Question

In the experiment of balancing moments, suppose the fulcrum is at the \(60 \mathrm{~cm}\) mark, and a known mass of \(2 \mathrm{~kg}\) is used on the longer arm. The greatest mass of \(\mathrm{m}\) which can be balanced against \(2 \mathrm{~kg}\) such that the minimum distance of either of the masses from the fulcrum is at least \(10 \mathrm{~cm}\). (Neglect mass of meter scale.) What will be the value of \(\mathrm{m}\) ? (A) \(12 \mathrm{~kg}\) (B) \(4 \mathrm{~kg}\) (C) \(6 \mathrm{~kg}\) (D) \(8 \mathrm{~kg}\)

Short answer

The greatest mass (m) that can be balanced against the 2 kg mass while maintaining a minimum distance of 10 cm from the fulcrum for both masses is \(6 \mathrm{~kg}\). Answer: (C) 6 kg.

Step-by-step solution

Identify the known parameters

The fulcrum is at the 60 cm mark and a mass of 2 kg is used on the longer arm. The minimum distance of either of the masses from the fulcrum is 10 cm.

Apply the principle of moment balance

The equation for moment balance considering the distance from the fulcrum is: \(Moment_{mass_1} = Moment_{mass_2}\) Where: - \(Moment_{mass_1} = Force_{mass_1} \times Distance_{mass_1}\) - \(Moment_{mass_2} = Force_{mass_2} \times Distance_{mass_2}\) - \(Force_{mass} = mass \times g\), where g is the acceleration due to gravity. In our case, \(mass_1 = 2 \mathrm{~kg}\) and \(mass_2 = m\). The minimum distance of 2 kg from the fulcrum is 10 cm. The distance of the unknown mass m from the fulcrum (X) can be calculated from the total length of the scale (100 cm) minus the fulcrum position (60 cm) and the minimum distance of 2 kg from the fulcrum (10 cm): \(X = 100 - 60 - 10 = 30 \mathrm{~cm}\).

Apply the moment balance equation

Now let's plug in the known values and solve for the unknown mass m: \((2 \mathrm{~kg} \times g) \times (10 \mathrm{~cm}) = (m \times g) \times (30 \mathrm{~cm})\) Since the acceleration due to gravity (g) is the same for both masses, we can cancel it out of the equation: \(2 \mathrm{~kg} \times 10 \mathrm{~cm} = m \times 30 \mathrm{~cm}\) Now, solve for m: \(m = \dfrac{2 \mathrm{~kg} \times 10 \mathrm{~cm}}{30 \mathrm{~cm}}\) \(m = \dfrac{20 \mathrm{~kg\cdot cm}}{30 \mathrm{~cm}}\) \(m = \dfrac{2}{3} \times 10 \mathrm{~kg}\) \(m = 6.666 \mathrm{~kg}\) Since it must be an integer value within the given options, the greatest m that can be balanced is: \(m = 6 \mathrm{~kg}\) Answer: (C) 6 kg

Key concepts

Fulcrum

The fulcrum is the pivot point or the support about which a lever rotates. It serves as the center of balance in the system. By placing the fulcrum at a particular location, you can control how different weights influence the lever balance. In our problem, the fulcrum is positioned at the 60 cm mark along the scale's length. This means that from this point, we can determine how each mass interacts with the scale when they are placed at different distances on opposite sides.
It's essential to precisely select this point to maintain or achieve equilibrium in a balancing experiment.

Moment Equation

The concept of a moment equation is central to solving leverage problems. It is grounded in the idea that the moment is a product of the force applied and the distance from the fulcrum. Mathematically, it is represented as:

• Moment = Force \(\times\) Distance

In our scenario, we have two masses on either side of the fulcrum. For the system to be in balance, the moments around the fulcrum must be equal. Thus, the equation becomes:

• \( \text{Force}_1 \times \text{Distance}_1 = \text{Force}_2 \times \text{Distance}_2 \)

Here, Force equates to the weight of the mass (mass \(\times\) gravity), and Distance is the length from the fulcrum to where the mass is placed. This balance leads us to determine unknown variables like the mass \(m\) in our problem.

Balancing Moments

Balancing moments is achieved when the clockwise moment equals the counterclockwise moment about the fulcrum. This principle helps determine the correct positioning of weights to achieve balance on a lever.
To achieve this balance in the given problem:

• The 2 kg weight on one side is placed 10 cm from the fulcrum.
• The unknown mass \(m\) is positioned on the opposite side.

By ensuring the moments are equal (the product of mass and distance are the same on both sides), one can calculate the balance condition:

• The equation becomes \(2 \times 10 = m \times 30\)

simplifying to give the value of \(m\), resulting in balance.

Mass Distribution

Mass distribution is key in understanding how different mass positions influence leverage. In simpler terms, it explains how the distribution of weight affects the equilibrium of a balancing experiment.

• In this exercise: The shorter the distance from the fulcrum for a heavier object, the more effective it is in balancing a heavier mass further away.
• The 2 kg mass is closer to the fulcrum (10 cm), applying a strong influence in a compact area.
• The unknown mass \(m\) is located further (30 cm away), thus needs to be larger to exert enough counterbalancing moment.

This principle allows us to understand why, to balance moments, the position and size of mass mutually dictate their effectiveness on the lever.