This is amazing. Thank you SO much. I could understand in 10 minutes what I couldn't for months in other physics classes. Teaching is a skill on its own.
Wanted to say I came here to say looking for help to solve a challenge problem for my physics class only until I realized half way through that it was a trick answer. Still stayed for the excellent lesson, you’ve definitely helped me on my path Mr. van Biezen. Many thanks.
I was the whiz when we did this at school (in the 60s) but I moved in to chemistry and not gone near since. Too many years later and m son needs help with his homework. French school as well so it's in French God help me. You might just have saved my life with this. Very well explained. Thank you.
I swear to my teacher’s explanation was so complicated and I didn’t understand anything until I saw this vid and I have a test tomorrow thank u so much
Some days ago i was ill and I couldn't come to school. I tried to study but didn't understand until I watched this video. Thank you so much for explaining. You saved my life
Thank you so much , i find all your videos very helpful since I’m taking AP physics and the teacher in school tries to explain things fast since we have a lot of material to cover and I often find myself lost in her lesson and it’s hard for me to keep up with her , one day i found your videos and ever since that day i depend on them whenever i need help . :) thank you very much 💕✨
Michel Biezen, i enjoyed watching and learning from your tutorials. Your Engineering tutorials are very good, and i like them. You did an excellent jobs of explaining concepts in a very entertaining , clear and understandable way . Many thanks for your tutorials.
Sir, you might hear this a lot of times or probably none but you're awesome sir. You have saved my life lol I hope you're doing well! And thanks a lot for this video!
We remember that when we were students we were at times very desperate to find anything that could help us understand the material. That is a big part of the reason why we are making these videos to help students today that are in the same position. Comments like yours tell us it is working. We are very glad they are of help to students. Thank you for your comment.
Yes here you can use Lami's theorem, but you can't use it when you have an unknown force and angle that's why it is better to know how to resolve forces.
Been watching your vids for a while, but just subscribed. I like the straightforward style you have. My only input would be based on how I like to do problems like this and that is to wait and obtain the final forms of the variables and equations and then do all the trig functions and other numbers and arithmetic. Great video however, I enjoy watching and seeing another style and explanation. :)
That is indeed the typical way in which one should solve physics problems (solve the equations first before substituting numbers), but there are a number of types of problems, where plugging in numbers earlier is advisable.
@@MichelvanBiezen I like the videos you've done where you take the moment of inertia of a pulley into account. I'd like to see one where you take the mass of the rope into account too.
great vid! im currently in my final year of highschool, and due to the coronavirus im very bored so im practising for physics in college. watching this makes me feel dumb and clever at the same time😂
You are The Boss , I am glad to watch your video ,you are amazing and i understood everything, even though My English is not good enough , thank you very much , made my day .
A 100 lb weight is suspended from 2 cables. The tensions T1 and T2 satisfy the following equations: 0.6T1+0.8T2=100 (1) 0.8T1−0.6T2=0 (2) Solve for T1 and T2.
You can work this problem in 2 ways. Simply solve it algebraically (2 equations and 2 unknowns). Multiply the bottom equation by (-6/8) and then add the result to the top equation to eliminate T1 so you can solve for T2.
The lesson is pretty amazing I appreciate, For further knowledge, I'd like to know what's contributes to the large tensions in T1 and T2, whereas in normal situations, we would expect that summation of tensions in T1 and T2 will be equal to the tension in T3.
Thank you Sir I tried to solve a problem in my high school text which was like this i tried to use Lami theoram for equilibrium but i failed to find the angles it helped alot Sir May God bless you
Interesting thing is that we have only one force pulling down, i.e. a cause of all tensions and this force is 4900N while all tensions caused by that force are bigger - both above 5000N
That is correct. If those angles approach zero degrees, the tension will approach infinity. That is why with hanging cables they don't try to stretch them too much or they will break.
