Mechanical Engineering: Particle Equilibrium (7 of 19) Tension of Cables Attached to Hanging Object

[@thierneylionneveit7545]

This guy is always saving my physics grades, because my teacher fails to explain anything. 👍

[@MichelvanBiezen]

Glad you find our videos helpful.

[@vivianmamboleo7538]

Truly a gift to all those of us who leave physics class falling on our faces.

[@MichelvanBiezen]

Hahahaha! We all been there. Glad we can help.

[@ankitghosal2839]

We learn this concepts in class 11 before engineering proud to be indian

[@mogambo9517]

@@ankitghosal2839 of course I am also indian. Our education system is one of the best education system in the 🌎

[@kartiknagrale1479]

@@ankitghosal2839 🍆

[@kartiknagrale1479]

@@mogambo9517 🍆

[@mvp5332]

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.

[@MichelvanBiezen]

You're very welcome!

[@egytian100]

Bless your soul you amazing man! I have a test tomorrow and i get everything you are doing, keep going.

[@jeffsonhillaire5705]

you passed?

[@alitr-tf1dc]

@@jeffsonhillaire5705 prolly no

[@ulysses11]

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.

[@MichelvanBiezen]

Thank you. Glad you found our videos. We have thousands of videos on physisc and engineering.

[@mustafaeltigani1203]

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

[@Amityz72323]

You kind of sound like gru and it’s amazing. Thank you so much for this 🙏

[@MichelvanBiezen]

More people have mentioned that. That you.

[@PhilJonesIII]

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.

[@thefridge5844]

So you had your kid when you were around 50?

[@PhilJonesIII]

@@thefridge5844 I'm 66 now, he is 18.

[@laylaali8890]

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 💕✨

[@madnad4284]

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

[@carvelbell181]

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.

[@Iseeyou0165]

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!

[@MichelvanBiezen]

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.

[@slugthetree2956]

thank you ever so much for this video. even 7 years later, people like me still find this useful! May Allah bless your soul.

[@MichelvanBiezen]

Thank you. Glad you found our videos! 🙂

[@arunavamajumder6823]

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

[@LW62_Gaming]

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

[@kahlesidelo9432]

thank you so much. From South Africa. This sure will help for tomorrow's physics exam

[@melissafleming7592]

fellow south african! i also have an exam tomorrow!!! ugh!

[@orodiopportunities4719]

from africa kenya thanks almost graduating come next year channel has helped alot

[@MichelvanBiezen]

Keep it going! Glad you found our videos.

[@mamont33]

Never studied mechanical engineering. But after watching your lessons, I feel that I missed a lot! So interesting! Tank you.

[@noheartcatchannel9063]

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

[@hsy_123]

you're a savior sir where i'm able to understand by your videos since my examination is incoming, wish me luck. 🥺

[@fizixx]

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. :)

[@MichelvanBiezen]

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.

[@fizixx]

@@MichelvanBiezen :)

[@fizixx]

@@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.

[@ElHal04]

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😂

[@oOr3x3rOo]

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 .

[@kkome1388]

I really love your lectures they are simple and understandable THANK YOU

[@MichelvanBiezen]

You are welcome. Glad you found our videos.

[@kidsnorlaxluvr]

best teacher i had i swear to jerry's future kids

[@sadacutely]

I dont get what he did in 8:56 can someone explain to me?

[@MichelvanBiezen]

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.

[@yosiefberhe4974]

A very helpful video, i was just reviewing my class lecture and this video explained a lot. Thank you so much.

[@AugustineJBenson]

You just summarized my entire semester in 10 minutes. Thank you, Sir🥲.

[@MichelvanBiezen]

Glad you found our videos,

[@carolinelapointe3627]

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 !

[@MichelvanBiezen]

Merci bien

[@DAT-OFFICIAL]

At 3:54, he meant to say T1x in the x-direction, not y. Just a happy little accident. :]

[@MichelvanBiezen]

"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 KZbin days.

[@DAT-OFFICIAL]

@@MichelvanBiezen Heh. I wish. Bob Ross was before my time.

