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THANK YOU VERY MUCH SIRRRR!!! AREA MOMENT METHOD IS SO MUCH BETTER since most of us are struggling with the equation methond cause we always assume all the time that x(length) is just equal to distance from the origin to the cut section and we put value in it. and that's why we have wrong results. Thank youuuuuuuuuuuuuuuuuu <3
Sir, thank you for your video! A problem baffle me for a whole day, at 4:48, there are 5 external forces exert on pin C, which are Cy, Cx, -Cy, -Cx, -600N. Four of them: Cy, Cx, -Cy, -Cx canceling each other in both x and y direction, leaving the -600N alone, which means, if I drawing the FBD of pin C, I can’t find any way to get equilibrium of the y-axis.
So the equation we found is with respect to the angle shown as θ. So when θ = 60 is when we need to find the length S since the question wants the angular velocity at θ = 60.
When we find the resultant force and couple about the origin, does that mean their line of action will cross the origin? And does changing the reference point affect the resultant force and couple?
My professor just really finished this topic in 40 mins or something, but, I was clueless what happened in those 40 mins. Gotta say, these 8 minutes were worth! Kudos, keep making such easy-to-understand videos! You gained a sub, sir!
I watched your content before going to write my exam , The result are not back yet and the level of confidence i have is way to high. Your videos helped me a lot thank you brotherman🤝
I hope you did great on your exam! Thanks for taking the time to write a comment like this, it's really nice. Keep up the great work and I wish you the best in your future endeavors.
Sorry I have so many questions video but for the spring position equation why did you include 2(Sa-a)? I’m still rusty on these pulley problems do you have any recommendations to really get the concept of when you count some sections vs when not? I’m practicing to no avail. 9:40
Hello, I am a little confused about the differences between the terms moments, couples, and torques. Is the total torque of a system equal to the sum of all the moments and couples acting on the system?
They are all the same, just interchangeable words when it comes to solving problems. The total summation of moments about a system, is the same as the total torque of a system, etc. Please see this video first: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-QNNnPZ68STI.html
Sorry, I don't understand your question. We didn't calculate force BC? Did you give me the proper timestamp? Regardless, all values shown are correct, except at 9:08, where force F_HI = 42.5 NOT 45.2.
How do u even know which table to refer to? Ive watched the properties diagram video multiple and i still dont get why some is superheated or saturated when its not stated in the question
So the only answer I can give you is, you may have watched the video on property tables but you haven't understood it. Unfortunately, it is absolutely fundamental that you take the time and watch it over and over until you understand it. If it isn't helping, and reading the textbook doesn't help either, I highly encourage you to go and speak with your professor or TA. It is a must that you understand and have the ability to distinguish superheated substances, compressed liquids and more. Hopefully, you are referring to this video? ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-rKbjRG4Y-HM.html I explain how to determine what is what, and go through a few examples. So please take the time to view it and follow along with the examples.
hey man, i just wanted to say thanks for these videos. I got a 97/100 on my first exam in dynamics and i learned the most information from your videos. Thank you.
@@QuestionSolutions thank you so muchhhhhh Sir for replying me all the time. I got it now. Also, I’m preparing for my civil engineering licensure exam this November. I’ll inform you Sir once I passed it. Thank you so muuuuchhhhhh
3:21 wait i understand the pattern and the logical conclusion, but isn't Fdccos60 is itself Fde? I know it will give a logical result (which is Fde=-Fdccos60, it's negative bcs of the direction), but what about the sigma Fx? Are those things (Fde and Fdccos60) two different forces, or they are just the same single force? I'm guessing they are two different forces since Fdccos60 is just Fdc pointing to the horizontal direction (it's one force broken up into two directions), while Fde is a force that always points to the horizontal direction. Is my guess correct?
In this case, yes, because there were no other forces at the pin. Also, I assume you meant cos60? Because the x-component of force FDC is equal to the magnitude of force FDE. Again, that's just in this instance. If, for example, there was another force pointing in another direction (as long as it wasn't straight up), FDE would not be equal to the x-component of force FDC.
@@QuestionSolutionsdont we usually do Tan(angle)=(height/base) ??? Our height is 1.5 and the base is 2… why in the video we did the opposite…like base over height…sorry for my bad explanation
@@6rozh9 That's incorrect and you should remove that thought from your mind. Instead, remember, tan = opposite over adjacent. So you look from the perspective of the angle, and look for the opposite side to the angle. Here, that's 2 m, and the adjacent side is 1.5 m. While we are at it, remember this too. Sine is opposite over hypotenuse and cosine is adjacent over hypotenuse. Always look from the perspective of the angle and look to see what side is opposite to it and what side is adjacent to it.
The weight is hanging straight down. The vector for weight is straight down, it's not slanted in any other direction. Which means, it has just a z-component.
@@PulengManchidi There isn't much to consider about zero force members. What I mean is, doing a few examples will allow you to instantly realize which members are zero force members, so you can easily write them off. Other than that, when you solve your problems using the method of joints, you will get zero force members pointed out in the solution.
@@readbhagwatgeeta3810 Where do you see wB/IC? You're just multiplying a distance labeled rB/IC by a variable labeled wBC. When you plug your values in, you get a solution for the variable wBC. I don't understand what you mean by "How wBC = wB/IC" where did we write that on the solution?
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!
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.
For the gear and rack example , why isn't F also 200N ? Isn't the rack applying a force on the wheel's teeth of 200N? Also , you made a mistake where the radius of gyration should be 125mm not 0.15m.
No mistake. You are confusing radius to radius of gyration. Radius of gyration was used at 7:13, and radius was used at 5:12. Also, we find the value of F at 8:03, and it's not 200 N.
This is the definition of radius of gyration: "A radius of gyration in general is the distance from the center of mass of a body at which the whole mass could be concentrated without changing its moment of rotational inertia about an axis through the center of mass." So where did you get the equation I =mk*k ? That seems to be the mass MOI for a hoop , not a solid gear.
You are not using the proper equation when the radius of gyration is given in a question. Go to any website, but for your convenience I got one for you www.pw.live/exams/gate/moment-of-inertia-and-radius-of-gyration/ and scroll to the bottom where it talks about the radius of gyration and how to apply that to find the moment of inertia. This is also in your textbook and I encourage you to review your notes. Always keep in mind, if the radius of gyration is given, the equation to use is I = mk^2. @@jackflash8756
So let's assume we write an equation for x-axis forces. We have Fcosθ-Fcosθ = 0. It doesn't really do anything for us (there is no way for us to solve it). So we write an equation of equation for the y-axis forces instead.
Does it apply to all problems that if it's only one cable, the tension is just the same to the starting point and the end point especially if it's a pulley.
@@QuestionSolutions thank youuuu! Just letting you know your example was my examination yesterday. Tension of the cable including the spring . You’re an angel!!!