Professer Biezen, Thank you so much for doing these static videos I am a Mechatronic Engineering student and I have been enjoying many of your videos to help me along. I just so happen to be taking statics this summer and this playlist is what i have needed. I am also taking Ordinary Differential equations I hope you get a chance to do this as well. Again thanks for the content.
Johnny Canazon Johnny, You are welcome. I started differential equations as well as statics. Now I just need time (still working more than full time besides doing these videos) Good luck with your studies this summer.
This is fantastic, this was poorly explained in my physics class and they kept the cos and sin instead of numbers but using numbers is far easier to learn
"Happy little acident..." Do you watch Bob Ross from "Joy of painting"? LOL. I (Mike van Biezen's wife) used to watch him on TV 25 years ago. Long before Bob Ross RUclips days.
This question came in our unit test. And now cbse ordered to make a mcq based question paper for half yearly😂🤩. And physics is so easy too. No. Of mcq directly propotional to my marks🤣🤣.
@@MichelvanBiezen thanx but tomorrow is my exam and i have 8 chapters to study. But i am still wasting time. Coz i never fail in any exam even if i study nothing. I have only failed once in my life in a maths exam😁. Maths is exception for me
@@MichelvanBiezen keep your mobile away from you for one day. Don't use it and wake up at 6 am. You will have a lot time. You will start doing things that will only benifit you. I tried it myself. I even started completing my homework bcoz i was getting bored of seating idle.
Ever try coating your Whiteboards with Renaissance Wax Polish ? Wheeo! Slick. The friction is noticably reduced! What is in this stuff. Had to Polish my swords to make room for math. But the friction between the towel and the metal blade changed dramatically so I though I'd try it on the Whiteboard and bingo gotta try the whole board it might smell a bit but what a difference.
This is an old video but since the results showed there are inbalance between the force, does that mean its not in equilibrium? Since it supposed to be at 0 however in the end we got like 6,012N T1 and 5,540N on T2? Plus the 4900N below
There is no net force on the object. (Otherwise there would be an acceleration --> F = ma) Therefore all the forces in the x-direction add up to zero and all the forces in the y-direction add up to zero.
Why do T3 and FG not cancel out? When you say that T3 is acting downwards when you are solving for T1 and T3, why is that? Why would that not nullify the tension upwards?
Pick the point where all 3 strings meet as the reference point. Since that point is stationary, all the forces acting on it in the x-direction and in the y-direction must add up to zero. (that is how the problem is solved as shown int the video). Your question can best be answered by drawing a free body diagram around the mass, which would then indicated that the net force on that block is indeed zero (T3 = mg) But that is not relevant to solving the problem.
This is a typical two equations for two unknowns type of problem. In the left equation we found T2 in terms of T1. This was then substituted into the second equation to eliminated T2 in the second equation so we could solve for the value of T1 (one equation with one unknown T1). Then we took the value we found for T2 and substituted it back into the first equation so we could solve for T2.
I'm just wondering about ty1 sin 30. Are we assuming that a right angle gets formed by the angle between t1 and tx2 above it so that we can imagine a 30/60/90 triangle with ty1 opposite?
The sign of vectors (and vector components) depend on the direction in which those vectors act, (not what region they are in). Since both T1y and T2y act upwards (relative to the point at which all 3 strings are connected), those components are positive.
Thank you Professor, I would like to ask about a part of a question I saw in an IG textbook, the question starts as "A smooth bead is threaded on a light inextensible string, the ends of the string are attached to the ceilling, the bead is acted on by a horizontal force F and the bead is in equilibrium..... ", the point is that the answer assumes that the tension on both sides of the string are equal, why is this true? Thanks
If the tension on the string was not the same on both ends, then there would be a net force acting on the string and the string would accelerate in the direction of the net force.
wires (or cables) made of metal will be rated with a maximum tensile strength expressed in force / cross secional area. Which means that your cable diameter must be sufficiently large to hold the load with margin. (The margin is usually a minimum of 300% meaning the you want a cable large enough to hold 3 times the inteded weight.)
