said if the acceleration is constant, which it isn't. Notice how we get acceleration with respect to "s" which means it can't be constant.
@@QuestionSolutions to avoid that why do not we substitute s=0.2 when we find the force . fs=120(s+0.327) and the acceleration no longer dependent on s. please please elaborate it more
@@mebawubeshet6729 You can't do that because you're assuming the acceleration is the same for the whole 0.2 m, which isn't the case. So if the acceleration is not a constant value, you cannot plug in the value at the final compression and assume that's the acceleration for the whole process. In other words, the acceleration is different at 0.01, 0.5, 0.1, and 0.2. So you can't plug in 0.2 and get a value, you need to integrate. The only time you can just plug in a value is if the acceleration is constant for the whole duration of a process.
@@desireemaebattaring6716 So you can use substitution or elimination. Isolate for a single variable, and plug it into the next equation, then isolate again, and plug it into the last equation. This website covers a simple example: www.mathsisfun.com/algebra/systems-linear-equations.html
Did you plug it into your calculator? So 360sin20 = 123.13, and then -80(9.81)=-784.8. So you get: N+123.13-784.8=0, N= 661.67. You're getting 456 because your calculator is in radians, change it to degrees.
@@QuestionSolutions Thankyou very much! i didnt know that it was on radians. thats why i failed my prelim exam in mechanics!!! damnnn. anyway thanks sir! it really helps us u answering in the c omment section
Come on,,, I am waiting for planar Kinetics of Rigid bodies. I am from India. Complete statics, and dynamics topic wise....with Exhaustive theory and number of numerical problems.
Vikram, then you will have to wait longer, this is something I do in my free time. When and if I get to that topic, then you will see a video on it. I am sure there are lots of videos on the topic you mentioned, UA-cam is full of educational videos that are completely free, so please take a look. I am sure you can find some :)
This is incredibly helpful. Thank you for taking the time to choose great examples, show everything in well designed animations, and discuss the topic in a complete yet succinct manner.
Thank you for taking the time to write a comment like this. It's always a pleasure to read and I really appreciate it. Best wishes with your studies :)
Can't believe that you made everything that easy!!!, Best content ever made about dynamics on UA-cam
Glad to hear it was easy to understand! Many thanks!
This channel is god sent, thank you 🙏🏻
You're very welcome 👍
these help so much!! mostly because unlike the fundamental or preliminary problems, the answers in the book don't include the steps so if you have a small mistkake it takes ages sometimes to find it because there are so many steps.. thank you sir and bless you!! appreciation all the way from South Africa💛
You are very welcome, I am really glad these videos helped you :)
Best explanation, but i am always here day before quiz or exam😅
Thank you very much! I wish you the best on all your quizzes, midterms and exams!
@@QuestionSolutions 😀
How did our fathers do engineering without UA-cam ??😭😭😂
Lots and lots of books!
Lets consider ourselves to be lucky otherwise😅
😂😂
You study engineering to ?
That time only few and the best did engineering but today in every second house there is an engineer😂😂😂
question on 9:45 , I was trying to get to the same answer that you got on your pully. I first did aA=-3aB then i plugged it in the equation below it to get T=640.5aB and when i plugged it in the last formula I got a huge number. I was wondering if you can explain what have you done?
It's hard for me to say exactly what went wrong without seeing your steps, but the answers shown on screen is correct. Here it is solved: www.symbolab.com/solver/step-by-step/a%2B3b%3D0%2C%20t-%5Cleft(50%5Ccdot9.81%5Cright)%3D-50a%2C%203t-%5Cleft(15%5Ccdot9.81%5Cright)%3D-15b?or=input
best dynamics channel on youtube
Thank you very much!
Great video. for the 7:30 problem, how did you find the angle for W, I'm struggling to understand that
It's mostly extending lines and seeing the angles the formed. But to make it easier, you can draw this on a big piece of paper with a ruler and see how the components make up the 30 degree angle. :)
I'm in love with the way you explain, big respect
Thank you very much, really appreciate it.
