Finding torque (vector cross product)

[vu10758]

A particle position is described by position vector r = 3i + 2j and the force vector i - 2j acts on the object.

1) Find the torque about an axis through the origin and perpendicular to the xy plane. Draw the two vectors to check your torque direction.

I used the right hand rule and found out that the torque will have a negative direction. Then I used cross product term by term.

My torque is -8 k NM.

2) Find the torque about an axis through the point x = 5m, y = 5m perpendicular to the xy plane. Draw the two vectors to check your torque direction. The force should produce a roation about (5,5) consistent with your answer.

How do I do this? The answer is 7 z Nm. Do I find the torque the same way as for part 1, and then somehow take the distance between them into consideration? How come the direction is now positive?

 

[Pyrrhus]

Ok the first question was pretty straightforward, they gave you the position vector from the origin to your force point of application. Now for the 2nd question you have another position. They give you the position vector for the other axis, and you have the position vector for your force point of application. Therefore, you can calculate the new position vector from the new axis position to your force of application. Remember is from the new axis position to your force point of application.