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?

Question

Physics

Armando

21 September, 02:05

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?

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Answers (1)

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Delilah Weiss · Answer 1

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.