General problems with coriolis force

[Dracovich]

Ok I'm having some general problems with solving Coriolis problems.

So the general way of writing up the Coriolis force is

$$F_c = -2*m*(\vec{\omega} \times \vec{v})$$

But most questions i get about the Coriolis force involve some information about it's latitude position on earth, but i fail to see where this comes into the equation. The chapter discussing the Coriolis force does not seem to touch on this, and the two examples in my book both use the equator, so they are no help.

Also i have some problems visualising which way the Coriolis force works, more in general i have a problem seeing which way a cross product points. Now it's perpendicular to the two vectors, but which way? (+ or - axis).

Just as an example, this is the most basic problem from my book, (which i can't solve since i can't figure out the latitude thing):

"A vehicle of mass 2000kg is traveling due north at 100km/h at latitude 60°. Determine the magnitude and direction of the Coriolis force on the vehicle."

**edit** forgot to smack a minus sign on the formula.

 

[MathStudent]

The right hand rule can be used to find the direction of a cross product of two vectors: I'll try to describe it as follows;

in the case of $\vec{\omega} \times \vec{v}$
use your pointer finger (of your RIGHT hand) to point in the direction of $\vec{\omega}$
your middle finger should then point in the direction of $\vec{v}$ (only the part of $\vec{v}$ that is perpindicular to $\vec{\omega}$)
stick your thumb out so that its perpendicular to both your pointer and middle fingers.
The direction of your thumb is the direction of $\vec{\omega} \times \vec{v}$

 

[Dracovich]

Ahh thanks, that should make life easier for me :) Now just the question of latitude ...

 

[Andrew Mason]

So the general way of writing up the Coriolis force is

$$F_c = 2*m*(\vec{\omega} \times \vec{v})$$

"A vehicle of mass 2000kg is traveling due north at 100km/h at latitude 60°. Determine the magnitude and direction of the Coriolis force on the vehicle.

The latitude is needed in order to determine the cross product:

$$\vec{\omega} \times \vec{v}$$

The velocity vector points to the north pole along the surface and the angular velocity is parallel to the Earth's axis. To do the cross product you need to know that angle, which is just the latitude.

AM