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Torque Equilibrium on a Pivot

by JamesEarl
Tags: equilibrium, pivot, torque
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JamesEarl
#1
Oct16-09, 08:12 PM
P: 9
1. The problem statement, all variables and given/known data

The two objects in the figure below are balanced on the pivot, with m = 2.4 kg. What is the distance d?

http://www.webassign.net/knight/p13-27alt.gif


2. Relevant equations

T=rf

3. The attempt at a solution

2.4kg + 4.0 kg= 6.4kg
6.4kg/2= 3.2kg on each side of pivot
Looking at the right half, 1.2kg+4kg= 5.2kg.
3.2kg/5.2kg * 1m = 0.615 m from right side
So 1.38 m from left side = d...THIS ANSWER IS WRONG

Basically, I understand all the main concepts, I got everything else on my homework right, but for some reason this problem has me stuck. Any help would be appreciated!
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rl.bhat
#2
Oct16-09, 11:33 PM
HW Helper
P: 4,433
Find the torque by the center of masses about the left end.
From the left end center of masses produce clockwise torque.
The reaction of the center of masses on pivot produce counterclockwise torque. Equate them to find d.


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