Statics: Lesson 16 - Equilibrium of a Particle, 2D Forces Around a Pulley

[@hallfamily2141]

Can't thank you enough for this videos. I've done several forms of online learning before the shutdown and I have to say this is by far one of the best series for self learning. I am a BSME and I am stuck working full time and going to doing engineering courses full time this semester so I've been using this month to get as far ahead as possible to reduce my study time. Highly entertaining stuff. You remind me of my honors intro to engr professor at Purdue. Thanks again for being a such a great and free source of material during these trying times.

[@domyniclamphere8973]

System-solver coming in clutch? nah you're coming in clutch!! thank you for these videos

[@gaberillman6778]

Thank you so much for doing this problem specifically, Jeff! My Statics Professor not only chose this problem for our chapter 3 particle equilibrium quiz, but he also included a problem just like this on our first exam.... for some reason I have a feeling this is the problem he will want to include for our final exam lol

[@georgesadler7830]

Professor Hanson, thank you for another solid explanation of Equilibrium of a Particle. Solving problem continues to be the key to understanding mechanics.

[@lilbeing133]

OK teacher, you won, I love mechanics engineering now. WOW MAGICAL, thank you so much deeply, really!!

[@helenday5031]

So helpful, thank you. It filled in a few gaps in my knowledge & thus helped to clarify questions that were puzzling me before. For example, the way the video explains how tension is the same on both sides of a pulley. All delivered with a great, individual and likeable style.

[@pavlypoktor3486]

why is the angle in BC and DE the same? it seems from a visual view DE would have a higher degree.

[@twinkle_pie]

Coz their positions are different, note that they are not on the same point , so when you shift the DE on BC, now their tails are coinciding and they have the same angle

[@billhensley9288]

Really great videos. I’m also an engineering student, however I got lost when Jeff summed the forces to come up with 1154.7 Newtons. Can someone provide some guidance on this?

[@LucasLeow]

He solved the simultaneous equations using a calculator. You can make either "T" or "F" the subject and solve the equation. From Summation of Fx Equation: -0.221T + 0.707F = 0 0.707F = 0.221T (0.707F / 0.707) = (0.221T / 0.707) F = 0.3125T From Summation of Fy Equation: 1.078T + 0.707F = 1500 But F = 0.3125T from above 1.078T + 0.707(0.3125T) = 1500 1.302T = 1500 T = 1152N (Some deviation from his answer due to the rounding) Hope this clarifies it for you.

[@ghostinfr5749]

@@LucasLeowthanks bro

[@babydoyoung96]

if you have calculator 570ms,you can use eqn for solution 😊

[@PulseandPower]

Jeff Hanson I love you

[@carlosnataren5334]

How does Jeff summed the forces to come up to 360.95 N? Can someone show me the steps, I appreciate it.

[@clearflow7925]

he used matrices on a scientific calculator. not sure which models have it but there should be youtube videos to show you how to input equations in matrices form. really cool

[@clearflow7925]

there might be another method on the calculator but i do know matrices

[@YsabellaCastillo]

@jeffhanson Hello! I just wanted to thank you for these videos, they are a life saver. However, I was wanting to know when you use your system solver feature on your calculator, how do you know x is the cable T(abc) and y is the Force (de) in this problem? Since they both have components from both the force and tension?

[@milascerenhoke8689]

Hi, love your work! But why did we assume the Tension Force around the pulley acts on the center, same with the Force PE? Shouldn't we at least mention that our solution is in equilibrium under this condition? Or is Tabc really acts on point D? Can you clarify please? Thank you.

[@jkgan4952]

my thoughts exactly

[@jackline8808]

I was also wondering the same thing.. it doesn't look like they have the same line of action

[@batulh5459]

You're amazing omg

[@smithyyow6348]

I dont get the last part can you explain it how is it equal to 1154.7 and 360.95???

[@deranged4610]

2 Equations 2 unknowns Using the Calculator (I'm using the old model fx-570MS)

[@samuelotono3367]

Thank you very much for these videos, but one question though, why are you assuming the forces are in 2-D space? In real life is this applicable?

[@farhadjoya7882]

There are two T. One is T shorter and one T is longer. My question is which TBC or TBA has more tension? Why the then is equal in TBC and TBA ?

[@mohammedalkhliwi445]

How did you come up with the final answer, not sure how to do it w/ the calculator

[@leahfisher2451]

On the TI-36X Pro, you want to use sys-solv. It's (2nd) --> (tan). Then you just enter your equations!

