College Help

Finn McCool

Finn McCool

Captain
Registered Member
adeptitus said:
Irvine is in South Orange County, we're about 1-2 hrs away by car to either Downtown LA or San Diego, pending traffic. The city is also next door to Newport Beach, so we're not that far inland.

Irvine is a fairly nice master-planned community, except the idiots that live here voted down the mass transit/light rail project, which is killing us with heavy car traffic on the freeways right now. If anyone is planning to attend UC Irvine, I recommend living close to campus (within reasonable surface street driving distance).

This is a very nice & safe community where you could leave your garage door open or front door unlocked and not worry about people coming in to steal stuff. But it's OC suburbia and a bit "bland" compared to, say, San Francisco. Also, most restaurants in this city is "so-so", you'd find better dining places in Tustin, Costa Mesa, or Newport Beach.

UC Irvine is known to have the largest % of Asian student population of all UC schools in California. We used to joke and call it "University of Chinese Immigrants" (UCI).
Click to expand...

Hey adeptius I didn't know you lived in Irvine. I live in Newport. I go to water polo practice at the UCI pool twice a week.

sumdud said:
Thought it was near Finn's with Riverside.
Click to expand...

Hey. Let's not be saying I live ANYWHERE near RIVERSIDE. We Newporters tend to look down on those from THAT far inland. :rofl:

Anyway $100,000 dollar scholarship to Berkely is about as good as it gets. One of the seniors on my high school's water polo team got a free ride to Berkley to play polo.
 
bd popeye

bd popeye

The Last Jedi
VIP Professional
Anyway $100,000 dollar scholarship to Berkely is about as good as it gets. One of the seniors on my high school's water polo team got a free ride to Berkley to play polo.
Click to expand...

You better believe it..That girl is something else. since she was in the 6th grade she has recieved all A's and A+++..She did get a B in the 9th grade and she cried......I'm very proud of her.:)
 
A

adeptitus

Captain
VIP Professional
Finn McCool said:
Hey adeptius I didn't know you lived in Irvine. I live in Newport. I go to water polo practice at the UCI pool twice a week.

Hey. Let's not be saying I live ANYWHERE near RIVERSIDE. We Newporters tend to look down on those from THAT far inland. :rofl:
Click to expand...

OMG you... rich Newport bum. =P

I just moved back to Irvine in Dec 2006, still have about 30+ boxes of junk in my garage to sort out. I managed to flip out of a condo in Placentia, sold it to a... rich Newport guy who bought it for his daughter (she was attending CSUF).

Most of the restaurants in my area are "meh", I have to drive toward your city to get good food.

Do you know Cafe Blanc on 17th in Costa Mesa? It's off Newport & 17th. I go there for desert every week:
Please, Log in or Register to view URLs content!


I used to go fishing by "the rail" in Newport after desert & ice cream, but kinda slacked off this year:
Please, Log in or Register to view URLs content!
(not my web page)
 
sumdud

sumdud

Senior Member
VIP Professional
Newport?! Whatever happen to Ontario?

Uh-oh, mistaking a rich kid for an inlander, RUN!!!!!

And before I ditch, Congrats on her attendance for Berkeley, Popeye.

*running to China*
 
Chengdu J-10

Chengdu J-10

Junior Member
Question: A 8.00 kg hanging weight is connected by a string over a pulley to a 5.00 kg block that is sliding on a flat table (see figure). If the coefficient of kinetic friction is 0.162, find the tension in the string.

Round your answer to three significant figures. Take the free fall acceleration to be 9.8 .

My working out:
Uk=Fk/N
0.162=Fk/49
Fk=7.938N

(m1 x g) - 7.938=a(m1 + m2)
(8 x 9.8) - 7.938=a(5 x 8)
70.462=a x 40
a= 70.462/40
a=1.762m/s^2

T-7.938=(8 x 1.762)
T=14.096 + 7.938
T=22.034N My Answer for solving the Tension in the string

But this answer is apparently wrong and i cannot figure out why my answers is wrong? Any help will be good people, in achieving the correct answer. Need help in getting the answer. Cheers people.
 
K

kunmingren

Junior Member
Chengdu J-10 said:
Question: A 8.00 kg hanging weight is connected by a string over a pulley to a 5.00 kg block that is sliding on a flat table (see figure). If the coefficient of kinetic friction is 0.162, find the tension in the string.

Round your answer to three significant figures. Take the free fall acceleration to be 9.8 .

