Asked by SONIA
The fishing pole below (F = 94 N , = 1.90 m) makes an angle of 20.0° with the horizontal. What is the torque exerted by the fish about an axis perpendicular to the page and passing through the fisher's hand?
i TRIED doing fdsinthetha=(1.90*94*sin20) this was not the right answer. pls help
i TRIED doing fdsinthetha=(1.90*94*sin20) this was not the right answer. pls help
Answers
Answered by
bobpursley
sine20? is the fish pulling horizontally? If the fish is pulling vertically, it is sin70.
Answered by
Anonymous
Not enough information is given. Needs 2 angles, not one, shown in image.
Torque is:
Force*cos[90-(20 degrees + x degrees)]*distance
Torque is:
Force*cos[90-(20 degrees + x degrees)]*distance
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