A 1-kg rock is suspended from the tip pf a uniform meter at 0 cm mark so that the meter stick balances like a see-saw when the fulcrum is at 20 cm mark. what is the mass of the meter stick?

To find the mass of the meter stick, we need to apply the principle of torques.

A torque is created by a force acting on a lever arm, causing rotation. The torque is calculated by multiplying the force exerted by the distance from the fulcrum. For an object to be balanced, the torques on both sides of the fulcrum should be equal.

In this case, the force is the weight, which can be calculated using the formula:

weight = mass x acceleration due to gravity

Given that the rock has a mass of 1 kg, we know its weight is 9.8 N (assuming the acceleration due to gravity is 9.8 m/s^2).

Since the meter stick balances at the 20 cm mark, the distance from the fulcrum to the 1 kg rock is 20 cm.

The torque produced by the rock can be calculated by multiplying the weight (9.8 N) by the distance (0.20 m):

torque_rock = weight x distance = 9.8 N x 0.20 m = 1.96 Nm

Since the meter stick balances, the torque produced by the rock must be counterbalanced by the torque produced by the meter stick.

If we assume that the meter stick's mass acts at its center (50 cm mark), then the distance from the fulcrum to the center of the meter stick is 30 cm.

Let's assume the mass of the meter stick as m_kg.

torque_meter stick = weight x distance = m_kg x 9.8 N/kg x 0.30 m

Since the meter stick is balanced, the torque produced by the meter stick must equal the torque produced by the rock:

torque_rock = torque_meter stick

1.96 Nm = m_kg x 9.8 N/kg x 0.30 m

Simplifying the equation:

1.96 Nm = 2.94 Nm x m_kg

Dividing both sides of the equation by 2.94 Nm:

m_kg = 1.96 Nm / 2.94 Nm

m_kg = 0.6667 kg

Therefore, the mass of the meter stick is approximately 0.6667 kg.