Personally, I wouldn't have used the decimal form of the trig functions until the end where an answer is required or I'm solving for one of the variables.
Thank you so much for the person in this video.Im getting to know something new about Force hanging case.Hope for more such a great video in the future.
This is so interesting to me (now)... now that I'm trying to build a backyard zipline! I'm trying to figure out what the tension load is on the anchor points (to estimate deflection based on a post materials of choice). With AnchorA at the different height from AnchorB (to gain the 3% slope)... adding a sag in the cable to create a low point (and use gravity to slow down/brake the zipline)... the low point will NOT be in the middle. With the low point NOT in the middle, then the angles at the anchors will be different, so I think the equations you presented here would be applicable to finding the tension at each anchor points based on the low point (of the sag). Is my assumption correct? Can I apply these equations to my problem? Or does this only apply to a cable with anchor points that are level (as in your example)? If so... my other problem remains, which is how do I know where my low point will be (based on the heights on the anchors and the distance between the anchors)? Is there an equation to find that?
Michel van Biezen yes... That’s a good approach too. While I do have a tree at the starting anchor... I don’t have a tree at the opposite end, so I’ll need to install a post. This will require a sizable post hole, concrete and post itself (I imagine). And while I’d prefer to over build it anyway... I’m trying to at least find an objective reference point. I certain don’t want to under estimate the size of the post because redoing a post installation like that would almost be unbearable. Anyway... I also thought this is a fun problem where I could actually apply math and physics to solve.
I would recommend doing a few calculations with various amount of sag and with various resulting angles. You would be surprised how much tension you will have when the sag is reduced.
Please help me Calculate this problem A car traveling 20m/s passes a street corner. the car maintains its speed even though the speed limit is 10m/s. The police car sitting at the corner begins to chase the car by accelerating at 2m/s/s. How long will it take for the police car to chase the speeder? How far from the corner is the catch-up point? How fast would the police care be traveled at that time?
Assuming that the police begins the chase as soon as the car passes the police car. When the police car catches the speeding car, they will have traveled the same distance. ( d = v t) (20) t = (1/2) (2) (t^2) solve for t. Once you find t, d = (20) (t) then you find the velocity v = a t = (2) (t)
Dear sir I have been watching your videos and they have been very helpful. I just have one question. What if the rope is not weightless(Like reality)? How can we incorporate the weight of rope into this equation if we know the weight/m?
This guy is always saving my physics grades, because my teacher fails to explain anything. 👍
Glad you find our videos helpful.
Bless your soul you amazing man! I have a test tomorrow and i get everything you are doing, keep going.
you passed?
This is amazing. Thank you SO much. I could understand in 10 minutes what I couldn't for months in other physics classes. Teaching is a skill on its own.
You're very welcome!
Wanted to say I came here to say looking for help to solve a challenge problem for my physics class only until I realized half way through that it was a trick answer. Still stayed for the excellent lesson, you’ve definitely helped me on my path Mr. van Biezen. Many thanks.
Thank you. Glad you found our videos. We have thousands of videos on physisc and engineering.
I was the whiz when we did this at school (in the 60s) but I moved in to chemistry and not gone near since. Too many years later and m son needs help with his homework. French school as well so it's in French God help me.
You might just have saved my life with this. Very well explained. Thank you.
So you had your kid when you were around 50?
@@thefridge5844 I'm 66 now, he is 18.
You kind of sound like gru and it’s amazing. Thank you so much for this 🙏
More people have mentioned that. That you.
I swear to my teacher’s explanation was so complicated and I didn’t understand anything until I saw this vid and I have a test tomorrow thank u so much
from africa kenya thanks almost graduating come next year channel has helped alot
Keep it going! Glad you found our videos.
Some days ago i was ill and I couldn't come to school. I tried to study but didn't understand until I watched this video. Thank you so much for explaining. You saved my life
thank you ever so much for this video. even 7 years later, people like me still find this useful! May Allah bless your soul.
