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One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. The first force has a magnitude of 2.00 N and is applied perpendicular to the length of the stick at the free end. The second force has a magnitude of 6.00 N and acts at a 42.9o angle with respect to the length of the stick. Where along the stick is the 6.00-N force applied? Express this distance with respect to the end of the stick that is pinned.

Answer :

Answer:

  x = 0.455 L

Explanation:

For this exercise we must use the rotational equilibrium condition

        Σ τ = 0

it has two forces, the first is perpendicular to the rod, so its stub is

         τ₁ = F₁ L

the second force is applied with an angle, so we can use trigonometry to find its components

          sin θ = F_parallel / F₂

          cos θ = F_perpendicular / F₂

         F_parallel = F₂ sin θ

         F _perpendicular = F₂ cos θ

torque is

         τ₂ = F_perpendicular x + F_parallel 0

the parallel force is on the rod therefore its distance is zero

           

we apply the equilibrium equation

          τ₁  - τ₂ = 0

          F₁ L = F₂ cos θ  x

          x = [tex]\frac{L}{cos \theta} \ \frac{F_1}{F_2}[/tex]

let's calculate

          x = [tex]\frac{L}{cos \ 42.9} \ \frac{2.00}{6.00}[/tex]

          x = 0.455 L

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