Q11-43P - University Physics with Modern Physics

Question

A box of negligible mass rests at the left end of a 2.00-m, 25.0-kg plank. The width of the box is 75.0 cm, and sand is to be distributed uniformly throughout it. The center of gravity of the nonuniform plank is 50.0 cm from the right end. What mass of sand should be put into the box so that the plank balances horizontally on a fulcrum placed just below its midpoint?

Step-by-Step Solution

Answer

The mass of sand is 20 kg.

Step 1: The given data

Given that a box of negligible mass rests at the left end of a 2.00-m, 25.0-kg plank. 

The width of the box is 75.0 cm, and sand is to be distributed uniformly throughout it. 

The center of gravity of the nonuniform plank is 50.0 cm from the right end.

Step 2: Formula used

Center of mass, $\mathit{x}\mathbf{=}\frac{{\mathbf{m}}_{\mathbf{1}}{\mathbf{x}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}{\mathbf{x}}_{\mathbf{2}}}{{\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}}$ where

${\mathit{m}}_{\mathbf{1}}\mathbf{,}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }{\mathit{m}}_{\mathbf{2}}$ are masses of the system and ${\mathit{x}}_{\mathbf{1}}\mathbf{,}\mathbf{ }\mathit{a}\mathit{n}\mathit{d}\mathbf{ }{\mathit{x}}_{\mathbf{2}}$ are positions of masses.

Step 3: Find the mass of the sand

Let the fulcrum be the origin.

Let $m_1=25 kg$

So $x_1=0.5 m$

Let m₂ be mass of the sand

So $x_2=-0.625 m$

Since the center of gravity is in the middle of the box

Therefore $x=0 m$

This implies $x=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}=0 m$

Thus 

$\begin{array}{rcl}\frac{\left(25 kg\right)\left(0.5 m\right)+m_2\left(-0.625 m\right)}{m_1+m_2}& =& 0 m\\ m_2& =& \frac{12.5 m·kg}{0.625 m}\\ m_2& =& 20 kg\end{array}$

Hence, the mass of sand required is 20 kg.