Question

A force f=2i-3j+k acts at the point (1,5,2). Find the torque due to f

Answer 1

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First find the displacement vector of the point (2, -3, 1) from the point (0,0,2). This is basically the 2nd point 'minus' the 1st point:

r = (2-0)i + (-3-0)j + (1 - 2)k

. = 2i - 3j - k

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Then use the cross product to find torque (about (0, 0, 2)):

τ = rXF

=

|i . .j . k| unit vectors go in the1st row

|2 -3 -1| components of r go in the 2nd row

|2 .3 -1| components of F go in the 3rd row

= [(-3)*(-1) - (-1)*3]i - [2*(-1) - (-1)*2]j + [2*3 - (-3)*2]k

= [3 - (-3)]i - [-2 + 2]j + [6 + 6]k

= 6i - 0j + 12k

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Then find the magnitude:

|τ| = √(6² + 0² + 12²)

. . = √180

. . = 6√5

(= 13.4 if approx. value needed)

(If the positions are in metres (m) and the force is in newtons (N), the units are Nm).