Hello, when drawing the FBD for the first problem @ 4:12 how did you know that the normal force N is acting on the can 30 degrees below the radial axis? You said that the normal force is always perpendicular to the tangent [of the path] but when did we find the tangent to figure out the position of N?
So it's actually where the wheel is touching, because wherever the wheel touches (it's the brown curve) would be the tangent point since it's a circle. The angle is already given, so it would be 30 degrees from the radial line. If you had a wheel on a flat surface, the normal force would be straight up, right? That's because that's the contacting point from the floor. In the same way, this wheel is touching 2 places, the arm and the curve. The normal force is simply tangent to that curve while the force being applied is tangent to the radial line. I hope that makes sense. 👍
mind if I take a crack at this from a different approach? It took me a while to come up with another explanation as to why the normal force had the direction of 30 degrees below the radial axis, but I have a theory. If you hypothetically move the rod all the way up to the very top of the circular path with radius of 0.6m, theta would be equal to 45 degrees, while the normal force would be completely vertical and pointing upwards for the barrel. If theta were approaching zero, the normal force would become almost horizontal even though the magnitude would be smaller (if theta were zero, the barrel would not even be on the circular path). So, to find the direction of the normal force for any given point along this circular path, it's basically theta = normal force direction in relation to the radial axis. To my understanding of the problem, this is why the angle is 30 degrees!
Origin of the mechanical arm to the middle of the mountain is 600mm, the radius of the mountain is also 600mm. This creates an isosceles triangle. And inside such a triangle, we know that the angle that 1 leg makes (with it's base) has to be equal to the other leg. The normal force on a circle surface goes through it's origin (like a radius line). Thus we know the r axis is the mechanical arm, we know that the normal force is with the radius line and we know that the arm, the radius line and the ground create a isosceles triangle (with the mechanical arm as the base) so the angles between the ground and the arm (which is 30 degrees) has to be the same angle the mechanical arm and the radius line creates. Just draw the triangle if you cannot follow the explanation.
In this problem, to figure out where is the normal force, we draw a circle its radius is 600 mm. The point C is on the circle. We all knew that the tangent line at any points on the circle will be perpendicular to the radius. So in this problem, the normal force is along a radius connecting the center and point C.( point C is also the intersection of the radial axis and a circle ). From the figure we will see the angle between the radial axis and the normal force = 30 o.
These videos are really nice . I have a question though . At 7:19 the normal force is included into the FBD following the idea that its normal to the tangent line. But then the force F is included in the direction of theta. Why is it that we dont always assume there is a force parralel and one perpendicular to the tangent line rather than making conclusions based on the structure? If I was to assume the first , would it be ok or is that incorrect ?
Well, let's say for the discussion's sake, the force is parallel to the tangent line. If we look at how that force would be applied, it doesn't make much sense right? Just by looking at how this pin sits in the slot and where contact occurs, it's not possible for the force to be parallel. It has to come straight down at the location shown on the diagram. Also, note that this is a top down view so it might be a bit hard to see. Let's switch to a simple box. Imagine a box sitting on an inclined plane, and I am pushing this box from the side of the inclined plane but at an angle of 45 degrees with respect to the incline. Is it right to assume that the force is parallel to the incline? No, right? The force is independent. But now what about the normal force? That's going to be perpendicular to the inclined plane. Another force that behaves similar is weight. It doesn't matter about the incline, that force will always point down (along y-axis). I can put the box on a 45 degree incline, a 30 degree incline, doesn't matter, the weight is straight down and the normal force is always perpendicular to the plane. So just like that, the normal force in this question is perpendicular to the tangent line. Force applied is independent because we can push our box from any direction we want, at any angle we want.
At 6:28, y does the FBD for the second question not have the weight of the cylinder acting on the rod in the transversal axis?? Since the weight was used in the FBD in the first question.
So in the question, we are told "Motion occurs in the horizontal plane" as in, it's like you're doing a top down view. So the weight would be acting into your screen.
That's a really vague question. First, why can't you solve dynamics problems? If you know the concepts, where are you getting stuck? I encourage you to try out the problems I solve in these videos. First, try it yourself, when you get stuck, see what steps are taken next in the video. That can maybe help you out. The only way to get better at solving problems is to really bite down and do as many as you can. Create a schedule where you allocate certain amounts of time to each course you take. So 1 hour for course one, 1.5 hours for dynamics, etc. Give more time for courses that you struggle most with. Then do as many questions as you can, over and over during that time period. This usually helps students get a good idea on how to go about solving a question.
If the question is given with rad/s, then it's better to find it in radians, you don't have to, but it makes it easier. You can also plot in degrees, but again, that's your choice, usually, if everything else is in radians, then its better to use radians.
so for the second example, if the force is pointed upwards, why isn't there another force that pushes the cylinder in the COUNTERCLOCKWISE direction, since the force you calculated turns out to push the cylinder around the clockwise direction?
I don't think I understand your question. Which other force will push it counterclockwise? Remember that it's looking at an instantaneous point of time. At that instant, so when theta is 180 degrees, we find that the force is upwards, rather than down. This changes based on where the cylinder is in the slot. There aren't any other forces to consider, you can think of friction, but we aren't given values to solve for it.
@@QuestionSolutions oh ok, I was wondering since I thought that the rod was continuing to push the cylinder around in a counterclockwise direction. Could it be that this force is causing the cylinder to slow down?
@@med7756 The normal force is always perpendicular to the tangent line. So your assumption is incorrect here, also, don't get too stuck on the diagram because it can lead to incorrect answers. Instead, remember, the normal force will always be perpendicular to the tangent line, because that's where they touch. I go through this in detail around 1:01.
Two questions: one is from example 2 @ 7:50 you said the force is neg bc its coming from the bottom, what dose that mean, since the F in the FBD is pointing towards the particles motion? Example 3 @ 8:54 when taking the 2nd and 3rd time derivative of r, why not use the product rule, I watched the video on time der. and it looked like the product rule is only used when sin and cos are in the term?
Thank you so much for the timestamps, makes things so much easier. So first question. Notice how we assumed that the force would be coming from the top in the FBD (the red arrow). Now we got a negative value, so that means instead of it coming down, it's actually pointing upwards. So the red arrow would be facing upwards, and be applied from the bottom of the cylinder. For the 2nd question, product rule is used whenever 2 functions are being multiplied together. In this problem, the 0.75 is a constant, so all we have is one single function. When you do normal derivatives, we don't really care about the integer in front right? So like the derivative of 2x^2 = 2(2)x = 4x. So we leave the 2 alone, and then at the end, we just make it look nice by multiplying it out. The same here. I just want to be very clear, the product rule is used whenever we have 2 functions being multiplied, not just in cases of sine and cosine. Going a bit deeper with this example, we have r = 0.75z, we take the derivative with respect to time. We get ṙ = 0.75 (dz/dt), then we take the derivative again, so we ignore the 0.75, and just care about the (dz/dt) giving 0.75 (d^2z/dt^2). Again, this is not 2 functions, but just a constant multiplying a function. 👍
when we take weight into account? for example in 2nd question weight wasn't included in the solution. that also happens in 12th edition dynamics hibbler book, example 13.10 & 13.12
Usually, in cases where we consider the collective forces in a certain axis, instead of individual ones, are places where weight doesn't need to be shown. It also depends on the axes we use, so if weight is "through" the page/plane, we don't need to draw it, or in other examples, cases with the z axis, will have weight. It depends on the question, and you usually gain a "feel" for when and when not to generalize forces by doing a few questions. Sorry I couldn't be more helpful.
Don't get the difference between the tangent line and the transverse coordinate at 6:10. So, we draw the transverse coordinate perpendicular to the radial, but how did you come up the tangent line at 6:12?
@@TTcreations77 Thanks for time stamp. So the equation should be in your textbook, along with the proof. But in essence, you're using a ratio to figure out the angle. The dr/dθ is the slope, because remember, it's a derivative. the r value is the "distance."
When you calculated psi in example 2. I get -89.68 degrees and that is when my calculator is set to radians if i do not use radians and only substitute -pi then I get 72.34 degrees. I'm not sure when to use radians and when not to???
If you used radians, you should get -1.26, which then converted to degrees will still give you -72.34 degrees. Please double check. Since we are trying to find an angle on a coordinate axis system, use degrees, it makes the math easier than using radians. The psi angle will be in degrees, not radians. 👍
@@QuestionSolutions thanks for replying so fast! If I use theta and then use tan(180 degrees) i also get - 89...sorry if I'm misunderstanding something
@@ruandre5619 Okay, I think I know what's happening. So you should be entering pi, exactly as is, 3.14. This 3.14 is just a ratio we found (conveniently for us, it turned out to be pi, but it could be any value like 6.28, 9.42, etc). So don't put in 180 degrees. When you take the tan inverse of the value, if your calculator is in radians, you will get -1.26, if it's in degrees, you will get -72.34 degrees. If you convert the -1.26 rad to degrees, you will get -72.34 degrees.
Could you elaborate which part you didn't understand if you have the time? For example, was it where the cos30 and sin30 came from? I'll do my best to help!
@@QuestionSolutions thanks for replying! , My problem is with the normal force at 4:12 Like shouldn't there be two normal forces one with the rod and the other with the surface ? And how did you determine the angle to be 30? Is the normal force parallel with radial line? And if it's perpendicular to the tangent line what is the relation between the tangent line and the radial and transverse line ?
@@Furfur-rj2tw To answer your first question, no, there is one normal force. Imagine you have a box on the floor and you push the box with your hand, (you apply a force from the left). Where do we draw the normal force? Straight upwards, right? That's where the box contacts the floor. In the same way, here, the rod simply applies a force that pushes the cylinder along the surface. So the normal force is tangent to the surface. For the second part of your question, no, the normal force is not parallel to the radial line. It is tangent to the surface, and it's 30 degrees below the radial line. We are given the angle in the question (it's the angle that's created with respect to the x-axis and the radial axis), you can draw out the normal force tangent to the surface, and with a bit of geometry, you will see that it's 30 degrees (if you want to prove it). To answer the third question, I am unsure of which tangent like you are referring to? Keep in mind that the normal force will always be perpendicular to surface of contact. Also, please take a look at the previous video with the x, y, z coordinates as well, as that builds on some of the stuff that is used in this chapter. Let me know if that clears up some stuff. Thanks!
@@QuestionSolutions Thanks for replying again ! I appreciate you giving me some of your time. What I was trying to say about the tangent line is that , we know that the normal force is perpendicular to the tangent line , so if we were to draw a tangent line in the F.B.D where would it cross? It's obviously not going to be perpendicular to the radial line , I would imagine it a little tilted so basically my question is If I were to draw the tangent line in the question @ 4:12 what shall I draw it with respect to or how to draw it basically? And thanks for your time again !
So in the question, we are told "Motion occurs in the horizontal plane" as in, it's like you're doing a top down view. So the weight would be acting into your screen.
The best way to get a feel for when you need to find specific forces is to do as many questions as you can. If you do about 10 per chapter, you should have a very good idea on what to find, when to find it, and so forth. 👍
Hello! Example 2: the weight is ignored, which is in the direction of θ. Considering the weight I got F = -5.8 N. I appreciate if anyone check this and reply.
@@willgggg900 This question is all about the wording. "Motion occurs in the horizontal plane" as in, it's like you're doing a top down view. So the weight would be acting into your screen.
@@mebawubeshet6729 Something that might help you is to draw a big diagram on a large piece of paper and use a protractor. This might help you visualize it better.
@@mebawubeshet6729 Unfortunately, I don't. It might help to refresh some geometry. I think a lot of the time, just doing as many questions as possible usually helps. Try drawing them on large pieces of paper and seeing if that helps you :)
When you need to figure out the angle between the normal force and the tangent line. If you can solve the question without using the psi angle, then you don't need to find it. But sometimes, you'll notice that you can't figure out what the angle is for the normal force, in that case, you need to find it.
Hi! Which example are you referring to? If it's the last one, we are looking at all the collective forces in the transverse, radial, and z axis which includes all forces that effect the car in the axis we are looking at.
Yes, that should have been drawn in the free body diagram of the pin. My mistake. Please remember that weight is always straight down. Even if it's not shown, it is there, since it's mass times the acceleration due to gravity.
@@Aooomr100 Yes, that should have been drawn in the free body diagram of the pin. My mistake. Please remember that weight is always straight down. Even if it's not shown, it is there, since it's mass times the acceleration due to gravity.
The way of explanation and the vedeo is something special , good job ,may allah protect you.❤❤
Thank you very much! Best wishes with your studies.
bro u literally helped mepass my exam 6 hrs..i salute u teacher!
Super happy to hear that! Very nice job :)
You can't imagine my happiness when I found out you also make videos of dynamics.
Love you bro
Keep it up ❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️
Interestingly, I actually made the dynamics videos before I made the statics videos 😅And thank you! ❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤❤
😅
Sorry we studied statics first 😅
Keep it up bro ❤️❤️
@@mohibjadoon3359 Yes, that is actually the way, since you need statics to do dynamics, but for some reason, I decided to do dynamics videos first 😅
Hello, when drawing the FBD for the first problem @ 4:12 how did you know that the normal force N is acting on the can 30 degrees below the radial axis? You said that the normal force is always perpendicular to the tangent [of the path] but when did we find the tangent to figure out the position of N?
So it's actually where the wheel is touching, because wherever the wheel touches (it's the brown curve) would be the tangent point since it's a circle. The angle is already given, so it would be 30 degrees from the radial line. If you had a wheel on a flat surface, the normal force would be straight up, right? That's because that's the contacting point from the floor. In the same way, this wheel is touching 2 places, the arm and the curve. The normal force is simply tangent to that curve while the force being applied is tangent to the radial line. I hope that makes sense. 👍
mind if I take a crack at this from a different approach? It took me a while to come up with another explanation as to why the normal force had the direction of 30 degrees below the radial axis, but I have a theory. If you hypothetically move the rod all the way up to the very top of the circular path with radius of 0.6m, theta would be equal to 45 degrees, while the normal force would be completely vertical and pointing upwards for the barrel. If theta were approaching zero, the normal force would become almost horizontal even though the magnitude would be smaller (if theta were zero, the barrel would not even be on the circular path). So, to find the direction of the normal force for any given point along this circular path, it's basically theta = normal force direction in relation to the radial axis. To my understanding of the problem, this is why the angle is 30 degrees!
Origin of the mechanical arm to the middle of the mountain is 600mm, the radius of the mountain is also 600mm. This creates an isosceles triangle. And inside such a triangle, we know that the angle that 1 leg makes (with it's base) has to be equal to the other leg. The normal force on a circle surface goes through it's origin (like a radius line).
Thus we know the r axis is the mechanical arm, we know that the normal force is with the radius line and we know that the arm, the radius line and the ground create a isosceles triangle (with the mechanical arm as the base) so the angles between the ground and the arm (which is 30 degrees) has to be the same angle the mechanical arm and the radius line creates.
Just draw the triangle if you cannot follow the explanation.
In this problem, to figure out where is the normal force, we draw a circle its radius is 600 mm. The point C is on the circle. We all knew that the tangent line at any points on the circle will be perpendicular to the radius. So in this problem, the normal force is along a radius connecting the center and point C.( point C is also the intersection of the radial axis and a circle ). From the figure we will see the angle between the radial axis and the normal force = 30 o.
@@QuestionSolutions but 30 degree is n't the angle from x axis
Thank you Sir.
May Lord bless you always.
Thank you very much :)
Thank you! the Hibbler text explanation for cylindrical coordinates is horrible, but this video is fantastic!!
I am glad to hear you found this video helpful! Best of luck with your studies.
These videos are really nice . I have a question though . At 7:19 the normal force is included into the FBD following the idea that its normal to the tangent line. But then the force F is included in the direction of theta. Why is it that we dont always assume there is a force parralel and one perpendicular to the tangent line rather than making conclusions based on the structure?
If I was to assume the first , would it be ok or is that incorrect ?
Well, let's say for the discussion's sake, the force is parallel to the tangent line. If we look at how that force would be applied, it doesn't make much sense right? Just by looking at how this pin sits in the slot and where contact occurs, it's not possible for the force to be parallel. It has to come straight down at the location shown on the diagram. Also, note that this is a top down view so it might be a bit hard to see.
Let's switch to a simple box. Imagine a box sitting on an inclined plane, and I am pushing this box from the side of the inclined plane but at an angle of 45 degrees with respect to the incline. Is it right to assume that the force is parallel to the incline? No, right? The force is independent. But now what about the normal force? That's going to be perpendicular to the inclined plane. Another force that behaves similar is weight. It doesn't matter about the incline, that force will always point down (along y-axis). I can put the box on a 45 degree incline, a 30 degree incline, doesn't matter, the weight is straight down and the normal force is always perpendicular to the plane. So just like that, the normal force in this question is perpendicular to the tangent line. Force applied is independent because we can push our box from any direction we want, at any angle we want.
At 6:28, y does the FBD for the second question not have the weight of the cylinder acting on the rod in the transversal axis?? Since the weight was used in the FBD in the first question.
So in the question, we are told "Motion occurs in the horizontal plane" as in, it's like you're doing a top down view. So the weight would be acting into your screen.
Thank you very much, so happy that you also have videos for dynamics
You're very welcome! best wishes with your studies :)
I cant solve dynamics problems.What can i do.I know concepts but cant solve.😢
That's a really vague question. First, why can't you solve dynamics problems? If you know the concepts, where are you getting stuck? I encourage you to try out the problems I solve in these videos. First, try it yourself, when you get stuck, see what steps are taken next in the video. That can maybe help you out. The only way to get better at solving problems is to really bite down and do as many as you can. Create a schedule where you allocate certain amounts of time to each course you take. So 1 hour for course one, 1.5 hours for dynamics, etc. Give more time for courses that you struggle most with. Then do as many questions as you can, over and over during that time period. This usually helps students get a good idea on how to go about solving a question.
It is straight necessary to calculate de psi angle in radians? Why we don't plot teta in degrees? Thanks for the video!
If the question is given with rad/s, then it's better to find it in radians, you don't have to, but it makes it easier. You can also plot in degrees, but again, that's your choice, usually, if everything else is in radians, then its better to use radians.
Thanks for the help. I really dislike this class, but you help make it bearable.
Sorry to hear that, but I can understand as well. Glad to hear these make it bearable, I wish you the best with your studies!
Thank you so much, Professors really became lazy after covid
You're welcome! It's a tough time. Hope for a better 2021.
very good explanation and nice understanding of the topic.subscribed
Glad to hear and thank you so much for subscribing!
so for the second example, if the force is pointed upwards, why isn't there another force that pushes the cylinder in the COUNTERCLOCKWISE direction, since the force you calculated turns out to push the cylinder around the clockwise direction?
I don't think I understand your question. Which other force will push it counterclockwise? Remember that it's looking at an instantaneous point of time. At that instant, so when theta is 180 degrees, we find that the force is upwards, rather than down. This changes based on where the cylinder is in the slot. There aren't any other forces to consider, you can think of friction, but we aren't given values to solve for it.
@@QuestionSolutions oh ok, I was wondering since I thought that the rod was continuing to push the cylinder around in a counterclockwise direction. Could it be that this force is causing the cylinder to slow down?
@@Mechomittencaterpillar That would make the most sense, especially since we are looking at it at 180 degrees. :)
in the second example why did we take the r and theta axes in the other direction that we usually do? is it because we found psi as -ve ?
Please kindly give me a timestamp so I know where to look.
The question on the horizontal plane, shouldn't the normal force be in z-direction which is the static direction?
