JohnKSmith
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Homework Statement
This image: http://i52.tinypic.com/35i0j1k.jpg, assume the table surface and pulley to be frictionless. Please ignore the given a=1.96 in the photo as I will show how I reached that further below.
Homework Equations
F=ma
The Attempt at a Solution
Well, my stab at the problem was like this, for the 12kg block we had it at rest on the vertical so I left it out as Fn+12g=0. I assumed the clockwise spin of the pulley to be positive so the horizontal of the 12g block's equation looked like
T-0=12a ∴T=12a
and I found the hanging block's equation to be 3g-T=3a so I take the T I found for the top block into this equation to find
3g-(12a)=3a
3g=15a
a=1.96 as the acceleration in meters per second squared.
With this acceleration I took it and placed it into the very first equation T=12a of the block resting on the table to deduce rope tension to be 23.5N assuming three significant digits.
Is this method the correct way to find the tension of a rope as seen in the given image above? I was told the answer is actually somewhere around 58.8N and I don't see the logic to reaching that conclusion. Comments, corrections, and explanations welcome.