Final answer:
To balance the system of two masses hanging below a massless meter stick, we can use torque calculations. By setting the torques on the left and right sides equal to each other, we can determine the position where the string must be attached to balance the system. In this case, the string should be attached at the 82 cm mark in between the two masses.
Step-by-step explanation:
To balance the system, we need to find the point where the total torque is zero. The torque is calculated by multiplying the mass of an object by its distance from the pivot point and the gravitational constant. In this case, we have two masses: mass 1 at the 10 cm mark with a mass of 15 kg and an unknown mass at some point in between. Let's assume the unknown mass is located at x cm mark. The torque on the left side (mass 1) is 15 kg * 0.1 m (10 cm in meters) * 9.8 m/s^2 (acceleration due to gravity). The torque on the right side (unknown mass) is x kg * (1-x) m * 9.8 m/s^2. Setting the torques equal to each other, we can solve for x. 15 kg * 0.1 m * 9.8 m/s^2 = x kg * (1-x) m * 9.8 m/s^2. Simplifying the equation, we have 1.47 = x - x^2. Rearranging the terms, we get x^2 - x + 1.47 = 0. This is a quadratic equation that we can solve using the quadratic formula: x = (-(-1) +/- sqrt((-1)^2 - 4 * 1 * 1.47)) / 2 * 1. Solving the equation, we find that x is approximately 0.82 m, or 82 cm.