91Ó°ÊÓ

Given that a uniform, 255-N rod that is 2.00 m long carries a 225-N weight at its right end and an unknown weight W toward the left end. When W is placed 50.0 cm from the left end of the rod, the system just balances horizontally when the fulcrum is located 75.0 cm from the right end.

Weight of rod $w_1=255N$

So $x_1=1m$

Weight $w_2=225N$

So $x_2=2m$

Let$w_3=W$

Center of mass $\mathit{x}\mathbf{=}\frac{{\mathbf{w}}_{\mathbf{1}}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{w}}_{\mathbf{2}}{\mathbf{x}}_{\mathbf{2}}\mathbf{+}{\mathbf{w}}_{\mathbf{3}}{\mathbf{x}}_{\mathbf{3}}}{{\mathbf{w}}_{\mathbf{1}}\mathbf{+}{\mathbf{w}}_{\mathbf{2}}\mathbf{+}{\mathbf{w}}_{\mathbf{3}}}$

Where ${{\mathbf{w}}_{\mathbf{i}}}^{\mathbf{\text{'}}}\mathit{s}$are the weight and ${{\mathbf{x}}_{\mathbf{i}}}^{\mathbf{\text{'}}}\mathit{s}$are the positions.

$$\begin{gathered} x=1.25m \\ x=\frac{w_1{x}_1+w_2{x}_2+w_3{x}_3}{w_1+w_2+w_3} \\ Thisgives{w}_3=\frac{\left(w_1+w_2\right)x-w_1{x}_1-w_2{x}_2}{x_3-x} \end{gathered}$$

Hence

$\begin{array}{rcl}W& =& \frac{\left(480N\right)\left(1.25m\right)-\left(225N\right)\left(1m\right)-\left(225N\right)\left(2m\right)}{\left(0.5m\right)-\left(1.25m\right)}\\ & =& 140N\end{array}$

Therefore weight is$W=140N$

W is now moved 25.0 cm to the right

Now $w_3=140N$

And $x_3=0.75m$

The new center of mass

$\begin{array}{rcl}x& =& \frac{\left(255N\right)\left(1m\right)+\left(225N\right)\left(2m\right)+\left(140N\right)\left(0.75m\right)}{\left(255N\right)+\left(225N\right)+\left(140N\right)}\\ & =& 1.31m\end{array}$

Hence W must be moved $\left(1.31m\right)-\left(1.25m\right)=0.06m$to the right.