Can you give a timestamp as to where you're talking about? Thanks! :)
@@med7756 The normal force is always perpendicular to the tangent line. So your assumption is incorrect here, also, don't get too stuck on the diagram because it can lead to incorrect answers. Instead, remember, the normal force will always be perpendicular to the tangent line, because that's where they touch. I go through this in detail around 1:01.
Two questions: one is from example 2 @ 7:50 you said the force is neg bc its coming from the bottom, what dose that mean, since the F in the FBD is pointing towards the particles motion? Example 3 @ 8:54 when taking the 2nd and 3rd time derivative of r, why not use the product rule, I watched the video on time der. and it looked like the product rule is only used when sin and cos are in the term?
Thank you so much for the timestamps, makes things so much easier. So first question. Notice how we assumed that the force would be coming from the top in the FBD (the red arrow). Now we got a negative value, so that means instead of it coming down, it's actually pointing upwards. So the red arrow would be facing upwards, and be applied from the bottom of the cylinder. For the 2nd question, product rule is used whenever 2 functions are being multiplied together. In this problem, the 0.75 is a constant, so all we have is one single function. When you do normal derivatives, we don't really care about the integer in front right? So like the derivative of 2x^2 = 2(2)x = 4x. So we leave the 2 alone, and then at the end, we just make it look nice by multiplying it out. The same here. I just want to be very clear, the product rule is used whenever we have 2 functions being multiplied, not just in cases of sine and cosine.
Going a bit deeper with this example, we have r = 0.75z, we take the derivative with respect to time. We get ṙ = 0.75 (dz/dt), then we take the derivative again, so we ignore the 0.75, and just care about the (dz/dt) giving 0.75 (d^2z/dt^2). Again, this is not 2 functions, but just a constant multiplying a function. 👍
@@QuestionSolutions thanks for the quick reply thats makes more sense, I've been subscribed since my semester started your videos are great
@@AverageNeighbor Glad to hear! I hope these videos help you and you do really well in your courses.
when we take weight into account? for example in 2nd question weight wasn't included in the solution. that also happens in 12th edition dynamics hibbler book, example 13.10 & 13.12
Usually, in cases where we consider the collective forces in a certain axis, instead of individual ones, are places where weight doesn't need to be shown. It also depends on the axes we use, so if weight is "through" the page/plane, we don't need to draw it, or in other examples, cases with the z axis, will have weight. It depends on the question, and you usually gain a "feel" for when and when not to generalize forces by doing a few questions. Sorry I couldn't be more helpful.
@@QuestionSolutions thank you!
Don't get the difference between the tangent line and the transverse coordinate at 6:10. So, we draw the transverse coordinate perpendicular to the radial, but how did you come up the tangent line at 6:12?
Oh never mind I might have confused the tangent line with the tangential acceleration
@@Mohamed-pj7sx Okay, glad to hear you got it cleared up. :)
could you please explain how to get the equation for tan value between tagential and radial direction?
Please kindly provide a timestamp so I know where to look. Many thanks!
@@QuestionSolutions sir about what I asked is in @ 6.22
@ 6:22
@@TTcreations77 Thanks for time stamp. So the equation should be in your textbook, along with the proof. But in essence, you're using a ratio to figure out the angle. The dr/dθ is the slope, because remember, it's a derivative. the r value is the "distance."
@@QuestionSolutions Thank you sir. Now, I understood now how it was taken.
Thanks for helping me ace my Dynamics test !
Awesome!!! I am so happy for you. I wish you the best in your future studies :) Keep it up. 👍👍👍👍
why on example 3, is there no Normal force acting on the car??
We didn't use a coordinate system that contained the normal axis, we used the z-axis instead.
When you calculated psi in example 2. I get -89.68 degrees and that is when my calculator is set to radians if i do not use radians and only substitute -pi then I get 72.34 degrees. I'm not sure when to use radians and when not to???
