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Posted: 10/4/2012 3:58:59 PM EDT
[#1]
This makes me glad I went to law school.
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Posted: 10/4/2012 4:00:45 PM EDT
[#2]
Frankly I don't see what this has to do with a biochem degree either.
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Posted: 10/4/2012 4:01:38 PM EDT
[#3]
a body at rest, stays at rest
a body in motion, stays in motion a body at rest is hit by a body in motion = someone got fucked up. |
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Posted: 10/4/2012 4:02:07 PM EDT
[#4]
Quoted:
Frankly I don't see what this has to do with a biochem degree either. It's all the same physics behind everything. |
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Posted: 10/4/2012 4:02:51 PM EDT
[#5]
What goes up must come down
E=mc2 Magnetohydrodynamic propulsion Moon walk |
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Posted: 10/4/2012 4:03:56 PM EDT
[#6]
It appears to be a rather simple F=mA problem.
Just figure out what the final velocity would be without friction. Then subtract the velocity with friction. Then calculate the force required to make-up that difference. |
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Posted: 10/4/2012 4:04:01 PM EDT
[#7]
I'm taking an intro to physics class this semester. The math is killing me. I have apps that do all this crap, why do I need to know how to do it?
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Posted: 10/4/2012 4:04:38 PM EDT
[#8]
Quoted:
It appears to be a rather simple F=mA problem. Just figure out what the final velocity would be without friction. Then subtract the velocity with friction. Then calculate the force required to make-up that difference. That's what I'm thinking. |
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Posted: 10/4/2012 4:04:44 PM EDT
[#9]
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Posted: 10/4/2012 4:05:30 PM EDT
[#10]
It looks like a simple enough problem, did you miss a class?
http://www.understandingforce.com/calculatingfriction.html Or do you have a specific question about the problem? |
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Posted: 10/4/2012 4:06:23 PM EDT
[#11]
Quoted: So this plane is on a treadmill... Kidding. Here's my homework for lab: One dimensional problem. F=-2N v(@t=0s)=1.5m/s m=2.2kg v(@t=5s)=-9.5m/s What is the force of friction? F=m*a. V= a*t Vinitial = 1.5 m/s Vfinal-Vinitial = Vnet Fnet = F - Fdrag. Think about it and if you are stumped, I'll be along to help you through it. I'm not getting the grade. Yes, you need to know dynamics because many processes in biochem are dynamic. |
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Posted: 10/4/2012 4:06:28 PM EDT
[#12]
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Posted: 10/4/2012 4:07:17 PM EDT
[#13]
ETA: Keith's letting you work from there is better than me giving it to you.
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Posted: 10/4/2012 4:07:37 PM EDT
[#14]
the apps might fail you one xay...xray i mean...learn it.
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Posted: 10/4/2012 4:07:41 PM EDT
[#15]
F = -2N
M = 2.2 Kg a = F/M = -2 / 2.2 = 0.91 Gs (ok, 0.909090909090909) V[0] = +1.5m/s V[t] = V[0] + a*t = (1.5 - 0.91*t) m/s V[5] = 1.5 -0.91*5 = -3.045 m/s now, if the mass were 2.2 pounds = 1Kg a = -2Gs V[5] = 1.5 - 2*5 = -8.5 With the given V[5] = -9.5 We know that the force of friction is negative––which is not sensible. |
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Posted: 10/4/2012 4:10:13 PM EDT
[#16]
GRAVITY
/THRODE |
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Posted: 10/4/2012 4:10:22 PM EDT
[#17]
Quoted:
Quoted:
So this plane is on a treadmill... Kidding. Here's my homework for lab: One dimensional problem. F=-2N v(@t=0s)=1.5m/s m=2.2kg v(@t=5s)=-9.5m/s What is the force of friction? F=m*a. V= a*t Vinitial = 1.5 m/s Vfinal-Vinitial = Vnet Fnet = F - Fdrag. Think about it and if you are stumped, I'll be along to help you through it. I'm not getting the grade. Yes, you need to know dynamics because many processes in biochem are dynamic. Here's my equations so far: assuming no friction: -2N=2.2kg(a) a=-.91m/s^2 with friction: -9.5m/s=1.5m/s + a(5s) a=-2.2m/s^2 with friction: F=2.2kg(-2.2m/s^2) F=-4.84N -4.84N-(-2N)=-2.84N friction force These fucking small details...give me a second. |
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Posted: 10/4/2012 4:11:28 PM EDT
[#18]
Quoted:
ETA: Keith's letting you work from there is better than me giving it to you. NOOO I was typing out the stuff I had so far. |
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Posted: 10/4/2012 4:12:47 PM EDT
[#19]
:D
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Posted: 10/4/2012 4:17:52 PM EDT
[#20]
Quoted:
So this plane is on a treadmill... Kidding. Here's my homework for lab: One dimensional problem. F=-2N v(@t=0s)=1.5m/s m=2.2kg v(@t=5s)=-9.5m/s What is the force of friction? Any chance this is the TTU physics quiz?? |
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Posted: 10/4/2012 4:19:42 PM EDT
[#21]
Quoted:
Quoted:
So this plane is on a treadmill... Kidding. Here's my homework for lab: One dimensional problem. F=-2N v(@t=0s)=1.5m/s m=2.2kg v(@t=5s)=-9.5m/s What is the force of friction? Any chance this is the TTU physics quiz?? Maybe... 