[@rehemachiona]

🥺🥺🥺🥺🥺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

[@johnnycanazon]

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.

[@MichelvanBiezen]

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.

[@sherigeorge5874]

YESS!! Thank you verymuch!! I couldn't understand this from my sir but u easily explainedit !!!!! THANKS ALOT!

[@vijay-dw1bl]

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

[@allanmendez4866]

it did not work. how?

[@tasneemayham974]

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.

[@nurmalas4021]

I hope I know this channel when I'm still an undergraduate student. 😌😌😌.

[@ЙОРИКОНСУЛТЕООД]

Thank you for your video! They are very useful first, and second you explain very well!

[@MichelvanBiezen]

Thank you. Glad you find these helpful.

[@wendimaginhabtemariambogal-q6l]

God bless bless my good teacher 🙏

[@MichelvanBiezen]

Thank you and God bless you as well.

[@cowgate]

You sir are a genius. May god bless you!

[@franzduckridge5069]

May God bless you, Sir. Your tutorial was really helpful.

[@running730]

8:22 - 10:00 I get confused Someone please explain! How does he get 6011 if he is just dividing .815/4900

[@MichelvanBiezen]

6011 = 4900 / 0.815

[@hadishandai1182]

It is 4900/0.8152

[@neiladventurestotheotherdi4052]

How did u get .8152 sir?

[@saikatroy4121]

Excellent tutorial. You make it so simple Sir.

[@Urawizardharryy]

why didn't you consider the force of gravity pulling it downward when adding all the forces in the y direction, and doesn't the tension in the 3rd cable supposed to point upward opposite to mg in order for the object to be at equilibrium?

[@MichelvanBiezen]

We did consider the force of gravity pulling downward. (F = mg)

[@MyDavidsen]

Thank you so much! After I saw some of your videos I got an A in Physic on the exam.

[@MichelvanBiezen]

Excellent. Good job!

[@kennsnthumbi6947]

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.

[@MichelvanBiezen]

When the angles appoach zero degrees, the tensions will approach infinity. 🙂

[@oscarww3]

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.

[@MichelvanBiezen]

Many thanks for your nice comment. We appreciate it.

[@mantakamahir]

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

[@MichelvanBiezen]

Great! Thank you. 🙂

[@imagine_laabs]

thanks Prof! love the way u easify the question.. and we know giving knowledge I's just doubling your knowledge

[@rt-viz954]

sonic boom are you American ?

[@allam8842]

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.

[@jimmatrix7244]

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.

[@MichelvanBiezen]

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.)

[@jimmatrix7244]

@@MichelvanBiezen Thank you vry much!

[@zheko7897]

Oh thank you so much now I solved all my problems really thank you, keep going

[@MichelvanBiezen]

Happy to help

[@harmantiwana6529]

You are truly brilliant...

[@luxill0s]

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?

[@MichelvanBiezen]

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.

[@arlieferguson3990]

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?

[@Reversely]

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

[@MichelvanBiezen]

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.

[@Tech-Aesthetic00]

thank you!!!! you saved my life

[@MichelvanBiezen]

Glad it helped!

[@edgaryepez3239]

what a god. I am passing my exam bc of this

[@MichelvanBiezen]

No god around here. But glad you found our videos and you found them helpful. 🙂

[@bradybenedejcic4577]

Your the goat michael

[@MichelvanBiezen]

Glad you found our videos 🙂

[@bobu5213]

Why is there a tension force equal to the the gravity force. Wouldn't that net in twice the force downward?

[@snursyafiqa6529]

thankyouuu!!! im now understand how to solve this type question!! Really worth to watch

[@MichelvanBiezen]

Glad it helped

[@kingsleyyusufu4920]

Best tutorial ever thanks 🙏

[@rachealkapembwa2791]

Excellent better than the people lecuring and tutoring me

[@ainz9387]

woww we need faculty like you

[@MichelvanBiezen]

Thank you.

[@hatdognasigang3041]

What if the mass is the one missing? And in that way we cant solve the T3. or for some instance lets say we have like 40138.86N but we have a missing angle?

