The meter stick in the drawing can rotate about an axis located at the 20.0-cm mark. The axis is perpendicular to the screen. A force F acts at the left end; the force is perpendicular to the meter stick and has a magnitude of 175 N. A second force, either F1 or F2, acts at the 80.0-cm mark, as the drawing shows. The meter stick is in equilibrium. Which force, F1 or F2, acts on the meter stick, and what is its magnitude?

 F₁ = 43.75 N

Explanation:

In this exercise you fear a meter that at 20 cm has a force of 175 N applied and at 80 cm has another F1 force applied, so that the system is in equilibrium.

Let's use the equation of rotational equilibrium,

               τ₁ - τ₂ = 0

               F d = F₁ d₁

               F₁ = F d / d₁

Let's calculate

               F₁ = 175   0.2 /0.8

               F₁ = 43.75 N

Since the two forces are applied on the same bar, they must have the opposite direction for the torque to be canceled.

The answer is

The force F₂, acts on the meter stick, with a magnitude of 71.21 N

Explanation:

To solve the question, we note that the sum of moment about a point = 0 or ΣM =0

That is sum of clockwise moments = sum of anticlockwise moments

Also we are required to maintain the sign convention when calculating for the moment of the force thus

Clockwise moment = -ve and clockwise = +ve

Therefore taking moment about the axis at the 20.0 cm mark we have

-175 N × 20, 0 cm -F₁y × (80 - 20) cm = 0 where F₁y is the y component of the the force at the 80.0 cm mark

or -175 N × 20, 0 cm = F₁y × (80 - 20) cm

From the above equation, it is seen that F₁y = - (175 N × 20 cm)/(60 cm) = 58.33 N

F₁y = -58.33 N, Hence the correct force at the 80.0 cm mark is

58.33 N in the opposite direction as F₁y or F₂y

Hence the magnitude of F₂ can be found by resolving the force into components thus

y component of the force F₂ = 58.33 N angle to the vertical  is given by

90 - 55 or 35 ° therefore the y which is the vertical component = 58.33 N at an angle 35 ° to the vertical, we have

F₂ × cos (35 °) = 58.33 N

∴ F₂ = (58.33 N)/(cos (35 °))

= 71.21 N

Therefore the force F₂ acts on the meter stick with a magnitude of 71.21 N

we need to know the time interval during which the force acts. impulse of force = momentum change

hi,

what occurs during a solar eclipse?

there are small tides across earth.

the moon is between earth and the sun.

p.s. im not sure if the first answer is correct because during a solar ecliphse there are spring tides and i dont know if they are the same.