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
2 answers
sine20? is the fish pulling horizontally? If the fish is pulling vertically, it is sin70.
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
3 answers