Question

(Optional) A uniform meterstick is in static rotational equilibrium when a mass of \(220 \mathrm{~g}\) is suspended from the \(5.0-\mathrm{cm}\) mark, a mass of \(120 \mathrm{~g}\) is suspended from the \(90-\mathrm{cm}\) mark, and the support stand is placed at the \(40-\mathrm{cm}\) mark. What is the mass of the meterstick?

Short answer

The mass of the meterstick is 75 grams.

Step-by-step solution

Understanding the Problem

We have a meterstick, which can be treated as a rigid rod in static equilibrium. This means the net torque around any point is zero. We have two masses attached to the meterstick and need to find the mass of the meterstick itself.

Define Torque

Torque (τ) is defined as the product of force and lever arm distance. In this case, the force is the weight of the mass hanging off the meterstick, which is calculated as mass times gravitational acceleration (g). Torque will be different for each mass due to their different positions on the meterstick.

Determine Lever Arms

The torque caused by each mass is calculated about the pivot at the 40 cm mark. The lever arm for the 220 g mass is 35 cm (40 cm - 5 cm), and for the 120 g mass is 50 cm (90 cm - 40 cm). The center of mass of the stick is at 50 cm, so its effective lever arm is 10 cm (50 cm - 40 cm).

Write Torque Equation

Because the system is in equilibrium, the sum of torques about the pivot point must be zero. Therefore, we have:\[\tau_1 + \tau_2 = \tau_3\]where \(\tau_1\) represents the torque due to the 220 g mass, \(\tau_2\) represents the torque due to the 120 g mass, and \(\tau_3\) is the torque due to the weight of the meterstick.

Calculate Forces for Torque

The forces due to the weights are: - For the 220 g mass: 220g = 220 * 9.8 cm/s² (force) at 35 cm. - For the 120 g mass: 120g = 120 * 9.8 cm/s² (force) at 50 cm.

Set Up Torque Balance Equation

According to the equilibrium condition, the clockwise torques must equal the counterclockwise torques:\[(220 \times 9.8 \times 35) + (mass_{stick} \times g \times 10) = (120 \times 9.8 \times 50)\]

Solve for the Mass of the Meterstick

Rearrange and solve the equation:\[mass_{stick} \times g \times 10 = (120 \times 9.8 \times 50) - (220 \times 9.8 \times 35)\]Simplify to find the mass of the meterstick:\[mass_{stick} = \frac{(120 \times 50) - (220 \times 35)}{10}\]Compute the result.

Key concepts

Torque Calculation

In physics, torque is a measure of the force causing an object to rotate. Torque is defined as the product of the force applied and the distance from the pivot point, also known as the lever arm. The formula for torque \( \tau \) is \( \tau = F \times r \), where \( F \) is the force applied and \( r \) is the lever arm's distance from the point of rotation. In the context of the meterstick problem, torque is generated by the weight of the masses hanging at specific points. Each mass applies a force (its weight) that generates a torque relative to the support stand. For instance, the mass of 220 g placed at 5 cm applies a torque with its force calculated as the product of the mass \( (220) \) and gravitational acceleration \( (9.8 \text{ m/s}^2) \). This force is then multiplied by the lever arm distance of 35 cm (from the 40 cm support stand), to compute the torque. Understanding torque is crucial for determining equilibrium in rotating systems, as it helps in balancing forces effectively.

Lever Arms in Physics

The lever arm is a critical component in determining torque. It is the perpendicular distance between the line of action of the force and the axis of rotation. The longer the lever arm, the greater the torque for the same amount of force. In the exercise, we determine different lever arms for each mass:

• For the 220 g mass, the lever arm is 35 cm (calculated as 40 cm - 5 cm).
• For the 120 g mass, the lever arm is 50 cm (calculated as 90 cm - 40 cm).
• For the meterstick itself, with its center of mass located at the 50 cm mark, the lever arm is 10 cm (50 cm - 40 cm).

These distances are used to calculate the torque exerted by each force, which in turn helps us to analyze rotational balance in the meterstick and determine its mass. Accurate assessment of lever arms ensures proper application of the torque equation.

Rotational Equilibrium Concepts

Rotational equilibrium occurs when a system experiences no net torque, meaning the sum of clockwise torques is equal to the sum of counterclockwise torques. This concept is essential to analyze systems where multiple forces cause rotation.In this meterstick problem, we looked at torques around the pivot point at 40 cm. For equilibrium:

• The sum of torques from the 220 g mass and the meterstick's mass must exactly balance the torque from the 120 g mass.
• This balance is expressed in the equation \[ (220 \cdot 9.8 \cdot 35) + (mass_{stick} \cdot 9.8 \cdot 10) = (120 \cdot 9.8 \cdot 50) \]

The key takeaway is that the torques on either side of the pivot should cancel each other out. By using this principle, the unknown mass of the meterstick can be determined, helping us ensure that all forces acting on a system are balanced.