Question

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?

Answer 1

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)