- be sure to include the number of the problem,
- State the problem the unknowns and knowns. Create a link to a picture.
- This will provide outside of the classroom to provide help and allow others including myself to plan and prepare discussions for class. (differentiated instruction)
Asteroid and Space Tug
Ex.
- 2 objects are tied together an pull on one another in deep space.
- They both accelerate toward one another
- but because they are different masses they accelerate at different rates
- where do the meet? what is their final position?
- How long does it take for this to occur
Application of Idea
Solution:
Bulleted list of steps for solving or some other representation or link for explaining the solution.
#58- "If the coefficient of static friction and the ground is .90, determine the largest angle that the crutch can have just before it begins to slip on the floor."
ReplyDeleteUnknown- maximum angle
Knowns- M=.9, g=(-)9.8m/s/s
Equation- FsMAX= (M)(Fn) I know that fsMAX depends on the applied force, and I am thinking that the angle would come from a component of Fn. I will try to link my picture... would Ff and Fa be in the (opposite) x directions?
This is how I solved 58.
ReplyDeleteI used the FUN equation and it ended being the inverse tan of (.9).
Fmax=uN
fN=fcosx
Fmax=fsinx
F(sinx)=u(F(cosx))
https://mail-attachment.googleusercontent.com/attachment/u/0/?ui=2&ik=11ff5a14b8&view=att&th=13b5d081878bad50&attid=0.1&disp=inline&safe=1&zw&saduie=AG9B_P-dc3YrHHVC1hpNFnOMKV3a&sadet=1354475542467&sads=tOdwfjWdpZC9Pap_wj1pyhAoIqk&sadssc=1
does anyone know what situation our FBD should be of, like a car or gravitron?
ReplyDeletejust kidding! i got it!
ReplyDeleteMONKEY PROBLEM: I was absent on Friday, so I might not have the right location for the final velocity.... but I assume it is as the monkey passes thru the lowest point of the swing after one of the ropes has been cut? At this point, there would be no motion in the y direction. Thus, EFx=Tx. For Tx, I got 86.6 N and I set that equal to "ma." Again, I might not even have the right location...
ReplyDeleteMONKEY: So, I tried using centripetal force to solve this problem, but it turns out that won't work because that was only assuming acceleration was zero, but it actually changes, and in that case we would be dealing with jerks.
ReplyDeleteMr. Crane's method of solving it involved using the formula vf squared = vo squared + 2 ax. He drew a picture along with it, but in the end I think we got a velocity of around 16 m/s.
ReplyDeleteFor the Monkey problem with the final velocity I got my Vf to be 7.24 by using the way he showed me in class...( Integrate ma from 30 to 0 which equals integral of mgsinx from 60 to 0. Cancel an m and integrate I got a= (g)cos(x) evaluated at 0 and 60 which equals 5m/s^2)
ReplyDeleteAnd also for my d I got 5.24 by using (2*pi*r)(30/360)=5.24m. Plug it all into Vf^2=Vo^2+2ad and i got 7.24 m/s.
ReplyDeleteMonkey problem: When we first tried to do it, we assumed that the acceleration was zero and that you used ma=mv^2/r - mg. Mr. Crane then explained that we can't assume acceleration is zero. And since we didn't know time, we had to use vf^2=vo^2+2ad. And initial velocity is zero so the problem just becomes vf^2=2ad, where acceleration is gravity. So we got about 16.9 m/s as the final velocity.
ReplyDeleteOnce you get the velocity in part c, how would you find the tension in vine B for part d? Are the only forces acting on the monkey at this point the force of tension and the force of gravity, and would the sum of the forces be equal to zero ...I didn't think so because the monkey was accelerating while swinging...
ReplyDeleteFor part d the sum of the forces=Fc=T-(mg). Rearrange and you get Fc +(mg)=T. Fc is v^2/r(=5.24m/s^2) and the you add 50N to that to get the Tension My final answer for this problem turned out to be 55.24N.