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Posted: 10/4/2012 4:24:30 PM EDT
[#22]
Quoted:
:D I had to recheck my work, I listed the information and still plugged in the wrong number. FML. |
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Posted: 10/4/2012 4:24:38 PM EDT
[#23]
Quoted:
Quoted:
So this plane is on a treadmill... Kidding. Here's my homework for lab: One dimensional problem. F=-2N v(@t=0s)=1.5m/s m=2.2kg v(@t=5s)=-9.5m/s What is the force of friction? F=m*a. V= a*t Vinitial = 1.5 m/s Vfinal-Vinitial = Vnet Fnet = F - Fdrag. Think about it and if you are stumped, I'll be along to help you through it. I'm not getting the grade. Yes, you need to know dynamics because many processes in biochem are dynamic. Yup.... Simple problem that can give a senior level engineering student fits b/c of lack of info (dynamics and 3D problems rattling around in their heads). 
ZA |
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Posted: 10/4/2012 4:27:28 PM EDT
[#24]
Tag.
KeithJ has it covered though. I wish Arfcom had an email alert that would send out a notice when it saw keywords...I always seem to be late to the party on physics threads. It makes me sad. |
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Posted: 10/4/2012 4:28:53 PM EDT
[#25]
Well, this is what I got:
assuming no friction: -2N=2.2kg(a) a=-.91m/s^2 with friction: -9.5m/s=1.5m/s + a(5s) a=-2.2m/s^2 with friction: F=2.2kg(-2.2m/s^2) F=-4.84N -4.84N-(-2N)=-2.84N friction force If anyone wants to confirm/reject my results. |
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Posted: 10/4/2012 4:42:45 PM EDT
[#26]
Quoted:
Quoted:
Quoted:
So this plane is on a treadmill... Kidding. Here's my homework for lab: One dimensional problem. F=-2N v(@t=0s)=1.5m/s m=2.2kg v(@t=5s)=-9.5m/s What is the force of friction? Any chance this is the TTU physics quiz?? Maybe... thought so. Haha, The key is that acceleration is not constant for the entire motion. say the right is the positive direction. you have a constant 2N to the left the entire time, however Vi=1.5 m/s ... so between v=1.5 and v=0, v is to the right and friction opposes velocity so friction force is to the left. sum(F)=ma ––> (2 + Friction)=(m)(a1) now from v=0 to v=-9.5, this means velocity is to the left and therefore friction force must be to the right .... so sum(F)=ma ––> (Friction - 2)=(m)(a2) the friction force has a constant magnitude, but changes direction @ v=0 due to the change in direction of velocity. Its almost like throwing a ball straight up in the air and having it fall, but sideways, and with friction. Hope that helps and im not just drunk |
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Posted: 10/4/2012 4:43:32 PM EDT
[#27]
Quoted:
Well, this is what I got: assuming no friction: -2N=2.2kg(a) a=-.91m/s^2 with friction: -9.5m/s=1.5m/s + a(5s) a=-2.2m/s^2 with friction: F=2.2kg(-2.2m/s^2) F=-4.84N -4.84N-(-2N)=-2.84N friction force Recheck your math here If anyone wants to confirm/reject my results. You're golden, I just goofed, edited to correct 
When summing forces, we do just that. We sum them, and attach signs to the individual forces. Fapplied + Ffriction = ma Fapplied = -2 -2 + Ffriction = (2.2kg)(-2.2m/s^2) Ffriction = -4.84N + 2N = -2.84N |
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Posted: 10/4/2012 4:44:40 PM EDT
[#28]
Something's wrong with the sign convention.
F=-2N v(@t=0s)=1.5m/s m=2.2kg v(@t=5s)=-9.5m/s What is the force of friction? If the force is negative then the velocity should increase in the negative direction, but it doesn't. This gives a negative value for coefficient of friction as someone else pointed out which makes no sense in the real world. Edit: damn, I missed the negative on the v(@t=5s)=-9.5m/s Gotta get a bigger screen or something.
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Posted: 10/4/2012 4:45:51 PM EDT
[#29]
AAHHHHHHH Physics, the class that converts engineering majors into business majors.
