A beam, balanced on a pin, has a \( 0.050\,\mathrm{kg}\) mass placed \( 0.32\,\mathrm{m}\) to the left of the center of the beam. Where should a \( 0.067\,\mathrm{kg}\) mass be placed, right of the center of the beam, so that the beam stays in equilibrium? Note that gravity is pushing down on the beam at the center point.
Fig. 4- A diagram depicting masses balanced on a beam.
Solution
Step 1: First, we need to determine all of the forces acting on the beam and each force's distance from the point of rotation. Remember that the force that each mass exerts on the beam is equal to the weight of the mass. Weight is calculated using the equation, \( F=mg\), where \(m\) is mass measured in \( \mathrm{kg} \) and the \(g\) is acceleration due to gravity which is a constant, \( 9.8\,\mathrm{\frac{m}{s^2}}\). Therefore, our calculations are as follows.
\[ \begin{align} F_1&=m_1{g},\\F_1&=(0.050 \,\mathrm{kg})\left(9.8\,\mathrm{\frac{m}{s}^2}\right),\\F_1&=0.49 \,\mathrm{N}. \\\end{align} \]
For the second mass, we have
\[ \begin{align} F_2&=m_2{g},\\F_2&=(0.067 \,\mathrm{kg})\left(9.8\,\mathrm{\frac{m}{s}^2}\right),\\F_2&=0.66 \,\mathrm{N}. \\\end{align}\]
Now, we know that the \( F_1 \) has a corresponding radius of \( 0.32\,\mathrm{m} \) as this information is given to us. However, we must determine the corresponding radius for \( F_2\).
Step 2: To determine the radius corresponding to \( F_2 \), we need to calculate the sum of all torques acting on the beam. Remember that for an object to be in equilibrium, the sum of all torques must be zero. However, before proceeding further, we need to identify which torque formula applies. The formula
\( \tau=rF\sin\theta \) applies and allows us to write the following.
\[ \begin{align}\tau_1+\tau_2&=0,\\r_1F_1\sin\theta+r_2F_2\sin\theta&=0.\\\end{align}\]
For \( \theta \), we know that both angles are 90 degrees due to the force of gravity forming right angles with each side of the beam. Furthermore, we know that the first torque is positive because of the right-hand rule. Pointing our fingers to the left and curling them down in the direction of gravity results in our thumb pointing outward. For the second torque, pointing our fingers to the right and curling them downward results in our thumb pointing inward, indicating a negative torque. Since \( \sin \left(90^{\circ}\right)=1\) we can simplify this equation \(r_1 F_1+r_2 F_2=0\).
Step 3: Rearrange our equation,\(r_1 F_1+r_2 F_2=0\), in terms of \(r_2\) and insert our given variables. Note that we are taking the positive direction to the right of the beam, resulting in
\[ \begin{align}r_1F_1+r_2F_2&=0,\\r_1F_2&=-r_2F_2,\\-\frac{r_1{F_1}}{F_2}&=r_2,\\-\frac{(-0.32\,\mathrm{m})(0.49\,\mathrm{N})}{0.66 \,\mathrm{N}}&=r_2,\\r_2&=0.24\,\mathrm{m}.\\\end{align} \]
The second mass should be placed \( 0.24\,\mathrm{m}\) to the right of the center of the beam.
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