Would you ever consider making a Thermodynamics set of videos? I'm taking it next semester and your dynamics videos helped me so much
Yes, it's going to start soon :)
yay!
Another fantastic video. These make preparing for my exams much easier! Thanks again!
You're very welcome! I wish you the best with your exams.
beautiful! I really liked the presentation and the way of writing it all down( the details ). Im glad I found this
Thank you very much :)
For the last problem at 12:00, why we cannot substitute s=0.2m at that step and calculate acceleration? I'm aware that this produces the wrong answer but I don't understand why.
Another question, what's a consistent way to assign signs to unknowns?
I noticed that in the pulley problem (9:29), the unknown accelerations of both block A and B were assigned to be negative, implying that they both go downwards. But this doesn't make sense to me since block B is supposedly being pulled upwards when block A falls.
However again in the spring problem (12:00), the unknown acceleration was assigned negative due to the spring being pushed downwards, which makes sense. However in my attempt I assigned it to be positive, thinking that in the end arithmetic would fix the sign convention for me, only to got stuck at solving the square root (12:49), where the radicand is negative my calculator couldn't give an real number.
@@kwtr So to answer your first question, you can't plug in 0.2 m because notice how F_S is not a constant, it's based on the distance the spring stretches or compresses which is also connected to the acceleration of the cylinder. This is why we need to integrate and go through the whole process of figuring out the velocity. This is sort of like applying a force to a box that's based on the displacement, something like
F = 500s where "s" is the displacement. And so, to find the work, you have to integrate it, same thing.
To answer your other question, when it comes to pulleys, it's a whole different way to assuming things. Once you draw out your datum, and we draw our position coordinates, like S_B and S_A, you must assume your acceleration to be in the same direction as your position coordinates. This is how pulley problems are solved, it's not really an assumption for each block, but following through what we did. If you assume the position coordinates to be down, then acceleration is also down, and vice versa. At the end, a negative answer will mean it's opposite to your assumption. I cover this in this video, I encourage you to take a look, it's not long: ua-cam.com/video/IudPPGIV5QM/v-deo.html Every pulley problem is solved in this manner.
With the spring question, it's more like an educated assumption. When you see the diagram, you know which way things will move, and you can guess which side is positive and negative. Over time, you will instinctively gain the ability to see how the question will progress so you know which sides to pick to be positive. In fact, being unable to solve the square root and getting an imaginary value is pretty much your que to assign a different direction for the acceleration. 😅
@@QuestionSolutions I'm greatly thankful for the in-depth explanation. Your videos really help a lot!
@@kwtr You're very welcome. I am really happy to hear that they are helpful :)
I love you question solutions this is ridiculously helpful
I am really glad to hear that :)
In case anyone else had doubts about this like I did, I'd like to clarify that the reason why in the last two problems he used f=m(-a) instead of f=ma is because he chose the positive direction for force to be opposite the positive direction for acceleration, if instead he had defined the positive direction for force to be downwards to match the acceleration, then the equations would have been f=ma but the final answers would be the same.
👍
Hello, at 9:28 when you solve newtons second law on each block, you took both Aa and Ab to be negative. How did you know that both blocks are moving in the upwards direction? Why isnt block A moving down and block B moving up since mass of A is heavier than B?
So I picked up to be positive for the equations and assumed both blocks to be going down, making their accelerations negative. This goes to a previous section on solving pulley problems and it's all to do with the position coordinates I picked at the beginning of the question. Both SA and SB are downwards, so that's my positive direction, I follow through with that assumption until the very end. If I get a negative answer, then I know the direction is opposite to my assumption.
