- #1

JohnKSmith

- 1

- 0

## 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.