If you used radians, you should get -1.26, which then converted to degrees will still give you -72.34 degrees. Please double check. Since we are trying to find an angle on a coordinate axis system, use degrees, it makes the math easier than using radians. The psi angle will be in degrees, not radians. 👍
@@QuestionSolutions thanks for replying so fast! If I use theta and then use tan(180 degrees) i also get - 89...sorry if I'm misunderstanding something
@@ruandre5619 Okay, I think I know what's happening. So you should be entering pi, exactly as is, 3.14. This 3.14 is just a ratio we found (conveniently for us, it turned out to be pi, but it could be any value like 6.28, 9.42, etc). So don't put in 180 degrees. When you take the tan inverse of the value, if your calculator is in radians, you will get -1.26, if it's in degrees, you will get -72.34 degrees. If you convert the -1.26 rad to degrees, you will get -72.34 degrees.
@@QuestionSolutions thanks man I appreciate!
@@ruandre5619 You're very welcome!
Hello! dude...Thanks uuuuu so much..U literally helped me.❤❤
You're very welcome! Glad to hear it was helpful. Keep up the good work and best wishes with your studies.
Thank you, cleared up my confusion!
I am very glad to hear that :)
Hey dude nice video , I hope you can make a video on how to calculate normal forces because I didn't really get how you calculated it in 4:12
Could you elaborate which part you didn't understand if you have the time? For example, was it where the cos30 and sin30 came from? I'll do my best to help!
@@QuestionSolutions thanks for replying! ,
My problem is with the normal force at 4:12
Like shouldn't there be two normal forces one with the rod and the other with the surface ?
And how did you determine the angle to be 30? Is the normal force parallel with radial line?
And if it's perpendicular to the tangent line what is the relation between the tangent line and the radial and transverse line ?
@@Furfur-rj2tw To answer your first question, no, there is one normal force. Imagine you have a box on the floor and you push the box with your hand, (you apply a force from the left). Where do we draw the normal force? Straight upwards, right? That's where the box contacts the floor. In the same way, here, the rod simply applies a force that pushes the cylinder along the surface. So the normal force is tangent to the surface. For the second part of your question, no, the normal force is not parallel to the radial line. It is tangent to the surface, and it's 30 degrees below the radial line. We are given the angle in the question (it's the angle that's created with respect to the x-axis and the radial axis), you can draw out the normal force tangent to the surface, and with a bit of geometry, you will see that it's 30 degrees (if you want to prove it). To answer the third question, I am unsure of which tangent like you are referring to? Keep in mind that the normal force will always be perpendicular to surface of contact. Also, please take a look at the previous video with the x, y, z coordinates as well, as that builds on some of the stuff that is used in this chapter. Let me know if that clears up some stuff. Thanks!
@@QuestionSolutions Thanks for replying again ! I appreciate you giving me some of your time.
What I was trying to say about the tangent line is that , we know that the normal force is perpendicular to the tangent line , so if we were to draw a tangent line in the F.B.D where would it cross?
It's obviously not going to be perpendicular to the radial line , I would imagine it a little tilted so basically my question is
If I were to draw the tangent line in the question @ 4:12 what shall I draw it with respect to or how to draw it basically?
And thanks for your time again !
@@Furfur-rj2tw The tangent line is parallel to the surface where the cylinder touches. The normal force is then perpendicular to that tangent line. 👍
7:11 Hey man in the FBD why we don’t have to out the weight of the cylinder. Thanks
So in the question, we are told "Motion occurs in the horizontal plane" as in, it's like you're doing a top down view. So the weight would be acting into your screen.
@@QuestionSolutions that explains it. Thanks a lot!
You're very welcome!@@Sigma_J001
This helps but I am still unsure of when I would need to find the normal force and angle φ in a problem
The best way to get a feel for when you need to find specific forces is to do as many questions as you can. If you do about 10 per chapter, you should have a very good idea on what to find, when to find it, and so forth. 👍
thanku so much for all your help sir .
You're very welcome!
in the second example I believe you didn't multiply dr^./dtheta^. by theta^.
Please give me a timestamp so I know where you're referring to. Thanks!
Thank you!
You're very welcome!