Thank you. Glad you found our videos! 🙂
Thank you so much , i find all your videos very helpful since I’m taking AP physics and the teacher in school tries to explain things fast since we have a lot of material to cover and I often find myself lost in her lesson and it’s hard for me to keep up with her , one day i found your videos and ever since that day i depend on them whenever i need help . :) thank you very much 💕✨
I haven't been able to understand this problems earlier but now everything is clear. It is only because of u. U r amazing. I also liked your bow
Michel Biezen, i enjoyed watching and learning from your tutorials. Your Engineering tutorials are very good, and i like them. You did an excellent jobs of explaining concepts in a very entertaining , clear and understandable way . Many thanks for your tutorials.
I missed this on my physics 1 exam. Looking back on it I have no idea how I didn't understand it. Its so easy now
you're a savior sir where i'm able to understand by your videos since my examination is incoming, wish me luck. 🥺
You just summarized my entire semester in 10 minutes. Thank you, Sir🥲.
Glad you found our videos,
Never studied mechanical engineering. But after watching your lessons, I feel that I missed a lot! So interesting! Tank you.
thank you so much. From South Africa. This sure will help for tomorrow's physics exam
fellow south african! i also have an exam tomorrow!!! ugh!
Sir, you might hear this a lot of times or probably none but you're awesome sir. You have saved my life lol I hope you're doing well! And thanks a lot for this video!
We remember that when we were students we were at times very desperate to find anything that could help us understand the material. That is a big part of the reason why we are making these videos to help students today that are in the same position. Comments like yours tell us it is working. We are very glad they are of help to students. Thank you for your comment.
I hope i found this earlier this is a great help you are amazing.....
If my prof is good as you i wont struggle understanding our lessons
Simply..u can use lami's theorem...
4900/sin(130)= T1/sin(110)=T2/sin(120)...
Solve this equation u can get T1& T2 value
it did not work. how?
Yes here you can use Lami's theorem, but you can't use it when you have an unknown force and angle that's why it is better to know how to resolve forces.
I really love your lectures they are simple and understandable
THANK YOU
You are welcome. Glad you found our videos.
Been watching your vids for a while, but just subscribed. I like the straightforward style you have. My only input would be based on how I like to do problems like this and that is to wait and obtain the final forms of the variables and equations and then do all the trig functions and other numbers and arithmetic. Great video however, I enjoy watching and seeing another style and explanation. :)
That is indeed the typical way in which one should solve physics problems (solve the equations first before substituting numbers), but there are a number of types of problems, where plugging in numbers earlier is advisable.
@@MichelvanBiezen :)
@@MichelvanBiezen I like the videos you've done where you take the moment of inertia of a pulley into account. I'd like to see one where you take the mass of the rope into account too.
great vid! im currently in my final year of highschool, and due to the coronavirus im very bored so im practising for physics in college. watching this makes me feel dumb and clever at the same time😂
best teacher i had i swear to jerry's future kids
You are The Boss , I am glad to watch your video ,you are amazing and i understood everything, even though My English is not good enough , thank you very much , made my day .
You sir are a genius. May god bless you!
I hope I know this channel when I'm still an undergraduate student. 😌😌😌.
A 100 lb weight is suspended from 2 cables. The tensions T1 and T2 satisfy the following
equations:
0.6T1+0.8T2=100 (1)
0.8T1−0.6T2=0 (2)
Solve for T1 and T2.
You can work this problem in 2 ways. Simply solve it algebraically (2 equations and 2 unknowns). Multiply the bottom equation by (-6/8) and then add the result to the top equation to eliminate T1 so you can solve for T2.
Excellent tutorial. You make it so simple Sir.
what a god. I am passing my exam bc of this
No god around here. But glad you found our videos and you found them helpful. 🙂
The lesson is pretty amazing I appreciate, For further knowledge, I'd like to know what's contributes to the large tensions in T1 and T2, whereas in normal situations, we would expect that summation of tensions in T1 and T2 will be equal to the tension in T3.