If you have a few minutes, please watch this video: ua-cam.com/video/IudPPGIV5QM/v-deo.html
It's the foundation for pretty much all the pulley problems you face. 👍
3:22 It should be noted, although probably obvious that the sum of forces in the y is still using f=ma, it's just given that there is no upwards acceleration, therefore m(0) = 0
Usually, dynamics is a course taken after statics, which deals with all stationary objects. So when you write equations of equilibrium, all you are saying is that all forces added together must equal 0 for that object to stay stationary. Even when you write summation of moments equation, you are not considering mass at all, but the sum of all moments added together is equal to 0. So writing m(0) = 0 is not done when it comes to equations to equilibrium and most students should know this because, again, this is a course taken right after statics, in fact, it's just a carried over course using the same textbook.
@@QuestionSolutionsI see, it was confusing to me initially so I wanted to leave a comment to clarify. Thanks for the response and for your excellent videos.
You're very welcome and I wish you the best with your studies :)@@swagodaman6320
Very clear and helpful. Thank you for making these videos.
You're very welcome!
You saved my life again thank you! Your videos are really helpful thanks again
You're very welcome! Keep it up and I wish you the best. Thanks again for taking the time to write a comment :)
7:08 How’d you know the angle is 30 when getting the y component of W?
So W is perpendicular to the base line (the brown line), so the angle it makes must be 30 degrees. If it's not intuitive, that's okay, you just need to consider alternate interior angles and co-interior angles.
Simple and excellent explanation. Grateful to you, sir
Thank you very much :) Keep up the good work!
very helpful video, i just have one question, what if a motor is attached to a load on pulley , does i need to do the equation of motion formula?
It depends on what you're trying to find. If it's similar to the problem with the pulley, then you can use the same methods shown in the video. If not, this video might help you out: ua-cam.com/video/IudPPGIV5QM/v-deo.html
9:59 Please explain it to me how did you solve the value of aA, aB, & T
So it's 3 equations with 3 unknowns. Isolate for one variable, and then plug that value into the second equation. Then isolate for the same variable in the 2nd equation and plug that into the 3rd equation. Then you can solve. Or you can use elimination or a matrix, up to you. Here is a video someone did with 2 equations, but the process for 3 equations is pretty much the same: ua-cam.com/video/ZSJ32Bq9sbQ/v-deo.html 👍
@@QuestionSolutions anyway thanks for responding huhu
in 9:37 why is acceleration b negative since its an upward acceleration a while u took up as positive
This goes back to how you would solve pully problems, but if you look, you can see that we have a_B acceleration drawn downwards. We do that because we assume everything is going down since our position coordinates are also downwards. At the end, if you get a negative value, then we know our acceleration is opposite to our assumption. So in essence, you solve pully problems by assuming a single direction, even if we know one goes up and one goes down. See: ua-cam.com/video/IudPPGIV5QM/v-deo.html
Hi, I have one questions regarding the video: 1) In 12:32, may I know why isn't the lower bound for the displacement is 0.327 because the original displacement for the spring is 0.327 due to the weight of the cylinder and afterwards it will become s when the force is applied.
btw, excellent video, I really appreciate you took the time for making such a great video and explaining everything patiently. Thank you so much!!!
Good question! So in our equation, we already accounted for the 0.327 since we wrote (s+0.327). Then when we write our integral, we just go from 0 to "s" since the 0.327 doesn't matter now, it's been accounted for. In other words, our equation took care of it already.
And thank you very much, I really appreciate your comment.
@@QuestionSolutions ohhhh understood. Thanks!!!
7:28 where did you find the -0.5P?
sin30 = 0.5
Another excellent video, thank you!
You're very welcome!
for the equations at 8:25 how’d you get the 2(sb-a) part
Please kindly see this video first: ua-cam.com/video/IudPPGIV5QM/v-deo.html
How do you get the acceleration and the force in 9:50? I don't get it.
