91影视

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 ${\mathrm{w}}_1=255\mathrm{N}$

So $x_1=1m$

Weight ${\mathrm{w}}_2=225\mathrm{N}$

So ${\mathrm{x}}_2=2\mathrm{m}$

Let $w_3=w$

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

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

x = 1.25 m

$\mathrm{x}=\frac{{\mathrm{W}}_1{\mathrm{x}}_1+{\mathrm{W}}_2{\mathrm{x}}_2+{\mathrm{W}}_3{\mathrm{x}}_3}{{\mathrm{W}}_1+{\mathrm{W}}_2+{\mathrm{W}}_3}$

This gives $w_3=\frac{\left({\mathrm{w}}_1{\mathrm{w}}_2\right)+\mathrm{x}-{\mathrm{w}}_1{\mathrm{x}}_1-{\mathrm{w}}_2-{\mathrm{x}}_2}{{\mathrm{x}}_3-\mathrm{x}}$

Hnece,

$\begin{array}{r}W=\frac{\left(480\mathrm{N}\right)\left(1.25\mathrm{m}\right)鈭�\left(225\mathrm{N}\right)\left(1\mathrm{m}\right)鈭�\left(225\mathrm{N}\right)\left(2\mathrm{m}\right)}{\left(0.5\mathrm{m}\right)鈭�\left(1.25\mathrm{m}\right)}\\ =140\mathrm{N}\end{array}$

Therefore weight is W = 140 N

W is now moved 25.0 cm to the right

Now ${\mathrm{w}}_3=140\mathrm{N}$

And ${\mathrm{x}}_3=0.75\mathrm{m}$

The new center of mass

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

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