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michellegu2894
Joined: 06 Jun 2008 Posts: 1
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Posted: Tue Jul 22, 2008 12:16 pm Post subject: Tension Problem |
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In the Tension example problem that uses the pulley system in the GS study manual, what equations were equated to each other to find Tension and how were they equated to each other? My study partner and I stayed at Barnes & Nobles and worked on that prob. for at least 40 min. We know that it is basic high school algebra but we just can't seem to retrieve it out of out brains!
GS Manual 2008 edition page 174-175.  |
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admin Site Admin
Joined: 08 Dec 2003 Posts: 2176
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Posted: Sat Jul 26, 2008 5:28 pm Post subject: |
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The equations are:
F = T - m1g = m1a
and
F = T - m2g = -m2a
There are several ways to solve these simultaneous equations.
Option 1: subtract one equation from the other in order to eliminate T (as mentioned in the book)
T - m1g = m1a
-(T - m2g = -m2a)
-------------------
-m1g + m2g = m1a + m2a
now isolate for a by dividing both sides by (m1 + m2) |
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pszopa1343
Joined: 05 Apr 2010 Posts: 6
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Posted: Sat Jul 10, 2010 11:09 am Post subject: |
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I'm having trouble solving for the Tension. The book doesn't show the work getting to the equation
T = g(2m1m2)/(m1+m2)
Any help with this one?
Thx. |
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DLBMCAT
Joined: 19 Jul 2006 Posts: 12
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Posted: Sun Aug 22, 2010 2:02 pm Post subject: |
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| Can anyone help and isolate for T? |
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jono6411
Joined: 23 Jun 2010 Posts: 1
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Posted: Sun Aug 22, 2010 11:16 pm Post subject: @DLBMCAT |
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Solve for a in the two eq's given by the admin, set them equal to each other, and solve for T.
T=m1a+m1g -> T-m1g=m1a -> T/m1-g=a
T=m2g-m2a -> T-m2g=-m2a -> -(T/m2)+g=a
T/m1-g=-(T/m2)+g -> T(m1+m2)/(m1m2)=2g -> T= 2g(m1m2)/(m1+m2) |
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