So here, we are solving the 3 equations with the 3 unknowns. You can use any method you're familiar with to do it, like substitution. Please see this link if you want to see a step by step solution: www.symbolab.com/solver/step-by-step/a%2B3b%3D0%2C%20t-%5Cleft(50%5Cleft(9.81%5Cright)%5Cright)%3D50%5Cleft(-a%5Cright)%2C%203t-%5Cleft(15%5Cleft(9.81%5Cright)%5Cright)%20%3D15%5Cleft(-b%5Cright)?or=input
For the spring question why did we have to calculate the displacement of the spring without the 60N force? could we not just do one calculation with the total displacement as one variable. Like Fspring = 120(s)
Please give me a timestamp so I know where you're referring to. Thanks!
@@QuestionSolutions 10:55
@@JordanPillay-g8g So we first need to figure out the displacement of the spring with just the weight of the cylinder and we need the displacement as a variable as a force is applied. This is the only way we can get an equation that can be integrated to get the acceleration.
What do we do if t = 4s is not given (exercise 5:40)
Is the question the exact same? The wording and all or are there other givens instead?
Hi, thank you so much for creating this video. I have a question of the pulley problem - why is the acceleration of B downward? won't it be moving upward?
Please see this video first: ua-cam.com/video/IudPPGIV5QM/v-deo.html
It's more to do with how we establish the initial directions.
on what instances do we need to check if the object will move before we compute the necessary parts?
@7:15 how can I tell that W had that cosine for a y component?
@@Hadeshinai1 So you have to be able to break forces into components, which should have been covered in statics before doing dynamics. Please see: ua-cam.com/video/NrL5d-2CabQ/v-deo.html
how did you find P at 7:45?
I used the substitution method. So isolate for N in the first equation and plug that in to the 2nd equation. You can also use cymath.com or wolframalpha to check your answers. I am thinking maybe you're confused about the Pcos30 part? Convert it to decimal form, so cos30 = 0.866. Then you get 0.866P instead of Pcos30.
Love the animation in your videos. You make engineering an art which makes it very easy on the eye. Mind sharing which programs you use in creating your content
Thank you very much! I really appreciate it. I use after effects for the animations and illustrator for the drawings. :)
You’ve really inspired me man. I intend on “opening” a UA-cam channel that focuses on college level engineering subjects and I had this idea of describing and explaining concepts using animation. That’s when I came across your channel. You’re a beast. Thank you for the prompt response.
@@zanenhlanhla_mthethwa That's awesome! Keep up the great work and I wish you the best with your future endeavors :)
thank you so much for your videos, way better than my professor
You're very welcome, best of luck with your studies!
How do you put the formulas in the calculator, if I do for example 360cos20 * 0.3 my answer is not 338,29N. The same for the Fy
Where are you referring to? Please use timestamps. Also, is your calculator set to deg and not rad?
Another question regarding the pulley problem. If A and B are released from rest, they would both accelerate in opposite directions. Why then do you assume both of their accelerations to be downwards in your free-body-diagram?
The position coordinates were picked to be downwards, meaning position, velocity, and acceleration are all downwards, that assumption is carried out until the end of the question. This is independent of the f=ma equation, this is to do with pulley problems. Hope that helps 👍
@@QuestionSolutions Hey once again haha, Hope youre having a great day. Would it be possible to send yyou a question and my solution to it (its the question in this video about pulleys), and have some sort of short discussion with regards to a small doubt? Prehaps by email or something?
@@radiatedbug Sure, send me an email at contact @ questionsolutions.com
Wow best explain every equations are clear
Thank you very much :)
for pulley how did you get 2(sb-sa) where does the two come from?
So this is to do with how you solve pulley problems. Please see this video: ua-cam.com/video/IudPPGIV5QM/v-deo.html it should clear it up :)
would you ever make one for fluid mechanics?
Yes, it's on my to-do list.
Why can't we directly input the value of s=0.2 into the acceleration equation and find the velocity using one of the suvat equations?
Please give me a timestamp so I know where to look. Thanks!