When the angles appoach zero degrees, the tensions will approach infinity. 🙂
Thank you Sir
I tried to solve a problem in my high school text which was like this i tried to use Lami theoram for equilibrium but i failed to find the angles it helped alot Sir
May God bless you
Great! Thank you. 🙂
Merci beaucoup !!! Vous m'avez vraiment aidé même si ce n'est pas ma langue ! C'est tout dire :) vous êtes très bon pour expliquer monsieur !
Merci bien
A very helpful video, i was just reviewing my class lecture and this video explained a lot. Thank you so much.
🥺🥺🥺🥺🥺this was so so so so helpful
Thank you very much
I just started my first year in university, and this didn't make sense until now
YESS!! Thank you verymuch!! I couldn't understand this from my sir but u easily explainedit !!!!! THANKS ALOT!
You are truly brilliant...
May God bless you, Sir. Your tutorial was really helpful.
Interesting thing is that we have only one force pulling down, i.e. a cause of all tensions and this force is 4900N while all tensions caused by that force are bigger - both above 5000N
That is correct. If those angles approach zero degrees, the tension will approach infinity. That is why with hanging cables they don't try to stretch them too much or they will break.
Dear respected sir, we can solve this problem conveniently by using Lami,s law as well
Yes, you can. Lami's Law is very interesting. We should create a video on it.
woww we need faculty like you
Thank you.
Professer Biezen, Thank you so much for doing these static videos I am a Mechatronic Engineering student and I have been enjoying many of your videos to help me along. I just so happen to be taking statics this summer and this playlist is what i have needed. I am also taking Ordinary Differential equations I hope you get a chance to do this as well. Again thanks for the content.
Johnny Canazon
Johnny,
You are welcome. I started differential equations as well as statics. Now I just need time (still working more than full time besides doing these videos) Good luck with your studies this summer.
Excellent better than the people lecuring and tutoring me
Thank you for your video! They are very useful first, and second you explain very well!
Thank you. Glad you find these helpful.
Thank you so much! After I saw some of your videos I got an A in Physic on the exam.
Excellent. Good job!
Other method
If all the forces are coplanar and concurrent and all angles are less than 180°, then Lami's theorem is applicable and its very simple.
God bless you and all your relative
This is fantastic, this was poorly explained in my physics class and they kept the cos and sin instead of numbers but using numbers is far easier to learn
thats an opinion, mine is the opposite, depends your method of learning
I have been watching all of your physics problems, very well done videos at a good pase. Much appreciated, thanks for your hard work and dedication.
Many thanks for your nice comment. We appreciate it.
thank you!!!! you saved my life
Glad it helped!
Oh thank you so much now I solved all my problems really thank you, keep going
Happy to help
thanks Prof! love the way u easify the question.. and we know giving knowledge I's just doubling your knowledge
sonic boom are you American ?
THIS EXACT QUESTION CAME OUT IN MY EXAM AND I AM ONLY FINDING OUT HOW TO DO THIS TODAY????? Y'ALL ARE TRIPPIN
Now you know where to find physics help.
Best tutorial ever thanks 🙏
Best explanation sir
Thanks and welcome
Your the goat michael
Glad you found our videos 🙂
(yes, 7 years later) I find it quite surprising that the individual tensions in the cables should greater than T3
As the angles become smaller, the tensions in the 2 cables will reach infinity.
@@MichelvanBiezen Oh right, I saw one of your other comments to that effect. Still ruffles my intuitions. Thank you for your answer!
thankyouuu!!! im now understand how to solve this type question!! Really worth to watch
Glad it helped
Very comprehensive instructions
At 3:54, he meant to say T1x in the x-direction, not y. Just a happy little accident. :]
"Happy little acident..." Do you watch Bob Ross from "Joy of painting"? LOL. I (Mike van Biezen's wife) used to watch him on TV 25 years ago. Long before Bob Ross RUclips days.
