Why is Torque the Cross Product Between r and F?

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Examine T= r x F (cross product),
where |T|=|r||F| sin t, where t is the angle between r and F

The intuitive idea of torque (let's only consider torque about the pivot for now) is that the stronger the force or the further away you are from the pivot point, the more the object will TEND to rotate.

The cross product formula above does match this intuitive idea. For example, larger force, the magnitude of the torque really does increase. It also matches the observation that no torque is generated when t is zero, and max torque when t is at a 90 degree angle.

My question is, what about the angles in between? Using the formula, does this really match with what nature is telling us? Does the formula truly reflect the intuitive idea of torque in terms of the tendency of making an object rotate? I don’t see a reason why it has to obey the sine relationship/curve other than the fact that at t =0 and t=90 we see min and max respectively. For example, why can't it be a linear relationship for angles in between?

Thanks!
 

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  • #2
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think of the sin(t) as a means to decompose the F vector into two components:

F sin(t) and F cos(t) where F sin(t) is perpendicular to R and F cos(t) is parallel to R then you can see that the F cos(t) doesn't have any effect on the rotation whereas F sin(t) does.
 
  • #3
jbriggs444
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My question is, what about the angles in between? Using the formula, does this really match with what nature is telling us? Does the formula truly reflect the intuitive idea of torque in terms of the tendency of making an object rotate? I don’t see a reason why it has to obey the sine relationship/curve other than the fact that at t =0 and t=90 we see min and max respectively. For example, why can't it be a linear relationship for angles in between?
Consider a long vertical shaft, like the drill shaft on an oil rig.

Now weld two handles on that shaft.

The first handle is one meter long and extends horizontally.

The second handle is 1.4 meters long and extends at a 45 degree angle diagonally upward so that its end is also about one meter away from the shaft.

The welds are solid and the handles strong.

Would you agree that pushing on the end of either handle provides the same rotational torque as pushing on the other?

Or would you claim that pushing on the longer handle provides only 70% as much torque as pushing on the shorter handle?
 
  • #4
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think of the sin(t) as a means to decompose the F vector into two components:

F sin(t) and F cos(t) where F sin(t) is perpendicular to R and F cos(t) is parallel to R then you can see that the F cos(t) doesn't have any effect on the rotation whereas F sin(t) does.
Thanks this was really helpful in seeing what was really happening. Yes so we indeed only consider the perpendicular component of the force which is accounted for in the cross product formula. There's no way it can be a linear relationship.

Thanks again :).
 
  • #5
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Once you have a notion of angular momentum, the relationship between that and torque is clear.

If you accept that [itex]\ell = r \times p[/itex] and that [itex]dp/dt = F[/itex] (force is the time derivative of momentum), then the time derivative of angular momentum is

[tex]\frac{d\ell}{dt} = \frac{dr}{dt} \times p + r \times \frac{dp}{dt} = v \times p + r \times F[/tex]

Of course, [itex]v \times p = 0[/itex], and we get [itex]d\ell/dt = r \times F[/itex]. We call this torque.
 

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