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basic physics - pulleys on a hill

Posted on 2010-08-25
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Last Modified: 2012-05-10
Hi

I was hoping someone would look at the following question for me. I have tried to derive equation 1 below (the show derivation link isn't working so i can't check) in their solution below. I have scanned in my working out. When i did the problem i got the correct answer a totally different way so I wanted to check i could do it  their way too.

It might look a lot but it will only take you experts a couple of minutes I'm sure :)

thanks
p1.jpg
p3.jpg
p4.jpg
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Question by:andieje
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9 Comments
 
LVL 84

Expert Comment

by:ozo
ID: 33526189
I would say force along the rope rather than force in x direction
0
 

Author Comment

by:andieje
ID: 33526252
Sorry what do  you mean?

Do you mean the derivation is wrong or that the derivation is ok but i shouldn't call the force Fx
0
 
LVL 84

Expert Comment

by:ozo
ID: 33526311
You can call it Fx, but it is only in the x direction at the very top of the pulley, so it may be confusing.
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Author Comment

by:andieje
ID: 33526327
ok, see what you mean. Any comments on the derivation?
0
 
LVL 3

Expert Comment

by:SantiagoA
ID: 33526728
Try starting like this:

Force for the rope is the same for both masses, and boxes don´t move, so:

gM1sin1=gM2sin2

M1=M2.(sin2/sin1)
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Author Comment

by:andieje
ID: 33530299
santiago, i do appreciate that you can get to the equation in different ways. I was just interested in any comments on the way i had done it.

Another way would have been to say that net force along x axis = M2.g.sin2 - M1.g.sin1

so acceleration = (M2.g.sin2 - M1.g.sin1) / m1 + m2

0
 
LVL 13

Accepted Solution

by:
Carl Bohman earned 500 total points
ID: 33534129
A piece of advice: You really ought to try to be consistent with your notation, specifically with the subscripts.  m1g is not the same as mg1.  The first implies more than one mass with a fixed value for g.  The second implies a single mass with multiple values for g.  While it is possible to realize that you mean the first when you write the second, such looseness has a much higher chance of leading to mistakes or miscalculations (which are even harder to find when the notation is inconsistent).

As to your derivation, I find it logical and relatively easy to follow (ignoring the notation inconsistencies).  In essence, you have defined equations for all elements of your system and then merged them into a single equation representing the entire system.  This basic approach is very repeatable, so it's a good approach to know.  Other approaches may be faster in some cases, but they often rely on a little more intuition (e.g., knowing how to quickly determine the net force as you did in ID:33530299) and/or are not generally applicable to all systems (e.g., consider if there were 3 or more masses or if the inclined plane was not a straight line or if the system was initially in motion).
0
 

Author Comment

by:andieje
ID: 33536802
I accept the point about the notation. I was just rushing when i wrote it out 'neatly' for this forum
0
 

Author Closing Comment

by:andieje
ID: 33536810
thanks - very useful advice
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