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

Answer:

The force exerted by the string connected at the 0 cm mark is 1.078 N

Explanation:

Given;

mass of the meter stick, m = 0.1 kg

weight of the meter stick, = 0.1 kg x 9.8 m/s² = 0.98 N

the weight attached at 70 cm mark = 0.2 kg x 9.8 m/s² = 1.96 N

0cm--------------------------50cm-----------70cm-------------100cm

↑                                     ↓                    ↓                     ↑

F₁                                 0.98N              1.96N                F₂

Take the moment about F₂: clockwise moment = anticlockwise moment

F₁(100 - 0) = 0.98(100 -50) + 1.96(100 -70)

100F₁ = 49 + 58.8

100F₁ = 107.8

F₁ = 107.8 / 100

F₁ = 1.078 N

Therefore, the force exerted by the string connected at the 0 cm mark is 1.078 N