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Posted: 10/4/2012 4:47:04 PM EDT
[#30]
Quoted:
Quoted:
Well, this is what I got: assuming no friction: -2N=2.2kg(a) a=-.91m/s^2 with friction: -9.5m/s=1.5m/s + a(5s) a=-2.2m/s^2 with friction: F=2.2kg(-2.2m/s^2) F=-4.84N -4.84N-(-2N)=-2.84N friction force Recheck your math here If anyone wants to confirm/reject my results. You're golden, I just goofed, edited to correct
When summing forces, we do just that. We sum them, and attach signs to the individual forces. Fapplied + Ffriction = ma Fapplied = -2 -2 + Ffriction = (2.2kg)(-2.2m/s^2) Ffriction = -4.84N + 2N = -2.84N That makes sense. I know the object changes directions, but I didn't know how that played into the problem. |
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Posted: 10/4/2012 4:47:37 PM EDT
[#31]
Quoted:
AAHHHHHHH Physics, the class that converts engineering majors into business majors. What about biochem/chem majors? Looking at your SN. |
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Posted: 10/4/2012 4:47:56 PM EDT
[#32]
Quoted: Well, this is what I got: assuming no friction: -2N=2.2kg(a) a=-.91m/s^2 with friction: -9.5m/s=1.5m/s + a(5s) a=-2.2m/s^2 with friction: F=2.2kg(-2.2m/s^2) F=-4.84N -4.84N-(-2N)=-2.84N friction force If anyone wants to confirm/reject my results. Double check your sign convention. If the acceleration of a 2.2 kg mass with a 2 Newton force WITHOUT friction is 0.91 m/s2, the acceleration WITH friction will be LESS than 0.91 m/s2. The force is 2 Newtons CONSTANT. A NET final velocity to 7 m/s over 5 second period is 7/52 or 7/25 m/s2 So the drag force is equivalent to 0.63 m/s2 or for this mass, 1.386 N. |
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Posted: 10/4/2012 4:48:07 PM EDT
[#33]
Quoted:
Quoted:
Well, this is what I got: assuming no friction: -2N=2.2kg(a) a=-.91m/s^2 with friction: -9.5m/s=1.5m/s + a(5s) a=-2.2m/s^2 with friction: F=2.2kg(-2.2m/s^2) F=-4.84N -4.84N-(-2N)=-2.84N friction force Recheck your math here If anyone wants to confirm/reject my results. You're golden, up until the last part (in my working of the problem.) Have you drawn a FBD of the problem? If you consider positive x to be to the right, then a force of -2N is 2N directed to the left. Since our initial velocity is going to be positive (to the right), and friction force always opposes the direction of motion, your friction force will be acting to the left as well. When summing forces, we do just that. We sum them, and attach signs to the individual forces. Fapplied + Ffriction = ma Fapplied = -2 -2 + Ffriction = (2.2kg)(2.2m/s^2) Ffriction = 4.84N + 2N = 6.84N almost, but the acceleration is to the left so it to is negative. |
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Posted: 10/4/2012 4:50:57 PM EDT
[#34]
The answer is 87.
It is always 87. |
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Posted: 10/4/2012 4:51:06 PM EDT
[#35]
Quoted:
So this plane is on a treadmill... Kidding. Here's my homework for lab: One dimensional problem. F=-2N v(@t=0s)=1.5m/s m=2.2kg v(@t=5s)=-9.5m/s What is the force of friction? It's the force opposing F = -2N that results in a change in v from 1.5 m/s to -9.5 m/s over a period of 5 seconds. Draw a pitcher with the forces applied to the mass. The frictional force operates opposite the direction of motion, i.e., opposes motion. |
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Posted: 10/4/2012 4:53:08 PM EDT
[#36]
Also make a FBD before and after the direction change (with a velocity vector drawn), that will make it more clear that the sum of the forces is not the same before and after, therefore different magnitudes of acceleration.
