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Tension Problem

 
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michellegu2894



Joined: 06 Jun 2008
Posts: 1

PostPosted: Tue Jul 22, 2008 12:16 pm    Post subject: Tension Problem Reply with quote

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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. Confused
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admin
Site Admin


Joined: 08 Dec 2003
Posts: 2176

PostPosted: Sat Jul 26, 2008 5:28 pm    Post subject: Reply with quote

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

PostPosted: Sat Jul 10, 2010 11:09 am    Post subject: Reply with quote

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

PostPosted: Sun Aug 22, 2010 2:02 pm    Post subject: Reply with quote

Can anyone help and isolate for T?
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jono6411



Joined: 23 Jun 2010
Posts: 1

PostPosted: Sun Aug 22, 2010 11:16 pm    Post subject: @DLBMCAT Reply with quote

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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