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Small block on a track - Finding Normal forces and....

 
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Nov2-11, 01:52 PM   #1
 
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Small block on a track - Finding Normal forces and....


1. The problem statement, all variables and given/known data

Here is the question. Can someone please help, since I missed the section of class that went over this, any my textbook does not cover this well. Look below at question 1, a) and b)





2. Relevant equations



3. The attempt at a solution
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Nov2-11, 01:57 PM   #2
 
Which are the forces at point Q?
Nov2-11, 01:59 PM   #3
 
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At point Q - Forces are Gravity down... Does it have tangential and radial?
Nov2-11, 02:03 PM   #4
 

Small block on a track - Finding Normal forces and....


The block is touching the track. Hence besides the weight there is also the ...
Nov2-11, 02:04 PM   #5
 
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Quote by grzz View Post
The block is touching the track. Hence besides the weight there is also the ...
Force of gravity and the normal force
Nov2-11, 02:05 PM   #6
 
Correct.
Now for a mass to perform circular motion, it needs a ...
Nov2-11, 02:06 PM   #7
 
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Quote by grzz View Post
Correct.
Now for a mass to perform circular motion, it needs a ...
Force?
Nov2-11, 02:09 PM   #8
 
What do we call this particular force?
Nov2-11, 02:10 PM   #9
 
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I am not sure. Momentum force?
Nov2-11, 02:12 PM   #10
 
Any mass performing circular motion needs a force towards the centre of the circle. This particular force is called ...
Nov2-11, 02:15 PM   #11
 
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Centripetal Force?
Nov2-11, 02:18 PM   #12
 
CORRECT.

what is the formula for this centripetal force in terms of the speed at Q?
Nov2-11, 02:22 PM   #13
 
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Quote by grzz View Post
CORRECT.

what is the formula for this centripetal force in terms of the speed at Q?
I am not sure on that... I cant find it in my book (awfull book). :(

?
Nov2-11, 02:25 PM   #14
 
This forum is for helping and not for doing the work for you.
Try looking up 'centripeta force' on internet search engine. That is the way to learn.
Nov2-11, 02:27 PM   #15
 
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Did not mean to imply that.

F = MAc = MV^2/r

Is that it? what is the c ?
Nov2-11, 02:36 PM   #16
 
It is ok. You are welcome to this forum!

You got the right formula.
A - acceleration, c- centripetal.

Now try to find the speed at point Q.
Nov2-11, 02:39 PM   #17
 
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This is where i have trouble.

When i plug everything in:

F = (.100)(2.00)^2 / (.150)

I get 2.66 ?
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