ANSWER
[tex]\begin{gathered} 1)\text{ }214.90\text{ }N \\ \\ 2)\text{ }0.09\text{ }m \end{gathered}[/tex]
EXPLANATION
First, let us make a sketch of the diagram showing the distances on the rod:
1) Since the fulcrum is balanced, the center of gravity of the system will be at the fulcrum.
The center of gravity (in the horizontal is given by:
[tex]x=\frac{W_1x_1+W_2x_2+W_3x_3}{W_1+W_2+W_3}[/tex]
where W1 = the weight on the right end = 225 N
W2 = the weight of the rod = 255 N
W3 = the weight place on the left = W
x1 = the position of W1 (taking the left as the origin) = 1.90 m
x2 = the position of the center of mass of the rod = x1/2 = 0.95 m
x3 = the position of W from the left end = 0.60 m
x = position of center of gravity of the rod from the left end i.e. at the fulcrum = 1.90 - 0.75 = 1.15 m
Now, substitute the values given in the question and solve for W:
[tex]\begin{gathered} 1.15=\frac{(225*1.90)+(255*0.95)+(W*0.60)}{225+255+W} \\ \\ 1.15=\frac{427.5+242.25+0.60W}{480+W} \\ \\ 1.15(480+W)=669.75+0.60W \\ \\ 552+1.15W=669.75+0.60W \\ \\ 1.15W-0.60W=669.75-552 \\ \\ 0.55W=117.75 \\ \\ W=\frac{117.75}{0.55} \\ \\ W=214.09\text{ }N \end{gathered}[/tex]
That is the value of W.
2) Now, W is moved 30.0 cm (0.30 m) to the right.
This implies that:
[tex]x_3=0.60+0.30=0.90\text{ }m[/tex]
Since the other values (including W) do not change, we can now solve for x, which is the new center of gravity:
[tex]\begin{gathered} x=\frac{(225\times1.90)+(255\times0.95)+(214.09\times0.90)}{225+255+214.09} \\ \\ x=\frac{427.5+242.25+192.681}{694.09}=\frac{862.431}{694.09} \\ \\ x=1.24\text{ }m \end{gathered}[/tex]
Therefore, the fulcrum must be moved:
[tex]\begin{gathered} 1.24\text{ }m-1.15\text{ }m \\ \\ 0.09\text{ }m \end{gathered}[/tex]
The fulcrum should be moved 0.09 m to the right (since the W is moved to the right).