12:14 sorry completely forgot
@@sameer9839 The acceleration isn't a fixed value. It's with respect to displacement, so whenever acceleration, or any other value for that matter, is not a fixed amount, equations of motion/suvat equations cannot be used.
hm how do we understand if the acceleration is not constant? the question doesn't say much about it.
never mind i got it, the force from the spring is a function of distance, since one force is changing while the other two forces are constant, the acceleration in f=ma is also changing.
how to know if we use kinematics eqn or not. i thought we use kinematics eqn only if the question mention about "constant acceleration"
Kinematics equations and when to use them are covered in this video: ua-cam.com/video/FsGBUM5o2-k/v-deo.html
Hey, I got a question. How do you identify whether it is sin or cos for the x and y component for force p? Thank you in advance
Please kindly watch this video: ua-cam.com/video/NrL5d-2CabQ/v-deo.html especially the first example, I go through how to pick sine and cosine. :)
Fantastic video, I have a quick question on the last example.
I was wondering why we don't just plug s into the equation of motion and solve for the spring force right away. I understand you had integration in mind so you didn't, but how do we know we need to integrate. Thanks so much.
Good question, I will answer with another question so I can understand your question better. Let's say we do exactly like you said, we plug in s and figure out the spring force. Using that, how will we find the velocity? Like what steps will you take? :)
@@QuestionSolutions Well I'd use the equation of motion in the y-direction to find my acceleration. Then I'd use the kinematic equation v^2 = (v0^2) + 2a(s - s0). Since it started moving from equilibrium s0 and v0 would be 0. Then v^2 = 2as would be all I have left, plug in the acceleration I found from the equation of motion and get my velocity. But, it gives the wrong value so I know that method isn't interchangeable with the integration we did.
Is it because the kinematic equation I used was for constant acceleration?
@@alaukwuanozie1415 That is absolutely correct. Well done! :) The kinematic equation you used was for constant acceleration, when this is not the case with this question. Whenever we have accelerations or velocities that are not constant, don't try with the kinematics equations, it will give the wrong values. Over time, you will read a question and know "okay, this won't have constant accelerations, so I have to integrate at some point." Best of luck with your studies!
I did exactly the same before watching the solution, thanks for the question
At the time around 12 min., it is mentioned that as the movement is toward negative direction, the acceleration is,also negative.
It does not sound correct, does it?
I think the acceleration is against the movement direction, that causes the movement to stop, and become reverse.
You are confusing positive and negative signs. In the question, we are assuming up to be positive, so any force upwards is going to positive and any force downwards is negative. The acceleration is facing down, which means it will be positive. So it is actually correct. You can tell it's correct because you get a positive answer at the end, if you get a negative answer, then the assumption is wrong. I hope that makes sense.
@@QuestionSolutions
Thank you so much but, it made more confusion.
Anyways, thanks.
@@arash4232 Okay, let me try and explain. So we are applying a force from top. That means the cylinder will start to move downwards, which means acceleration has to be downwards as well. If the acceleration is upwards, the cylinder will not move down, rather, it would just go up even with our force applied. I think what's happening is, you are thinking near the end of the force, so we compressed it so much that it's now slowing down the spring and trying to go back up. This is not the case, we are looking at the initial part, where we are just applying a force. Imagine you take your finger and you press down, if the cylinder moves down, the velocity is down and so is the acceleration, it's picking up speed as we push down until we get to a certain point where the resistance of the spring is now higher than the force we apply, creating a negative acceleration. Let's look at the equation we get, we got a = 15-30s. So from this, we see that acceleration becomes negative if "s" is greater than 0.5 m. But we are looking at values when s = 0.2 m. So unless we look at the question after 0.5 m, our acceleration is still down. After we go beyond 0.5m, the acceleration is upwards. Does that now make sense?
@@arash4232 Also, you said "acceleration is against the movement direction, that causes the movement to stop, and become reverse." The acceleration is dependent on the force being applied and the resistance of the spring, not the acceleration itself. If the resistive force of the spring is higher than the force we are applying, then acceleration will be negative. Keep in mind that acceleration is not independent here.
For F=ma is acceleration always taken to be positive (9.8m/s^2)? Does it depend on our chosen axis?
