Q11-47P - University Physics with Modern Physics

Question

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. (a) Find W. (b) If W is now moved 25.0 cm to the right, how far and in what direction must the fulcrum be moved to restore balance?

Step-by-Step Solution

Answer

1. $W=140 N$ N
2.  W must be moved to $0.06m$ to the right.

Step 1: The given data

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 $w_1=255 N$

So $x_1=1 m$

Weight $w_2=225 N$

So $x_2=2 m$

Let $w_3=W$

Step 2: Formula used

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.

(a) Step 3: Find W

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

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 $W = 140 N$

(b) Step 4: Find a new center of mass

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.