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Posted: 10/4/2012 4:53:34 PM EDT
[#37]
Quoted:
Quoted:
Well, this is what I got: assuming no friction: -2N=2.2kg(a) a=-.91m/s^2 with friction: -9.5m/s=1.5m/s + a(5s) a=-2.2m/s^2 with friction: F=2.2kg(-2.2m/s^2) F=-4.84N -4.84N-(-2N)=-2.84N friction force If anyone wants to confirm/reject my results. Double check your sign convention. If the acceleration of a 2.2 kg mass with a 2 Newton force WITHOUT friction is 0.91 m/s2, the acceleration WITH friction will be LESS than 0.91 m/s2. The force is 2 Newtons CONSTANT. A NET final velocity to 7 m/s over 5 second period is 7/52 or 7/25 m/s2 So the drag force is equivalent to 0.63 m/s2 or for this mass, 1.386 N. The deltaV is 11 m/s, though. 1.5 to -9.5, and his initial acceleration without friction is -.91, not positive. With friction now included, your negative acceleration will be greater, since the object will slow down faster. |
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Posted: 10/4/2012 4:56:59 PM EDT
[#38]
Quoted:
That makes sense. I know the object changes directions, but I didn't know how that played into the problem. It's one of those problems that plays with you, simply BECAUSE analytically, you know it changed directions, and get confused about how to apply that. Acceleration, mass, applied force, and force of friction are all constant here, so an analysis at any given time will yield the same results. |
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Posted: 10/4/2012 4:57:03 PM EDT
[#39]
i feel smarter after reading all of that
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Posted: 10/4/2012 4:58:20 PM EDT
[#40]
Tired ME here who isn't interested in more math after work
The sum of all the forces acting on the object = it's mass * acceleration. You have a known force, a mass, and enough to determine the acceleration. That leaves you one equation for one unknown. Write it out then plug and chug. |
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Posted: 10/4/2012 5:02:31 PM EDT
[#41]
Quoted:
Quoted:
Quoted:
Well, this is what I got: assuming no friction: -2N=2.2kg(a) a=-.91m/s^2 with friction: -9.5m/s=1.5m/s + a(5s) a=-2.2m/s^2 with friction: F=2.2kg(-2.2m/s^2) F=-4.84N -4.84N-(-2N)=-2.84N friction force If anyone wants to confirm/reject my results. Double check your sign convention. If the acceleration of a 2.2 kg mass with a 2 Newton force WITHOUT friction is 0.91 m/s2, the acceleration WITH friction will be LESS than 0.91 m/s2. The force is 2 Newtons CONSTANT. A NET final velocity to 7 m/s over 5 second period is 7/52 or 7/25 m/s2 So the drag force is equivalent to 0.63 m/s2 or for this mass, 1.386 N. The deltaV is 11 m/s, though. 1.5 to -9.5, and his initial acceleration without friction is -.91, not positive. With friction now included, your negative acceleration will be greater, since the object will slow down faster. That makes sense to you too? Me too! You guys are making me doubt myself. Which is probably a good thing. What about: the friction force is -2.84N while v>0, and 2.84N when V<0? Because of the direction change. |
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Posted: 10/4/2012 5:04:09 PM EDT
[#42]
As a toolmaker/model maker/inventor/designer:
Emperical data rules especially when considering friction/force stuff. I build the shit, with an educated guess, try it out, rebuild it considering the results and it usually works pretty damn well in the end. Or, An engineer can spend a shitload of time, in front of his computer, have me build a model, and figure out how his equasions are different from/affected with reality and I'll build another model and it will eventually work. I do have a love and appreciation for math,physics, etc etc, though. |
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Posted: 10/4/2012 5:05:01 PM EDT
[#43]
Quoted:
That makes sense to you too? Me too! KeithJ normally makes me feel like mental midget, but in this case he just didn't see the negatives. Put your glasses on, Keith 
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Posted: 10/4/2012 5:05:20 PM EDT
[#44]
We were taught that the coefficient of friction was called mu. Yes Dynamics, I took it in 88'.
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Posted: 10/4/2012 5:05:49 PM EDT
[#45]
Quoted:
Quoted:
That makes sense to you too? Me too! KeithJ normally makes me feel like mental midget, but in this case he just didn't see the negatives. Put your glasses on, Keith The minus signs are pretty small, like the Koreans. |
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Posted: 10/4/2012 5:08:32 PM EDT
[#46]
Do you own damned homework.
Or at the very least, go ask your TA. |
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Posted: 10/4/2012 5:10:22 PM EDT
[#47]
Quoted:
Do you own damned homework. Or at the very least, go ask your TA. GD has an average processing speed of 8 gigaflops. |
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Posted: 10/4/2012 5:13:45 PM EDT
[#48]
Free body diagram
currently taking statics & dynamics for ME
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Posted: 10/4/2012 5:14:19 PM EDT
[#49]
Quoted:
We were taught that the coefficient of friction was called mu. Yes Dynamics, I took it in 88'. mu * normal force = force of friction. A 2.2kg object (not accelerating in the y direction) will have a normal force of 2.2kg*9.81m/s^2 mu = force of friction / normal force = 2.84 / 21.58 = 0.13 Not really relevant for the problem, since Ff is known, and mu is not asked for. |
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Posted: 10/4/2012 5:15:23 PM EDT
[#50]
I don't think mu is really important for this question here. But what do I know? I'm the one asking for help.
And normal force would be mg=2.2kg(9.8m/s^2) |
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