[@MichelvanBiezen]

If the mass is missing, then you cannot determine the tensions in the cables. Sometimes you are given the tension and you have to find the angle. You would solve it in exactly the same way, by setting up the equations the exact same way. The only difference is that you have to solve for the angle instead of the tension.

[@hatdognasigang3041]

How?

[@hatdognasigang3041]

Can you give the formula?

[@MichelvanBiezen]

It is the exact same formula. You will be solving for a different variable in the same formula.

[@ALPHONSEKara-g2l]

Good🎉❤ Physics Dr..

[@MichelvanBiezen]

Glad you found our videos. 🙂

[@omarabu-khalaf6043]

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

[@MichelvanBiezen]

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.

[@littlereptilian7580]

I cant understand how an hanging object that weights 5000kg puts force on two points that fairly over its weight. For example if this object was hanging from a single rope then the point where its hanging will have the exact same force. If i would put two ropes i thought i would divide the force in two so the hanging could we weaker to save money. Its like saying with no matter how many ropes you connect to a hanging object the force will be more? Im confused

[@MichelvanBiezen]

It is because of the angle at which the ropes are stretched. It is impossible to pull the ropes with a large enough force so that the ropes would be horizontal (without any sag). That is why the power cables have a considerable amount of sag, (even though it takes more cable length to do so), otherwise the tension in the cables would be too high and they would break.

[@littlereptilian7580]

@@MichelvanBiezen kzbin.info/www/bejne/h4fFpnqGesmBe9E for instance in this video of yours with the pulleys. The object weights 100N, each rope gets 50N. If the ropes had an angle on this video would the force from the ropes exceed 100N?

[@littlereptilian7580]

Now i get it. To get the forces on the Y axis i just need to multiply by sinθ. And then when i add them up the mass of the object comes out. Thank you very much

[@MichelvanBiezen]

Yes, if the ropes on the pulleys were pulled at an angle it would require additional force.

[@mikerotch_tx]

This came in very useful!

[@MichelvanBiezen]

Glad it was helpful!

[@muhdsyamirul6689]

how you get 6011N? what do you mean by ''Plus point 5' and inverse?????

[@MichelvanBiezen]

4900 N / 0.8152 = 6011 N

[@KillU99]

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

[@CarlosSanchezMM]

thats an opinion, mine is the opposite, depends your method of learning

[@taffychiwozva8325]

Thanks so much sir, this stuff actually jumped in😁

[@MichelvanBiezen]

Glad it helped!

[@gatlatwal8499]

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?

[@MichelvanBiezen]

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)

[@gatlatwal8499]

@@MichelvanBiezen thanks

[@danielcastaneda8254]

woah, very helpful i must say. Definitely subscribing

[@AshuKumar-ow8sg]

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.

[@MichelvanBiezen]

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.

[@NEET-tt6sy]

Best explanation sir

[@MichelvanBiezen]

Thanks and welcome

[@Penghuoth]

hello sir can you explain me why when we add the results of T1 and T2 back to 0.9397T2 - 0.8660T1=0 and we can not get zero??????

[@MichelvanBiezen]

Since T1, T2, and T3 are vectors, they must be added like vectors (add all the x components together and all the y components together) If you do this correctly they will add up to zero.

[@Penghuoth]

@@MichelvanBiezen thank you very much teacher

[@sandeepv3232]

Sir can you explain equation relating to rope drum and load. Our project had faced issue since we supplied crane where it function on top of hill due to gravity parameters we couldn't find the exact load we carry out by crane. It runs on 25hp motor, which could lift 10ton , but at 5 ton itself the gears rotated opposite direction since the load seems heavier due gravity pull near Nepal hydro power generation project. It suppose to drill and lift deep to 78mts into mountain but at 9mts itself issue faced

[@lindanireymcing9014]

good work it was hard but you made it very simple

[@MichelvanBiezen]

Glad it helped

[@alanjoy7564]

thank you so much your videos are really appreciated

[@preciousegbunah731]

Thank you very much professor. Really amazing lesson.