ReplyDeleteBut again I also got a diffrent velocity in the previous part so your answer is probably much larger than mine.
ReplyDeleteI'm confused how to find the distance for the final velocity part. Do we need to assume its a semi circle and not just a straight line? Also, since from what Dan did it seems to be a circle, how do we determine how far the vine swings?
ReplyDeleteCleo, for the question where it asked for the tension when the monkey is at rest, I assumed the sum of the forces was 0 and was able to get that the tension was 85.81 N
ReplyDeleteI solved using two ways for the Vf
ReplyDeleteThe long way:
http://tinypic.com/view.php?pic=orib0h&s=6
the short(er) way:
ReplyDeletebut I dont understand why we would use g for the acceleration so I solved using g as my acceleration and the other component for the movement of the monkey. Thus, a total of three answers since I didn't know what I was supposed to plug in. So I plugged in everything.
http://tinypic.com/r/v5e8a1/6
@Caity, if you can, look at the pictures in the first link I made. The question asks for velocity at the lowest point on the swing. Therefore the vine would be perpendicular to a (theoretical) surface and let you see a right triangle there the vine swings 30 degrees. From there, use the arc length of a circle equation to find the distance traveled.
ReplyDeleteI used the Force of Tension in rope B from the last question to solve it, which I found to be 43 newtons
ReplyDeleteSo for B:
P: a money swinging on one rope. Sort of like a half circle.
U: vf=?
K: angle B: 60*, angle A: 30*, distance of B: 10m, distance of A: 10m, Tb: 43 N
E: ∑Fx: Tbcos60=ma=0
substitution makes it--> ∑fx: 43n(cos60*)=5a
21.5=5a
a=4.3, the acceleration of the monkey
Since its the lowest point I used 10sin60 to find the Y value. I think this would be easier to explain with a picture. Anyway, I made half the half circle the monkey travels into a right triangle with 10m the hypotenuse. 10sin60 is the side straight down from the vine.
10sin60=8.7
-I used this formula for the next few steps: Vf^2=2ad
Vf^2=2(4.3)8.7
The Vf ended up being 8.66, but I know that's the right answer. Something along the way of my process is off....
Thanks everyone! So next question...where is the homework/take home problem Mr. Crane was talking about in class? I searched but can't find it.
ReplyDeleteIt doesn't exist yet... At least I hope so because I couldn't find it either.
ReplyDeleteSorry guys thought I posted. The picture you need is on twitter.
ReplyDeleteNot sure if we're allowed to talk about the picture and how to solve it but could anyone do/find anything besides the FBDs and Fgs? I may be missing something but it didn't seem like we had enough information..
ReplyDeleteFor the monkey problem: Did it matter what type of monkey?
ReplyDeleteThat last comment was not from me.
ReplyDeleteI'm struggling too. I did the sum of the forces in the x and y for block A, and in the y i had it equal zero. I'm unsure though of how to get the Fgx and Fgy without an angle.
ReplyDeleteDoes anyone know which 2 tensions are "both tensions"?
ReplyDeleteI think we need to know the angle of inclination to be able to solve for the tensions (at least 1 of them anyway...). And the tensions being referred to are between a and b and between b and c
ReplyDeleteI was able to get the tension for B/C but I can't seem the get the others without an angle... I would check the textbook to see if there is a formula that doesn't need the angle.
DeleteATTENTION apparently the angle is 30 degrees
ReplyDeleteon who's authority?
Deletemr cranes......
DeleteI'm still confused as to what tensions we're finding. Like if we drew the FBD for all three objects, all three of them have tension. So are there only two tensions because the ones on A and B are the same?
ReplyDeleteyeah both objects are being pulled with the same force.
DeleteThanks!
DeleteOh never mind. I just realize that was a dumb question.
ReplyDeleteAnyone remember what we needed to do for the Pigs lab?
ReplyDelete