Question

A meterstick is balanced with the fulcrum at the 50 cm mark. If a 200 g mass is placed at the 20 cm mark, at which mark should a 500 g mass be placed so that the system is balanced?

Answer 1

To balance a meterstick with a 200 g mass at the 20 cm mark and fulcrum at 50 cm, a 500 g mass should be placed at the 62 cm mark, applying the principle of torque balance.

Step-by-step explanation:

The student is asking how to balance a meterstick with different masses, which involves the concept of torque balance in Physics. To solve for the mark at which a 500 g mass should be placed to balance the meterstick, we need to apply the principle that for a lever in equilibrium, the sum of the torques about the fulcrum must be zero.

In this case, we can find the position for the 500 g mass (m2) using the equation m1 × d1 = m2 × d2, where m1 is the 200 g mass, d1 is the distance from the fulcrum to m1, and d2 is the distance from the fulcrum to m2. The mass m1 is placed at 20 cm, which is 30 cm from the fulcrum at the 50 cm mark. Thus, we have 200 g × 30 cm = 500 g × d2, solving for d2 gives us d2 = (200 g × 30 cm) / 500 g = 12 cm. Hence, the 500 g mass should be placed at 50 cm + 12 cm = 62 cm mark on the meterstick to balance it.