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

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
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