Two forces are applied to a pipe, as shown in the diagram below. if f1 is 200 n, and f2 is 30 n, what is the resultant torque around the point 'o', in units of nm?

Summary of solution method

Sum horizontal and vertical components of forces about A, and use these components to find torque about O.  This is straight-forward and lends to systematic calculations.

A. resolve forces F1=200 and F2=30 into horizontal and vertical components, using a table

Force, F    Fx=Fcos(theta)   Fy=Fsin(theta)

F1=200     200*(4/5)=160   200*(3/5)=120

F2=30       30*(1/2)=15       -30*(sqrt(3)/2=-25.981

SUM         175                     94.019

B. Take moments about O  (clockwise positive)

Torque, To

=Fx*y -Fy*x

=175*0.25 -94.019*0.425

=+3.792 N-m (clockwise)

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