@@MichelvanBiezen Heh. I wish. Bob Ross was before my time.
good work it was hard but you made it very simple
Glad it helped
Good🎉❤ Physics Dr..
Glad you found our videos. 🙂
This came in very useful!
Glad it was helpful!
Excellent professor
So nice of you
Nice video, clears my confusion.
This question came in our unit test. And now cbse ordered to make a mcq based question paper for half yearly😂🤩. And physics is so easy too.
No. Of mcq directly propotional to my marks🤣🤣.
All the best
@@MichelvanBiezen thanx but tomorrow is my exam and i have 8 chapters to study. But i am still wasting time. Coz i never fail in any exam even if i study nothing. I have only failed once in my life in a maths exam😁. Maths is exception for me
I am a big time waster myself. I am trying to learn the discipline to make the best of my available time.
@@MichelvanBiezen keep your mobile away from you for one day. Don't use it and wake up at 6 am. You will have a lot time. You will start doing things that will only benifit you. I tried it myself. I even started completing my homework bcoz i was getting bored of seating idle.
I already get up at 5:30 AM every morning. :)
Ever try coating your Whiteboards with Renaissance Wax Polish ? Wheeo! Slick. The friction is noticably reduced! What is in this stuff. Had to Polish my swords to make room for math. But the friction between the towel and the metal blade changed dramatically so I though I'd try it on the Whiteboard and bingo gotta try the whole board it might smell a bit but what a difference.
Why is there a tension force equal to the the gravity force. Wouldn't that net in twice the force downward?
This is an old video but since the results showed there are inbalance between the force, does that mean its not in equilibrium? Since it supposed to be at 0 however in the end we got like 6,012N T1 and 5,540N on T2? Plus the 4900N below
There is no net force on the object. (Otherwise there would be an acceleration --> F = ma) Therefore all the forces in the x-direction add up to zero and all the forces in the y-direction add up to zero.
Thank you very much professor. Really amazing lesson.
Thank you so much for this explanation
Welcome 😊
thank you so much your videos are really appreciated
Why do T3 and FG not cancel out? When you say that T3 is acting downwards when you are solving for T1 and T3, why is that? Why would that not nullify the tension upwards?
Pick the point where all 3 strings meet as the reference point. Since that point is stationary, all the forces acting on it in the x-direction and in the y-direction must add up to zero. (that is how the problem is solved as shown int the video). Your question can best be answered by drawing a free body diagram around the mass, which would then indicated that the net force on that block is indeed zero (T3 = mg) But that is not relevant to solving the problem.
Really helpful advice
Glad it was helpful!
I dont get what he did in 8:56 can someone explain to me?
This is a typical two equations for two unknowns type of problem. In the left equation we found T2 in terms of T1. This was then substituted into the second equation to eliminated T2 in the second equation so we could solve for the value of T1 (one equation with one unknown T1). Then we took the value we found for T2 and substituted it back into the first equation so we could solve for T2.
I'm just wondering about ty1 sin 30. Are we assuming that a right angle gets formed by the angle between t1 and tx2 above it so that we can imagine a 30/60/90 triangle with ty1 opposite?
Sir, I would like to ask why your T1y is positive, T1y lies in the negative region. Thanks for the answer.
The sign of vectors (and vector components) depend on the direction in which those vectors act, (not what region they are in). Since both T1y and T2y act upwards (relative to the point at which all 3 strings are connected), those components are positive.
Thank you sir
Thank you Professor, I would like to ask about a part of a question I saw in an IG textbook, the question starts as "A smooth bead is threaded on a light inextensible string, the ends of the string are attached to the ceilling, the bead is acted on by a horizontal force F and the bead is in equilibrium..... ", the point is that the answer assumes that the tension on both sides of the string are equal, why is this true?
Thanks
If the tension on the string was not the same on both ends, then there would be a net force acting on the string and the string would accelerate in the direction of the net force.