It depends on your chosen axis. Up to you, but make sure to follow through with your positives and negatives until the very end of the question.
It's better to make the positive direction the direction of motion then you'd say?
If it's going down then make the positive direction that if it's going up then make positive up?
@@jesussaquin6266 Most of the time, yes, pick the direction where an object is going to be positive.
@@jesussaquin6266 If it's like a projectile problem, I'd pick up to be positive, however, you can do it either way.
Can you explain how you figured the acceleration is constant?
NVM I got it
Please kindly use timestamps so I know where to look. Many thanks!
ı dont understand question at 10.56. why Fy=0 firstly?
Initially, we are calculating it at rest, so when there is no other vertical force being applied.
For finding 'a' in the second example, you use a kinematic equation. However, for this equation to be valid, 'a' has to be constant. How do you know 'a' is going to be constant?
Two reasons, first, we aren't worrying about static friction, so there is nothing to "overcome" meaning it'll be linear. Also, the force P is a constant. And in most cases, if it's not a constant, the question will surely say it. :)
Why is it not possible to solve for the normal using the y-component of the weight? That is using mgsin theta instead of substituting in an equation for N.
So that would work if there is only weight affecting an object. That does not work if there are other forces effecting the object. For example, if there is a force pulling upwards, then the normal force wouldn't be "mg."
shouldn't you have used coefficient of friction to be 0.4 instead of 0.3 in the first question?
It was used, you can see it at 3:54
hello, hope you can see this comment. I just want to ask how you got both accelerations, Aa, and Ab on the pulley problem.
I look at all the comments 😅 Are you asking how I solved the 3 equations to get an answer? Or how we came up with Aa and Ab? If it's the first, you can use substitution or elimination. If you are tight on time, you can also just graph them equations. If it's the latter, please take a look at this video: ua-cam.com/video/IudPPGIV5QM/v-deo.html, I go through how to get velocity and acceleration when it comes to pulleys.
@@QuestionSolutions k
In the third question, when I worked it out, I go -9.0 as acceleration for A and 3.0 as acceleration for B. And 50 N for the tension. I plug exactly what you have, but for some strange reason I am not getting the same answer you have on screen.
Please see: www.wolframalpha.com/input/?i=a%2B3b%3D0%2C+t-%2850%289.81%29%29%3D%28-50%29%28a%29%2C+3t-%2815%289.81%29%29%3D15%28-b%29
Do you have the same equations as I have on screen?
@@QuestionSolutions thank you. And On the spring question. The equation for Hooke’s
Law is Actually F = -kx, but I see why you skipped the negative sign. Stationary springs: F=0 and x=0. Stretching: F < 0 and x > 0, Compression: F > 0 and x < 0. Because we ignored the applied force on top of the cylinder, we can say that the cylinder is falling do to gravity and therefore cause the spring to compress. So Hooke’s Law will be positive in this case.
@@darrylcarter3691 Hooke's law is actually F = kx. Hooke's law for a spring is sometimes, but rarely, stated under the convention that Fs is the restoring force exerted by the spring on whatever is pulling its free end. In that case, the equation is F = -kx
You can read more here: en.wikipedia.org/wiki/Hooke%27s_law
U are amazing god bless u
Thank you very much, I appreciate it.
4. A body whose mass is 735 kg, is kept in equilibrium suspended from two ropes, as indicated in the figure, one of the ropes pulls in a horizontal direction to the east, the other makes an angle of 37° with the ceiling holding it, calculate the tension in the strings. please help
Sorry but I don't solve problems like this because then I have to solve everyone's problems 😅 Please consider using a forum to get help with problems you get stuck with.
Hello . Your videos are very helpful . Can you please make videos on magnetism ? Thanks in advance .
