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

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

0cm50cm70cm100cm

↑                                     ↓                    ↓                     ↑

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

the minimum amperage that the current in this wire would have to have so that its weight is completely supported by the magnetic force due to earth's field, assuming that no other forces except gravity act on it

talk back to him about the sport and show him your expertise or leave the room