# Torque question

Please look at picture for details...

Is the reason why Force (F) does not create a torque when applied at the given point on the tire because it is pulling against the axis with a Force (F) towards the right and at the same time the axis pulls back with an equal force magnitude but opposite direction thus resulting in a zero net force??

#### Attachments

• Torque.jpg
9.5 KB · Views: 427

cepheid
Staff Emeritus
Gold Member
Please look at picture for details...

Is the reason why Force (F) does not create a torque when applied at the given point on the tire because it is pulling against the axis with a Force (F) towards the right and at the same time the axis pulls back with an equal force magnitude but opposite direction thus resulting in a zero net force??

Draw a line between the centre of the circle (i.e. the rotation axis) and the point on the circumference where the force is applied. There's no torque in this situation, because only the component of the force perpendicular to this line (that you drew) will contribute to the torque. In this case, the force is entirely parallel to that line, and hence there is no torque. This result comes from the definition of torque as a vector cross product:

τ = r × F

where r is a position vector going from the axis of rotation to the point where the force is being applied. If this position vector and the force vector are parallel, then the cross product is 0.

To relate it to your everyday experience: it should be pretty intuitive to you that if the line of action of the force passes through the centre of the tire, then the tire will not rotate about its centre. In order to get the tire to rotate about its centre, you have to apply a force whose line of action is offset from the centre. Here's another example: have you ever tried opening a door by pushing on it AT its hinges? If you have, it probably didn't work out too well for you. The hinges are aligned with the axis of rotation, and if you apply the force AT the axis of rotation, then the distance r between the axis of rotation and the point where the force is applied is 0.

EDIT: The tire situation is more analogous to trying to rotate a door around its hinges by pushing inward or pulling outward on the edge of the door (i.e. along a line direction towards or away from the hinges). Since there is no component of the force perpendicular to this line, the door will not rotate.

Code:
Door, as viewed from the top:

____________________________

hinge->O----------------------------->r <--------------------------- F
____________________________

This force F will not rotate the door, because it is entirely parallel to the door (and hence parallel to r). There is no component of it that is perpendicular to r.

Code:
Door, as viewed from the top:

____________________________

hinge->O -------------------------->r
____________________________
^
|
|
|
|

F

This force F will rotate the door around its hinges, because there is a component of F that is perpendicular to r (in fact, in this case, F is entirely perpendiular to r).

Last edited:
thank you for the explanation, very helpful

cepheid
Staff Emeritus
Gold Member
thank you for the explanation, very helpful

I added another example and a drawing.

is torque a vector?

I don't see the picture..?

cepheid
Staff Emeritus
Gold Member
is torque a vector?

For the purposes of introductory physics, yes, torque is a vector. By convention, it points in the direction given by the right hand rule (curl your fingers in the direction that the torque wants to rotate the object, and your thumb points in the direction of the torque). So, by convention, a torque that would tend to want to cause a clockwise rotation points "into the page" and a torque that would tend to want to cause a counter-clockwise rotation points "out of the page."

All of this is a consequence of its definition as a cross product of two vectors.

More formally, I think that the cross product produces something called a "pseudovector", which is used to represent these sorts of quantities that have a "handedness" (rotational directionality) to them. So, technically, torque is a "pseudovector", but like I said before, for your purposes, you can just think of it as being a vector that always points perpendicular to the plane of the rotation it causes.

I don't see the picture..?

I meant that I edited my first post in order to add it in.

Last edited: