Question

a meter stick is balanced at its center (i.e. at the 50 cm mark). at the 16 cm mark is placed a 135 gram mass. where should a 276 gram mass be placed, so that the stick remains balanced. write your answer as the cm mark on the meter stick (round to one place of decimal).

Answer 1

To balance the meter stick, the 276 gram mass should be placed at the 16.7 cm mark for a meter stick is balanced at its center (i.e. at the 50 cm mark). at the 16 cm mark is placed a 135 gram mass. where should a 276 gram mass be placed, so that the stick remains balanced.

Step-by-step explanation:

To balance the meter stick, we need to consider the torques acting on it.

The torque due to a force is given by the product of the force and the distance from the fulcrum. Since the meter stick is balanced at the center, the torques on both sides of the fulcrum should be equal.

Let's calculate the torque produced by the 135 gram mass at the 16 cm mark.

The distance from the fulcrum is 34 cm (50 cm - 16 cm). So, the torque is (135 g) * (34 cm).

To keep the stick balanced, the torque produced by the 276 gram mass at the other end of the stick should also be (135 g) * (34 cm).

Let's call the distance from the fulcrum to this mass x. We can set up an equation like this: (276 g) * (x) = (135 g) * (34 cm).

Solving for x, we find x = (135 g) * (34 cm) / (276 g)

= 16.7 cm (rounded to one decimal place).