Thanks you sir for this kind of videos
You're very welcome!
Hello!
Example 2: the weight is ignored, which is in the direction of θ. Considering the weight I got F = -5.8 N.
I appreciate if anyone check this and reply.
You don't have to consider the weight for this problem, unless your problem was different?
@@QuestionSolutions why dont you have to consider the weight but in other problems you do ?
@@willgggg900 This question is all about the wording. "Motion occurs in the horizontal plane" as in, it's like you're doing a top down view. So the weight would be acting into your screen.
@@QuestionSolutions okay thanks i just realised i really appreciate your videos it has helped me alot
@@willgggg900 You're very welcome! :)
how did you get the 15.19N?
Please let me know of a timestamp so I know where to look. Thanks!
please be more clear for the explanation on how do you find angles?
I don't know where you're referring to, please use timestamps.
@@QuestionSolutions 4:08 and do not get how the angle from radial is 30 degree
@@mebawubeshet6729 Something that might help you is to draw a big diagram on a large piece of paper and use a protractor. This might help you visualize it better.
@@QuestionSolutions ok and please recommend me videos. or do you have any videos on this topic????
@@mebawubeshet6729 Unfortunately, I don't. It might help to refresh some geometry. I think a lot of the time, just doing as many questions as possible usually helps. Try drawing them on large pieces of paper and seeing if that helps you :)
Thank you very much :)
You're very welcome! :)
When we take the sigh angle and when we don't ?
When you need to figure out the angle between the normal force and the tangent line. If you can solve the question without using the psi angle, then you don't need to find it. But sometimes, you'll notice that you can't figure out what the angle is for the normal force, in that case, you need to find it.
@@QuestionSolutions Thanks you sir
@@osama669 You're welcome!
Why is the Normal Force in example not included ?
Hi! Which example are you referring to? If it's the last one, we are looking at all the collective forces in the transverse, radial, and z axis which includes all forces that effect the car in the axis we are looking at.
@@QuestionSolutions yeah I meant the third one. My keyboard was on numlock that's why the number didn't show up. But now I understand it thanks xD
@@abrahamsweetvoice7687 You're very welcome! Best of luck with your studies :)
why didnt use 600mm at 4.45 question?
Sorry, where would we use it?
ar = -8,73 will be m*ar=0.5(-(-8.73)) =0.5*8,73 because left is negative ? at 7.32
No, the acceleration values are independent since they were pre-calculated.
why the weight is not marked in 2nd example ?
Yes, that should have been drawn in the free body diagram of the pin. My mistake. Please remember that weight is always straight down. Even if it's not shown, it is there, since it's mass times the acceleration due to gravity.
@@QuestionSolutions Thanks for the reply this channel is very helpful
@@ayyaash_munzir You're very welcome. Best of luck with your studies!
Sir send this ppt for revision
I don't have one, sorry.
@@QuestionSolutions ok no problem
In the Third question. Fz, in that case, is our normal force! 😅
👍
why you forget the wight it force ?
Please let me know a timestamp so I know where to look.
@@QuestionSolutions at 7:02 in this question when you simplify the forces you didt take care of wight
@@Aooomr100 Yes, that should have been drawn in the free body diagram of the pin. My mistake. Please remember that weight is always straight down. Even if it's not shown, it is there, since it's mass times the acceleration due to gravity.
I don't understand psi and how its relevant at all
Sorry, where are you referring to?
3:55 because the question is simple 😭
I only meant that it's simple in the sense that the angle was already given to us, so no extra step needed 🥺
@@QuestionSolutions i'm just kidding actually,
after going through the whole f=ma things, i agree it's simpler, thanks mate 😄
@@khairulanwar7642 You're welcome! Keep up the good work 👍
Give example while teaching. Then only goes in head.
Not sure what you mean? There are multiple examples covered step by step.
how are determining the angles as 30 for normal and weight
Please give a timestamp so I know where you're referring to. Thanks!
I cant get sign of forces right
Please give me a timestamp of a place where you're not getting the signs correct. I can help you out better that way. Thanks!