A fishing pole is 2.00 m long and inclined to the horizontal at an angle of 20.0∘. What is the torque exerted by the fish about an axis perpendicular to the page and passing through the hand of the person holding the pole?

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d2saint0 · Accepted Answer

Given that  Force F=100 N  Length of fishing pole L=2.00 m    The above diagram, angle θ=20∘+37icrc  θ=57∘  The parallel component of the force is F11=Fcos⁡θ  =(100 N)cos⁡(57∘)  =54.5N  The perpendicular component of the force is F1=Fsin⁡θ  (100N)sin⁡(57∘)  83.9N  Therefore, the torque exerted by the fish about axis perpendicular to the page and passing through the hand of the person holding the pole is  τ≠t=τ⊥⊥−τ⊥  =F11(0)sin⁡0−F⊥(L)sin⁡90∘  =0−(83.9N)(2.0m)(1)  =−168 N⋅m

A fishing pole is 2.00 m long and inclined to the horizontal at an angle of 20.0^\circ. What is the torque exerted by the fish about an axis perpendicular to the page and passing through the hand of the person holding the pole?

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