Question

A$0.120-\mathit{k}\mathit{g}$, $50.0-\mathit{c}\mathit{m}\mathbf{-}\mathit{l}\mathit{o}\mathit{n}\mathit{g}$uniform bar has a small 0.055-kg mass glued to its left end and a small $\mathbf{0}\mathbf{.}\mathbf{110}\mathbf{-}\mathit{k}\mathit{g}$mass glued to the other end. The two small masses can each be treated as point masses. You want to balance this system horizontally on a fulcrum placed just under its center of gravity. How far from the left end should the fulcrum be placed?

Short answer

Thus, the center of mass is right shifted having length 29.8 cm.

Step-by-step solution

Step 1: Center of Gravity

There is a point inside any object where the whole weight of that body is concentrated, balancing all other parts of that body. This point is considered to be the Centre of Gravity.

Step 2: Find the distance

Given, a rod with mass $m_1=0.120kg$having length $/=50cm$.

This implies, $x_1=\frac{50}{2}=25cm$

Also, the left and right masses with respective lengths taken from left side are:

$$\begin{gathered} m_2=0.055kg,x_2=0 \\ {m}_3=0.110kg,x_3=50cm \end{gathered}$$

Now, the distance of the point where the fulcrum must be placed will be:

$$\begin{gathered} x=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3} \\ =\frac{\left(0.120\right)\left(25\right)+\left(0.055\right)\left(0\right)+\left(0.110\right)\left(50\right)}{0.120+0.055+0.110} \\ =29.8cm \end{gathered}$$

Hence, this is the required answer.