How do we calculate the tension and wire diameter required to fix a vertical wobble of a bunk bed? Assuming load is 400kg and cross bracing.
wires (or cables) made of metal will be rated with a maximum tensile strength expressed in force / cross secional area. Which means that your cable diameter must be sufficiently large to hold the load with margin. (The margin is usually a minimum of 300% meaning the you want a cable large enough to hold 3 times the inteded weight.)
@@MichelvanBiezen Thank you vry much!
Personally, I wouldn't have used the decimal form of the trig functions until the end where an answer is required or I'm solving for one of the variables.
I like this explanation, it has help me a lot, how can we post questions for you for help?
We try to answer questions like this when time permits.
@@MichelvanBiezen so is there how we can post them to you?
Why not use Lami's theorum
I love u thanks :) you are a great teacher
Thanks so much sir, this stuff actually jumped in😁
Glad it helped!
Ok i have a question
How to find mass
weight = mass x acceleration due to gravity. w = mg
Thank you so much for the person in this video.Im getting to know something new about Force hanging case.Hope for more such a great video in the future.
This is so interesting to me (now)... now that I'm trying to build a backyard zipline! I'm trying to figure out what the tension load is on the anchor points (to estimate deflection based on a post materials of choice). With AnchorA at the different height from AnchorB (to gain the 3% slope)... adding a sag in the cable to create a low point (and use gravity to slow down/brake the zipline)... the low point will NOT be in the middle. With the low point NOT in the middle, then the angles at the anchors will be different, so I think the equations you presented here would be applicable to finding the tension at each anchor points based on the low point (of the sag). Is my assumption correct? Can I apply these equations to my problem? Or does this only apply to a cable with anchor points that are level (as in your example)? If so... my other problem remains, which is how do I know where my low point will be (based on the heights on the anchors and the distance between the anchors)? Is there an equation to find that?
When it comes to a backyard zipline, trial and error will work the best. The less sag you have to higher the tension in the cable.
Michel van Biezen yes... That’s a good approach too. While I do have a tree at the starting anchor... I don’t have a tree at the opposite end, so I’ll need to install a post. This will require a sizable post hole, concrete and post itself (I imagine). And while I’d prefer to over build it anyway... I’m trying to at least find an objective reference point. I certain don’t want to under estimate the size of the post because redoing a post installation like that would almost be unbearable.
Anyway... I also thought this is a fun problem where I could actually apply math and physics to solve.
I would recommend doing a few calculations with various amount of sag and with various resulting angles. You would be surprised how much tension you will have when the sag is reduced.
Please help me
Calculate this problem
A car traveling 20m/s passes a street corner. the car maintains its speed even though the speed limit is 10m/s. The police car sitting at the corner begins to chase the car by accelerating at 2m/s/s.
How long will it take for the police car to chase the speeder?
How far from the corner is the catch-up point?
How fast would the police care be traveled at that time?
Assuming that the police begins the chase as soon as the car passes the police car. When the police car catches the speeding car, they will have traveled the same distance. ( d = v t) (20) t = (1/2) (2) (t^2) solve for t. Once you find t, d = (20) (t) then you find the velocity v = a t = (2) (t)
@@MichelvanBiezen thanks
Thank you so much ... It's very useful but the board is hazy
Yes, we have been working on our lighting.
thank you from the bottom of my stupid heart
you're an angel
Dear sir I have been watching your videos and they have been very helpful.
I just have one question. What if the rope is not weightless(Like reality)? How can we incorporate the weight of rope into this equation if we know the weight/m?
Then you need to look at this playlist: MECHANICAL ENGINEERING 10: FORCES ON CABLES
how you get 6011N? what do you mean by ''Plus point 5' and inverse?????
4900 N / 0.8152 = 6011 N
Sir can you share how to add vectors by parallelogram method
Thanks. You’re a life saver
woah, very helpful i must say. Definitely subscribing
Great video, thanks man.
Thank you very much! Helped me out a lot.
you are a good man, i love you
Thanks!