Question

A 0.1-kg meter stick is supported at both ends by strings, and there is a 200-g mass attached to it at the 70 cm mark. What is the force exerted by the string connected at the 0 cm mark, if the system is in static equilibrium

Answer 1

The force exerted by the string connected at the 0 cm mark is 1.96 N.

Step-by-step explanation:

To find the force exerted by the string connected at the 0 cm mark, we need to analyze the forces acting on the meter stick.

Since the system is in static equilibrium, the sum of the forces and the sum of the torques must be zero.

Let's consider the forces. There are three forces acting on the meter stick: the force exerted by the string connected at the 0 cm mark, the force exerted by the string connected at the 70 cm mark, and the gravitational force acting on the 200 g mass.

Since the system is in static equilibrium, the sum of the forces in the vertical direction must be zero. This means that the force exerted by the string connected at the 0 cm mark must be equal in magnitude and opposite in direction to the gravitational force acting on the 200 g mass. Therefore, the force exerted by the string connected at the 0 cm mark is equal to the weight of the 200 g mass, which can be calculated using the formula:

Force = mass x acceleration due to gravity

Force = 0.2 kg x 9.8 m/s^2 = 1.96 N.