Yes, I will try, though not for some time :(
for the last problem why do not we just substitute s =0,2m to find the acceleration and use the constant acceleration equation s=vˆ2/2a to find the velocity
how to figure out if the given acceleration is constant
@@mebawubeshet6729 So you can only do what you said if the acceleration is constant, which it isn't. Notice how we get acceleration with respect to "s" which means it can't be constant.
@@QuestionSolutions to avoid that why do not we substitute s=0.2 when we find the force . fs=120(s+0.327) and the acceleration no longer dependent on s. please please elaborate it more
@@mebawubeshet6729 You can't do that because you're assuming the acceleration is the same for the whole 0.2 m, which isn't the case. So if the acceleration is not a constant value, you cannot plug in the value at the final compression and assume that's the acceleration for the whole process. In other words, the acceleration is different at 0.01, 0.5, 0.1, and 0.2. So you can't plug in 0.2 and get a value, you need to integrate. The only time you can just plug in a value is if the acceleration is constant for the whole duration of a process.
Thanks, sir examples are very helpful ❤❤❤❤❤❤❤❤
I am really glad to hear that! I wish you the best with your studies. ❤
@QuestionSolutions so proud with your reply .I'm Egyptian and I got many lessons from you ❤️❤️❤️❤️.finally, I love your accent so much❤️❤️❤️❤️🥺
@@israahelal2427 Thank you very much! You are much too kind :) ❤❤
how did you solve the value of a?
Which question are you referring to? Please kindly use timestamps in the future so I know where to look.
@@QuestionSolutions 9:40
@@desireemaebattaring6716 So you can use substitution or elimination. Isolate for a single variable, and plug it into the next equation, then isolate again, and plug it into the last equation.
This website covers a simple example: www.mathsisfun.com/algebra/systems-linear-equations.html
can you do a dry friction video? Pleaseee?
It's definitely on my to-do list. Usually, first year statics courses don't cover dry friction explicitly, which is why I didn't cover it.
To much helpful ❤️❤️❤️
Glad to hear! ❤️❤️❤️
thank you so much that was really helpful
You're very welcome!
helped a lot thanks man
Glad it helped! Keep up the great work.
In the second problem, why did we put cos30 as the angle at which the weight is inclined instead of sine 30
Please see: ua-cam.com/users/shortsvynnKlJD_Jo?si=BBsbwbiIVbRLq6Lx
great work, ty :)
You're very welcome! Best of luck with your studies :)
in the first problem i get N= 456.139.... howd u get 661.67? i tried shift solve and manual and still lead me to 456.139.. thankyou in advance!
Did you plug it into your calculator? So 360sin20 = 123.13, and then -80(9.81)=-784.8. So you get: N+123.13-784.8=0, N= 661.67. You're getting 456 because your calculator is in radians, change it to degrees.
@@QuestionSolutions Thankyou very much! i didnt know that it was on radians. thats why i failed my prelim exam in mechanics!!! damnnn. anyway thanks sir! it really helps us u answering in the c omment section
@@flee3695 Yes, easy mistake to make, always check if it's in radians or degrees :) Best of luck with your studies!
It is very helpful
Glad to hear!
thank you
You're very welcome!
it was helpful
Glad to hear!
well done 🌹
👍 Thank you!
tysm
You're very welcome!
the pulley problem where you just show the answers on a of A and a of B and T didn't show the process, please show the process
Where are you referring to? Please use timestamps.
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Come on,,, I am waiting for planar Kinetics of Rigid bodies. I am from India. Complete statics, and dynamics topic wise....with Exhaustive theory and number of numerical problems.
Vikram, then you will have to wait longer, this is something I do in my free time. When and if I get to that topic, then you will see a video on it. I am sure there are lots of videos on the topic you mentioned, UA-cam is full of educational videos that are completely free, so please take a look. I am sure you can find some :)
Is this for ualberta?
For anyone at any university/school that wants to learn 👍
Why you used second equation of motion in second example to find acceleration when acceleration is not constant???
It is constant, it's 0.75 m/s^2.
@@QuestionSolutions Thank you! Sir
Thank You
You're welcome :)