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

Paula Poole

Physics

31 January, 11:40

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)

Know the Answer?

Alexzander Ray · 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.