Coriolis Force on a Race Car at 45 Degrees North

[OmegaKV]

Homework Statement

Find the magnitude and direction of the Coriolis force on a racing car of mass 10 metric tons traveling due south at a speed of 400km/hr at a lattitude of 45 degrees north.

Homework Equations

$$F_{cor}=-2m\omega\times v$$

The Attempt at a Solution

$$\omega=2\pi/(24*3600 \quad seconds)$$
$$v=400km/hr=400/3600 \quad km/second$$
$$m=10000$$
$$-2 m\omega\times v=-2m\omega v sin(45) \quad Newtons \quad east = -114.27 \quad Newtons \quad east = 114.27 \quad Newtons \quad west$$

The answer in the back of my book says "41 nt west"

 

[jbriggs444]

$$v=800km/hr=800/3600 \quad km/second$$

v = 400 km/hour according to the problem statement.
I do not see you account for the mass of the race car.
Math not checked.

 

[OmegaKV]

v = 400 km/hour according to the problem statement.
I do not see you account for the mass of the race car.
Math not checked.

Updated my post to fix those mistakes. I mistyped 800km/hr in my post but I used 400km/hr in my calculation. Taking mass into account my answer is scaled by a factor of 10000, so it's still different from the answer in the back of the book.

 

[jbriggs444]

My result matches yours. Note that the race car mass is only given to two significant figures. Our result should be reported as 110 N (or 1.1x10²N)