[@xxcamicakexx4727]

How can i finish the problem at 9:41 without a system solver? My calculator doesn't have one.

[@zunaidhassan9662]

It's just simultaneous equations. You make one variable of one equation the subject and substitute it in the other equation to get the value of the second variable. Put that value in the first variable and that's it. But, you should consider getting a scientific calculator if you are studying engineering.

[@batulh5459]

Thankkkkkkkkkkk youuuuuuu

[@craig7878]

Hi Jeff, can you do a video on a similar problem to this? Can I email you the problem?

[@syeduzair7798]

Sir, doesn't the weight lifted downward multiplied by 9.81 as mentioned in the book hibbeler?

[@dajjelightblade]

When you measure in Newtons, gravity is already accounted for because Newtons are a unit of weight. If the object was given a value in kg, that is a value of mass and you would need to multiply by gravity to then get Newtons.

[@irvingpolanco]

Phil 4:13 I can do all things through Christ who strengthens me AMEN!!

[@abdulahadhussain4548]

Plz explain why dod we take both the tension of ropes T like why did we assume that both parts of the rope have same force T?

[@QualityOQ]

We assume that they have the same force because the question does not take into account friction. If friction was to be in play, then there would be 2 different force measurments for each side.

[@abdulahadhussain4548]

@@QualityOQ oh ok thanks 👍👍

[@dontworry1568]

What is a system solver? How do I figure this out on paper?

[@nathan8472]

he has a video just type his name and calculator// its worth the time in my opinion

[@ashlynoconnor5876]

I think on paper, you use one formula to solve for T or F and then input that in the other formula. This will give you one variable which you can solve out, and then use that to solve for the other variable.

[@fiona.e1026]

Idk why the force p has the same angle as the tension can someone explains to me

[@alwynekersonvillamil2851]

force p is assumed to be parallel with the tension T, thus, making their angles congruent

[@tyyamnitz8408]

what was the system solver used? was it just an app that could solve for 2 variables?

[@1234jhanson]

TI 36xPro has a system solver. See my videos on how to use this calculator.

[@TixocLoza]

If you have TI 84 and allowed to use on tests. You can use the Matrix option, setup 2x2 and 2x1 then A inverse B and you get the same answer.

[@jonathanlorens2312]

Or you can just set up a 2x3 matrix and just RREF since the solutions is trivial

[@StressedGrinch]

Hey does anyone know why he used tan instead of sin or cos for tan^-1(1/2.5)

[@gannondrums2741]

Soh cah Toa identifies that opposite over adjacent would be what 1 (the opposite) divided by 2.5 (the adjacent) is relative to the theta being used. Inverse sin or cosine would work, you would need to use the hypotenuse correctly, which solving for the hypotenuse is just another, needless step in order to calculate theta. If you’re confused about what the adj, opp, and hyp sides are, I advise looking up a diagram for clarity. The opposite side is opposite to theta, the adjacent side is the side along theta that is perpendicular to the opposite side, and hypotenuse is the longest in length and the one that is at a diagonal. Theta doesn’t have to be the same in every situation on the triangle, which is why you also need to assess the right triangle and where theta is, relative to the different sides. Making the 90° angle theta is pointless (as far as I know) since it’s known as 90°

[@franshiscadelgado1383]

Wait how do we use solve on the calculator? I got lost at that point. I have a TI-84 Plus. Any suggestions. Is there another option to solve this without using the solver?

[@balajihariharan7175]

just 2 eqn with 2 variable, solve using substitution method

[@gabrielbaradel8969]

or you can use inverse matrix

[@kininnat3189]

hello pls everyone here explained me why he got 45 degree between the tension(T) and downward weight of 1500N?

[@ikennao5299]

Same, why did he get both 45degrees between tension F and T

[@benjaminamankwaah4893]

He used tan theta = opposite/adjacent. Then tan inverse is 45

[@jiajunsun3845]

when you have a right triangle with two sides with the same length, the angle is always 45 degrees

[@catarinastoever7463]

Why does he assume that both Tensions are the same?

[@juliandinoshiho3678]

Can’t we solve them simultaneously?

[@zunaidhassan9662]

Yes.

[@afifazhar6058]

lets make it 100k subscribers

[@destinyegbuliese1851]

why is this course too hard .......... UNIBEN why