Physics 10-19-11 Equilibrium Posted on October 19, 2011 by Ms. Skinner PHYSICS: So how is the balance in your life? 🙂 Here’s the lecture from Thursday on net force, equilibrium, and equilibrants. Were you experiencing a little déjà vu? Let’s explore that a bit more tomorrow, and then ACT prep on Friday. Image source phoenix.fanster.com/…/2009/08/tug-o-war1.jpg Share this:TwitterGooglePrintFacebookPinterestMoreTumblrRedditLinkedInPocket

Ms. Skinner Concerning the demonstrations in class today, would there be an angle at which there is no downward force, thus meaning that one would be unable to hold himself up; perhaps horizontally? Reply

Will someone please do the equilibrium lab with me? Ms.Skinner said someone needs come to help me with the force table. Let me know who can help. Reply

Jackson! Courtney, David, and I need to finish doing that lab… we hardly started. Come do it with us! Reply

Is the mass of the woodpecker’s head on that second worksheet just a red herring? Our group couldn’t find where it factored into the problem. Reply

I was wondering the same thing, because I worked the problem without using the mass of the head anywhere in the problem and still got the correct answer. Reply

When doing these equilibrant force problems does it matter whether we convert kg to n at the beginning or the end of the problem? Will our rounding make our answers different Reply

On the newest worksheet, does the length of the wires factor into the equation maybe to give one twice the resultant? Reply

I figured out zorba after messing with if for about 3 hours. The right angle would be at the the top of the ceiling and the two cables are come down forming the legs of the triangle from there(just like this symbol: ^ ) Since one cable is twice the tension of the other, it is a 1 2 sqrt 5 triangle. So by using tangent, u can determine that the two angles are 27 degrees and 63 degrees [arctan(2) and arctan(1/2)]. Then just plug in the angles and the f2 = 2f1 equation into the equation we had in class and you get one of the resultants. Then you can plug in that into the f2=2f1 and get the other resultant. That’s how I did it. I don’t know if there is an easier way, but this one seemed to work for me. I didn’t get 1.36×1.36×10^3 though. I think that may be a typo. I got 680 and 1360 though which seems very close Reply

Well, Cory, you can explain this to me tomorrow sometime… I am still way confused about this one… Reply

You can try a right angle like we did in class but then the cos and sin are switched. Idk if you’ll get the same answer. Like I said this is how I did it. There may be other ways of doing it. Reply

Ms. Skinner

Concerning the demonstrations in class today, would there be an angle at which there is no downward force, thus meaning that one would be unable to hold himself up; perhaps horizontally?

Do we have SITN this week? I didn’t see it on the syllabus.

We don’t have SITN because of the ACT. She’s giving us a break.

so equilibrium is a state of stillness even though forces are acting upon an object.

i am pretty sure we do not!

Will someone please do the equilibrium lab with me? Ms.Skinner said someone needs come to help me with the force table. Let me know who can help.

Jackson! Courtney, David, and I need to finish doing that lab… we hardly started. Come do it with us!

Megan – Thanks! 🙂

Is the mass of the woodpecker’s head on that second worksheet just a red herring? Our group couldn’t find where it factored into the problem.

I was wondering the same thing, because I worked the problem without using the mass of the head anywhere in the problem and still got the correct answer.

Could we go over both homeworks tomorrow in class?

Charlie – We are!

When doing these equilibrant force problems does it matter whether we convert kg to n at the beginning or the end of the problem? Will our rounding make our answers different

Cory – 1) just convert when/where ever you need it. 2) yep!

On the newest worksheet, does the length of the wires factor into the equation maybe to give one twice the resultant?

Is the balloon take home lab due tomorrow?

I figured out zorba after messing with if for about 3 hours. The right angle would be at the the top of the ceiling and the two cables are come down forming the legs of the triangle from there(just like this symbol: ^ ) Since one cable is twice the tension of the other, it is a 1 2 sqrt 5 triangle. So by using tangent, u can determine that the two angles are 27 degrees and 63 degrees [arctan(2) and arctan(1/2)]. Then just plug in the angles and the f2 = 2f1 equation into the equation we had in class and you get one of the resultants. Then you can plug in that into the f2=2f1 and get the other resultant. That’s how I did it. I don’t know if there is an easier way, but this one seemed to work for me. I didn’t get 1.36×1.36×10^3 though. I think that may be a typo. I got 680 and 1360 though which seems very close

Well, Cory, you can explain this to me tomorrow sometime… I am still way confused about this one…

how do you know that it’s a right triangle hanging from the ceiling?

how do you know that it’s a right triangle hanging from the ceiling? Good job though

You can try a right angle like we did in class but then the cos and sin are switched. Idk if you’ll get the same answer. Like I said this is how I did it. There may be other ways of doing it.