Dynamics - Lesson 13: Additional Relative Motion Problem
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- Опубліковано 8 вер 2024
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Goood video! Thanks! One note for viewers: on the Ti-36X pro you can press " 2nd num-solv " Enter in " sin(x)=2/3 + 4/3cos(x) " and press enter a couple times. It will solve x = 76.71, if it is set to degree mode. :)
For the TI-84 Plus CE, press math and then its the last option in the list (Numeric Solver).
For me it was important that the 2/3 was in parentheses, otherwise i got syntax errors.
Thank you so much Jeff! Also, you can solve that trig equation at the end by substituting sqrt(1-cos^2) for sin, then square both sides to obtain a quadratic equation which you can graph and solve
You could make use of the pythagorean identity sin^2(x) + cos^2(x) = 1 for all x to solve this. Square the first Vm equation to get sin^2(theta) on one side, square the second to get cos^2(theta) on one side, then add the equations, set it equal to one and solve
hi another suggestion is square both sides, you get sin squared = 4/9 +16/9cos^2 + 16/9cos. then use the trig idenitity of sin^2 = 1 - cos^2... then you have all the only cos... which will be 0 = -5/9 + 2.77cos^2 + 16/9cos... using the quadratic formula you get a value of 0.23 & -0.87.... the use cos^-1(0.23) to get 76.702 degrees
Hi, thank you for your solution. I think you added an extra term in the beginning 16/9. After squaring both sides, you should have sin^2= 4/9 + 16/9cos^2.
@@sagebird759 Nope, you'll get sin^2@=(4/9)+(16/9)cos^2@+(16/9)cos@.
Hi everyone I fixed it... I will notify Jeff
Ty for this
That’s correct
DR. Hanson, thank you for another incredible lecture that analyze Additional Relative Motion. This is tricky problem for students in dynamics.
Awesome videos Jeff!!! These videos have helped me out so much in my Statics class. Taking Dynamics over the summer so getting a start on it during spring break!
P.S. Do you know anyone who puts out Thermodynamics type videos like yours?
Thanks again and keep them comeing!!!
Derek LaFosse I got some dynamics problems on my channel :)
UA-camr Baahir Jinadu has a series on Thermodynamics I. Quite helpful.
awesome video...i watch your video before i go to class and guest what,you teach better than my lecturer...please keep making video about it...i like the way you teach us...my dynamics might be RIP if i don't watch your video since i got a really bad lecturer in teaching me and my friends
This course is amazing! Please upload more videos, they will definitely become as popular as your statics course.
3:20 How is it possible to use the Pythagorean theorem on a triangle that is not a right triangle?
My understanding of how he drew the problem is, it's a right angle triangle, he just made it slightly skewed to emphasize more of a 3D view of this otherwise 2D problem.
you can also easily solve for theta by using similar angles and shift solve in any calculator 90+(90-theta)+theta=180
You can get the angle by using sin law. Arrange the vectors to form a triangle.
Professor Jeff, why can't the man just swim in the y-direction at a constant velocity of 2.67 ft/s? That way he would be in the water for 15 seconds, the 2 ft/s current of the river will push him 30 feet horizontally and 2.67 ft/s * 15 seconds = 40 feet. Therefore he ends up at point B, 30 feet in the x direction and 40 feet in the y direction. In that case, he would be swimming at theta = 90°, or along the y-axis as defined in your solution.
Edit: I guess I might've missed the fact that the man MUST swim 4 ft/s in the water, I think this question could be reworded for more clarity because I assumed the man was able to control his speed and just had to swim UNDER 4 ft/s. This problem makes more sense now, I will leave my comment here in case anyone else has a similar question lol
Dear Mr. Handsome
I would like to thank you for helping me pass 😁
I don't understand how the x-component of the relative velocity of the man w.r.t. the river is 4cos(theta). Aren't we supposed to take the Vr = 2i into account? The question states that the man can swim @4 ft/s in still water, which means we perceive his velocity as 4 ft/s when our velocity is 0. This is not the case for the river flowing in the x-direction. Could someone please show me where I am mistaken?
in relation to the river the man will always move at 4ft/s - if he goes straight up he travels 4ft per second in relation to a particle of water. he moves at an angle relative to the ground but the water doesnt care. if he swims downstream he moves 4 ft/s second relative to the particle of water. from the outside he will appear faster, but once again - the water doesnt care. from the water's perspective the man swims in still water but the earth around it is moving at -2ft/s. Hope that helps
@@MoX914 Thanks! beautifully explained btw :D
It would've been easier if you drew a vector diagram. Everything is getting clear with that. What you get is a triangle between vm, vr and vm/r. From that one can deduce the angle at which the swimmer has to take to get from A to B and with the law of cosines one can calculate the speed.
Btw: you made a mistake with the equations cause if you write a vector sum at one side of the equality sign then the other side should also be a vector.