My working out:
Uk=Fk/N
0.162=Fk/49
Fk=7.938N

(m1 x g) - 7.938=a(m1 + m2)
(8 x 9.8) - 7.938=a(5 x 8)
70.462=a x 40
a= 70.462/40
a=1.762m/s^2

T-7.938=(8 x 1.762)
T=14.096 + 7.938
T=22.034N My Answer for solving the Tension in the string

But this answer is apparently wrong and i cannot figure out why my answers is wrong? Any help will be good people, in achieving the correct answer. Need help in getting the answer. Cheers people.
Click to expand...

the answer i got was 43.2N

first you analyze the force for the 5kg block, which i will refer to as M1, and the hanging mass will be M2, the the friction force acting on it is coeficient x Normal force, which equals to uM1g. The other force is tension, which i will call T. We know that the blocks will move at the same acceleartion becuz of Newton third law of motion, so we can come up with 2 equations:

F=ma:

((M2)g-T)/((M1)+(M2)) = a = (T-u(M1)g)/((M1)+(M2))

cancel out the bottom: (M2)g-T = T-u(M1)g

(M2)g+u(M1)g = 2T

(8)(9.8)+(0.162)(5)(9.8) = 2T
86.338=2T
43.2N=T

i think on ur second step u multiplied the mass instead of adding them
 
Chengdu J-10

Chengdu J-10

Junior Member
kunmingren said:
the answer i got was 43.2N

first you analyze the force for the 5kg block, which i will refer to as M1, and the hanging mass will be M2, the the friction force acting on it is coeficient x Normal force, which equals to uM1g. The other force is tension, which i will call T. We know that the blocks will move at the same acceleartion becuz of Newton third law of motion, so we can come up with 2 equations:

F=ma:

((M2)g-T)/((M1)+(M2)) = a = (T-u(M1)g)/((M1)+(M2))

cancel out the bottom: (M2)g-T = T-u(M1)g

(M2)g+u(M1)g = 2T

(8)(9.8)+(0.162)(5)(9.8) = 2T
86.338=2T
43.2N=T

i think on ur second step u multiplied the mass instead of adding them
Click to expand...
Thanks for your help in the question but the answer is still apparently wrong when I enter it into the computer. Appreciate the help and time though. But I still can't seem to solve this question.
 
szbd

szbd

Junior Member
ehhhhh, why I think this is a very easy problem? may be I am wrong though.

Let's see the two objects (M1=5kg and M2=8kg) seperated. 5kg is moving with an accelaration, caused by the Force in string: Fs, and the friction, Fr.

So you have:

aM1=Fs-uM1g

and simiar, the M2 is moving with the same accelaration by the forces of its weight and the force in the string, so

aM2=M2g-Fs

So sum the two equations,

a=(M2g-uM1g)/(M1+M2)

therefore

Fs==M1g(M2-uM1)/(M1+M2)+uM1g=35.04

You can put a and Fs back to the equations for a double check
 
szbd

szbd

Junior Member
kunmingren said:
the answer i got was 43.2N

first you analyze the force for the 5kg block, which i will refer to as M1, and the hanging mass will be M2, the the friction force acting on it is coeficient x Normal force, which equals to uM1g. The other force is tension, which i will call T. We know that the blocks will move at the same acceleartion becuz of Newton third law of motion, so we can come up with 2 equations:

F=ma:

((M2)g-T)/((M1)+(M2)) = a = (T-u(M1)g)/((M1)+(M2))

cancel out the bottom: (M2)g-T = T-u(M1)g

(M2)g+u(M1)g = 2T

(8)(9.8)+(0.162)(5)(9.8) = 2T
86.338=2T
43.2N=T

i think on ur second step u multiplied the mass instead of adding them
Click to expand...

It's M1a or M2a, not (M1+M2)a for the whole. If you look at the force of (M1+M2)a, then it's

(M1+M2)a=(M2)g-u(M1)g

When you look at an object, you don't consider the "internal force", T is the internal force of the object (M1+M2)
 
Chengdu J-10

Chengdu J-10

Junior Member
szbd said:
It's M1a or M2a, not (M1+M2)a for the whole. If you look at the force of (M1+M2)a, then it's

(M1+M2)a=(M2)g-u(M1)g

When you look at an object, you don't consider the "internal force", T is the internal force of the object (M1+M2)
Click to expand...
I re-attempted myself with the correction you made and yes I get the same answer as you 35.04 when rounded to 3 significant numbers its 35.0. Entering this into the computer the answer came up as correct. Very much appreciate szbd & also kunmingren. Thankyou all people for your help.
 
You must log in or register to reply here.
Top