A uniform bar $AB$ weights $12$N ,rests with one part $AC$ of length $8$m on a horizontal table

Question

A uniform bar $AB$ weights $12$N ,rests with one part $AC$ of length $8$m on a horizontal table and the remaining part $CB$ projecting over the edge of the table. If the bar is on the point of over-balancing when a weight of $5$N is placed on it at point $2$m from $A$ and a weight of $7$N is hung from $B$. Find the length of $AB$ ?

Answer 1

Let $L$ be the length. With C as the balancing point, the following equilibrium can be established,

$$(8-2)\cdot 5+4\cdot W_{AC} = \frac{L-8}2\cdot W_{CB} + (L-8)\cdot7\tag 1$$

where the segment bar weights are,

$$W_{AC}=\frac8L\cdot12,\>\>\>\>\>W_{CB}=\frac{L-8}L\cdot12\tag 2$$

Substitute (2) into (1) to obtain the length

$$L=14m$$