Dynamics - Lesson 13: Additional Relative Motion Problem

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[zippySquirrelface]

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. :)

[jonathandespeaux]

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

[georgesadler]

DR. Hanson, thank you for another incredible lecture that analyze Additional Relative Motion. This is tricky problem for students in dynamics.

[heavyymetal]

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

[dereklafosse]

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!!!

[Rowtara]

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

[aidahusna]

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

[taylorocker]

This course is amazing! Please upload more videos, they will definitely become as popular as your statics course.

[HASANBAQIRY]

3:20 How is it possible to use the Pythagorean theorem on a triangle that is not a right triangle?

[reneumeh]

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

[montanerpauljohnv.]

You can get the angle by using sin law. Arrange the vectors to form a triangle.

[kadenielsen]

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

[lestergaela]

you can also easily solve for theta by using similar angles and shift solve in any calculator 90+(90-theta)+theta=180

[akshatverma]

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?

[dirtmoto]

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.

[sohaibal-hadheri]

Dear Mr. Handsome
I would like to thank you for helping me pass 😁

[georgipendurkov]

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)

[ernestschoenmakers]

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.

[timdevos]

You can solve it without guessing using the R-method (linear combination of sin and cos):

R sin( θ + α ) = A sin(θ) + B cos(θ)

R = √(A²+B²)

tan(α) = B/A
or
α = arctan( B/A )


in this problem A = 3, and B = -4

Solving for R and α, and substituting:

5 sin( θ + (-53.13°) ) = 2
sin( θ - 53.13° ) = 2/5
θ - 53.13° = arcsin( 2/5 )
θ = arcsin( 2/5 ) + 53.13°
θ = 76.71°

arctan and arcsin are the same as tan^(-1) and sin^(-1), it's just a different notation.

[thenatmann]

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.