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Old 10-18-2004, 09:51 PM   #1 (permalink)
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Physics, it moves me (help me please)

to tears sometimes

I'm supposed to find the acceleration of each block. The acceleration should be the same for each block because of the direction they are moving and the fact that they are attached with a string. My answer is off by just a little bit, so I know I'm on the right track. Who knows, maybe it's another algebra mistake. All relevant information is in the scan.

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Old 10-18-2004, 10:27 PM   #2 (permalink)
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Oh man...if only I could remember my cosine/sine/tangent stuff. Sheesh; this used to be hella easy for me.
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Old 10-19-2004, 07:12 AM   #3 (permalink)
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thanks anyway
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Old 10-19-2004, 08:21 AM   #4 (permalink)
Oh shit it's Wayne Brady!
 
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Location: Passenger seat of Wayne Brady's car.
Ok this is just me breaking it down for myself. Please add on anything I left missing:
Force = mass * acceleration
Kinetic Friction = Coefficient of Kinetic Friction * Normal Force
Coefficient of Kinetic Friction = .25
I'm gonna read up on my sine/cosine etc and try to do this on paper.

More breaking it down:
For the 7kg block, there are four different forces on it:
Normal Force pushing up and left,
Tension1 Pulling up and right,
Weight*sin37+Kinetic Friction pulling down and left, and
Weight*cos37 pulling down and right.
*Now I assume you use 9.8 m/s/s for the acceleration of gravity? If not, then substitute whatever your class uses instead of 9.8 m/s/s.*
Weight (mg) = 68.6 Newtons
sin37 = .601815023
>>>68.6 * .601815023 = 41.28451059
>>>Force pulling left and down = 41.28451059 Newtons
cos37 = .79863551
>>>68.6 * .79863551 = 54.78639599
>>>Force pulling right and down = 54.78639599 Newtons
Due to the equal and opposite reactions, Normal force is also = 54.78639599 Newtons.
So, for the 7kg block, we have the following equation for the total force (assuming that the Normal force and the force pulling down and right cancel each other out): F = 41.28451059 - Tension1

For the 12 kg block, there are only two forces:
Tension2 Pulling up,
and weight (mg) pulling down.
mg = 117.6 Newtons
So this gives us the following equation: F = Tension2 - 117.6

Now this is where I get fuzzy...We have to find both Tension1 and Tension2. So is the tension equal to the pulling force from each block? PLEASE correct me if I'm wrong. But if so, then Tension1 = 117.6 Newtons, and Tension2 = 41.28451059.

So for the 7kg block, we have:
F = 41.28451059 - 117.6
>>> F = -76.31548941
>>> F = ma
>>> a = -76.31548941 / 7
>>> a = -10.90221277

For the 12kg block, we have:
F = 41.28451059 - 117.6
>>> F = -76.31548941
>>> F = ma
>>> a = -76.31548941 / 12
>>> a = -6.359624118

Alright...so what did I do wrong?
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The words "love" and "life" go together. It is almost as if they are one. You must love to live, and you must live to love, or you have never lived nor loved at all.

Quote:
Originally Posted by Zeraph
...the best way to keep a big secret would be to make it public with disinformation...

Last edited by CityOfAngels; 10-19-2004 at 09:01 AM..
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Old 10-19-2004, 09:44 AM   #5 (permalink)
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Location: Singapore
dieseldog, if you have been mulling over this for a long time and getting frustrated, you are going to kick yourself.

You wrote:

mgsin37 + Fk = 41.28 + 13.70 = 44.98

I think you accidentally left out the "1" when pressing 13.70....

You will get the answer when you correct this mistake.

CityofAngel: Tension 1 and Tension 2 are equal. They are equal and opposite force acting on the same piece of string.
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Old 10-20-2004, 02:34 AM   #6 (permalink)
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Location: Sydney, Australia
The best way to do this class of problems is to ignore the concept of tension all together: don't find "T1" and "T2".

Instead transfer the forces to a free body diagram on one block, the force the string applies will be the sum of the forces on the other block but the direction will be in line with the string (obviously).

As they are connected and the string is under tension they will have the same numerical value of acceleration.

Edit: Thanks by the way, I have a dynamics test on this kind of thing on friday and this has put a few concepts straight in my head.

Last edited by molloby; 10-20-2004 at 02:36 AM..
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Old 10-20-2004, 06:21 AM   #7 (permalink)
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Location: Poland, Ohio // Clarion University of PA.
[QUOTE=molloby]The best way to do this class of problems is to ignore the concept of tension all together: don't find "T1" and "T2".

Instead transfer the forces to a free body diagram on one block, the force the string applies will be the sum of the forces on the other block but the direction will be in line with the string (obviously).
QUOTE]

I'd advise doing this, in my physics class we had the EXACT same project as you and had to figure out the
exact same things, forces, acceleration, etc, and free-body diagrams are the best things ever: saves alot of
space and clutter, and making them as large as human possible helps too.

Also, not to say what you're doing is bad, but, I find it also helps to do VERY NEAT problems in physics,
especially problems where you need pictures, etc...
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