Question

Three forces are applied to an object, as shown in the diagram below. F1 = 30 N, F2 = 34 N, F3 = 20 N, d1 = 25 mm, d2 = 20 mm, d3 = 10 mm, θ3 = 500. What is the resultant torque around the point 'x', in units of Nm?

Answer 1

Torque of F1 = 30*0.025 = 0.75Nm anticlockwise
Torque of F2 = 34 * 0.020 = 0.68Nm clockwise
Horizontal component of F3:
cos 50 = x/20
x = 12.856 N

Torque of F3 horizontal component = 12.856 * 0.01 = 0.129Nm anticlockwise

0.129 + 0.75 - 0.68 = 0.199Nm anticlockwise

sorry I'm not sure is it correct