finding torque with an axis perpendicular to the page?

Question

A fishing pole is 2.00 m long and inclined above the horizontal at an angle of 20.0°. A fish is pulling on it with a force of 100N 37-degrees below the horizontal. 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?

should I just use Torque = r * FsinX where X is 57-degrees

thanks!

Answer 1

You are on the right track .

T= 2 X100 sin(57)=168N


and this is why

Rx= -2 cos(20)=1.88 m
Ry= 2 sin(20) = 0.68 m

Fx= 100 cos(37)=79.9 N
Fy= 100 sin (37) =60.2 N

T=(RxFy - RyFx) into z- axis (the difference in sign since they appose each other. First torque is counter clock wise (CCW) and second is CW)



T= -1.88x60.2 - .068 x 79.9=
T=-2 cos(20)100 sin (37) - 2 sin(20)100 cos (37) )
T= -200(cos(20) sin (37) + sin(20) cos(37))

since sin (α+β) = cos β sin α + sin β cos α

T= -200( sin (20 + 37) )=
T= -200(sin(57))
T= -168 Nm (in a negative z -axis)

Have fun