You don't need guess and check sir! you square both sides and use Pythagorean trig identity
easiest way i found theta is to use graphing calculator (I have ti-84 plus) hit y= button and plug in for y1=2+4cos(x)-3sin(x). Hit graph and then use trace to see where it crosses on the graph on the first positive x value, get as close as you can to that value and just use the tblset and table to narrow in to closest value to zero. keep using tblstart to start at the closest you see on the table and then use (delta)tbl at .01 and then .001 to hone in on the closest point to zero. I was able to figure out theta=76.708 degrees in about 60 seconds using this method.
I just figured out an even easier way to find the exact zero (using ti-84 plus), hit y= and plug in 2+4cos(x)-3sin(x) and then use the 2nd calc button, hit enter on zero (which is number 2 on the list) then select a point above the x axis (for first positive x crossing of line) and then hit enter, then select just below the x axis (where line crosses x axis) hit enter and hit enter again and it goes right to the zero.
We only allow FE approved calculators on the exam so sadly this method wouldn’t work in class. But I love the creativity and this is definitely the method you would use in the real world. Good job!
Hello Professor Jeff, can't the problem be represented in a vector triangle and using the cosine rule, you can solve for theta and the velocity? That is what I did and I got the same answer
Cant u just use the trianguler laws and substetude cos(x) with some kind of a formula of sin(x) and make it solvable without guessing ?
for example cos(x)= square/root ( (sin(x))^2 - 1)
Thanks for the videos i really appreciate it a lot better than my teachers
I really enjoy your videos, however I believe you got this one wrong. The time it would take would be longer than 10 seconds due to the fact that the distance he has to swim is now longer than 40. I'm not sure what the answer is, this answer just doesn't make any sense to me. If I am wrong can someone please correct me.
also using trig identity: sin^2+cos^2= 1 => (.6vm-2/4)^2+(.8vm/4)^2 = 1 to solve for vm - then vm=5sin(theta) for theta
Well then vm according to you equals 5.38 m/s and theta cannot be calculated cause you get sin(theta)=vm/5=1.075.
TIME ......??!!
When will you complete this course
this question easliy solvable with guess check at first sight cause the swimmer needs to swim just 10 m to the left to reach to the point he wants to so you can go for 10 sqare+ 40 square root hypothemeus there for your nuber and you can divide those to to find the tangent and therefor the angle
my english might bleed your eyes. Caution!
Would the speed of the river add to the speed of the man? I added 2ft/s to the x-component of the 4ft/s of the man because hes swimming at 77 degrees and got that he could cross the river in 8.5s.
Why can't he just swim at 8/3 ft/s in a straight line. He will need 15 seconds to get across, so the current will take him 30ft downstream, exactly to point B.
you can solve this by using tangent function to , cause tangent = sin / cos
Yeah but then you'll get a quadratic equation in sec@ in stead of cos@.
I noticed he never solves this problem, what is the speed of the man and what is the time of the crossing?
Is the answer 5sin(76.1)=4.866ft/s and then 50ft/4.866ft/s = 10.28 seconds?
@@thenatmann1 the time is just ten seconds because technically hes still swimming straight across for 40 ft. you can think of it as him swimming through still water, and the ground moving at 2 m/s, it doesnt change his speed. for the velocity you just plug in the angle to the Vm/r equation and then solve for his velocity from there. so the 4.866 ft/s is his speed moving down the river, not along the hypoteneuse. i might be wrong so dont quote me on it but I think that's how it works.
@@TheM.F. if he was to swim straight across for 40ft he would end up at x=20ft though. Either you use the total speed and use the 50ft hypoteneuse or you calculate the distance he needs to travel at 76.71° (40/cos(12.3)) => 41.101ft and divide that by his 4ft/s speed. Either way you end up at a time of 10.275s
@@MoX914 I think it’s a question clarity thing. My understanding, since it is relative motion in the title, means that you omit the speed that the river adds on to him. Like I mentioned before, you can imagine the man swimming in a straight line while the ground moves along.
It’s like if I lined up a bunch of treadmills and told you to run across them. You would have a velocity going forward, but the treadmills would be shooting you off to the side.
You can make a triangle of three things,
Side one being his pure speed as if it was still water,
Side two being the speed of the water,
the hypotenuse being his combined speed.
I’m not sure but I think the aim of this exercise is to find his pure speed, not the one of the hypotenuse, like I think you are saying.
Mind you I took this course last year and don’t really remember exactly how I did this question. Anyways. Hope this helps
This is problem 12-230 in Engineering Mechanics: Statics & Dynamics 14th edition
the answer in the book and on slader show V_m = 4.87 ft/s and t = 10.3 s
Can someone help me explain how did he derived the Vm=3/5Vm + 4/5Vm..I don't know to get it...thanks
I might be wrong but the triangle made by the river is essentially a 3,4,5 triangle. using that, the 3/5Vm gives you his X direction speed, and the 4/5Vm gives you the Y
Afuuftuu Jaarsaa
Buddy better get some swimming lessons because 5sin(76.71)=4.87ft/s which he can't even do in still water.
Jeff Hanson, will you be my father?