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

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards