A particle, mass 0.25 kg is at a position (-7i + 7j + 5k) m, has a velocity (6i - j + 4k) m/s, and is subject to a force (-5i + 0j - k) N. What is the magnitude of the torque on the particle about the origin?
47.94Nm
The torque (τ) on a particle subject to a force (represented as force vector F) at a position (represented as position vector r) about the origin is given by the cross product of the position vector r for the point of application of a force and the force F. i.e
τ = r x F
Given:
r = (-7i + 7j + 5k) m
F = (-5i + 0j - k) N
| i j k |
r x F = | -7 7 5 |
| -5 0 -1 |
r x F = i(-7 - 0) - j(7+25) + k(0+35)
r x F = i(-7) - j(32) + k(35)
r x F = -7i - 32j + 35k
Therefore the torque τ = -7i - 32j + 35k
The magnitude of the torque is therefore;
|τ| = [tex]\sqrt{(-7)^2 + (-32)^2 + (35)^2}[/tex]
|τ| = [tex]\sqrt{49 + 1024 + 1225}[/tex]
|τ| = [tex]\sqrt{2298}[/tex]
|τ| = 47.94Nm
The magnitude of the torque on the particle about the origin is 47.94Nm