[@JayPatel-im4mq]

Nice video, clears my confusion.

[@lorrainegalaura6340]

Thanks for your videos!!!! 🎉🎉🎉👍

[@claudehermineakette9222]

God bless you and all your relative

[@mormar8608]

I love u thanks :) you are a great teacher

[@ChaldoEman]

i dont understand what you plugged into the calculator that gave you the answer for t1 towards the end! please help!!

[@MichelvanBiezen]

T1 = 4900N / 0.8152

[@dreamhousejewels158]

Sir can you share how to add vectors by parallelogram method

[@yareddejene9705]

Ok i have a question How to find mass

[@MichelvanBiezen]

weight = mass x acceleration due to gravity. w = mg

[@CamViesky]

Can someone please explain to me why it is that the force of tension in both of the wires are more than the force of the actual weight itself?

[@MichelvanBiezen]

When you add up the vertical components of the tension, they will add up to the weight of the object. The horizontal components of the tension cannot help in supporting the weight of the object, but they cannot be zero since they pull the object from a complete vertical hanging position.

[@CamViesky]

@@MichelvanBiezen Oh wow! Thank you so much. I understand it now.

[@sabrinaabraham6527]

Can someone explain why for T1 & T2x COS was used and for T/ & T2y Sin was used?

[@dbbyleo]

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?

[@MichelvanBiezen]

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.

[@dbbyleo]

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.

[@MichelvanBiezen]

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.

[@josephshaff5194]

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.

[@JAMESANDREWBAUTISTA]

Sir, I would like to ask why your T1y is positive, T1y lies in the negative region. Thanks for the answer.

[@MichelvanBiezen]

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.

[@jamesandrewbautista4485]

Thank you sir

[@nickthemememan3453]

you are a good man, i love you

[@MichelvanBiezen]

Thanks!

[@anirudh5320]

Proof that I need more brain cells

[@alemkiru2012]

Thank you so much for this video!! It helped a lot.

[@benjaminwise9058]

Great video, thanks man.

[@shpresamuca2911]

Excellent professor

[@MichelvanBiezen]

So nice of you

[@leannetan5119]

THIS EXACT QUESTION CAME OUT IN MY EXAM AND I AM ONLY FINDING OUT HOW TO DO THIS TODAY????? Y'ALL ARE TRIPPIN

[@MichelvanBiezen]

Now you know where to find physics help.

[@worldhasmenot7928]

Hello. Can you advise how much tension would be needed to pull a steel wire or any capable wire so that there is no droop, over 7 miles?

[@MichelvanBiezen]

There will always be a droop. Look at the hanging cable videos in ME.

[@worldhasmenot7928]

Michel van Biezen darn. We are trying to conduct an experiment to detect the curve of the earth. Visual tests have possible excuses (light refraction, gravity bending light) so we need a physical test. the idea is to have two boats drive away from each other to 7 miles and have a wire or cable pull completely tight with no droop. According to pythagorean mathematics there should be a hump of water between the boats due to the earth curvature, so if the cable is pulled completely tight and on each boat the cable is about 3 feet above the sea level, if at the middle point between the boats (3.5 miles from either) the cable does not touch the water then the earth is not a sphere or it is larger than we have been told.

[@MichelvanBiezen]

You should be able to conduct such an experiment using a laser.

[@worldhasmenot7928]

Michel van Biezen the problem with a laser is that refraction and gravity can be used to explain a redirection of visible light, but with a physical cable (if it can be droopless) cannot be explained away with vision.

[@worldhasmenot7928]

For example...if a laser should not be visible because it should be hitting the bump of water between the laser and the observer, It could just be said the laser light is bouncing off the water or the air so the observer sees it. Or if the laser should be 10 feet high at a target but it instead is only 1 foot high, it can just be argued that the gravity or refraction directs the light downward. So a visible test is not concrete because of these possible reasons

[@colloupdated]

Why not use Lami's theorum

[@harrytolley7686]

Thank you very much! Helped me out a lot.