See if this answers the question from your comment. Sorry it’s a bit long.
You take 40mph (=17.88 m/s) as the speed of the car and everything in it. The pair of dice are *stationary relative to the car*, so the dice are also moving at 40mph in a circular path.
To move in a circular path requires a force called the centripetal force (which equals mv²/r). For the dice (moving in a horizontal circle), the only horizontal force is the horizontal component of tension (Tsin(23º)). So Tsin(23º) is the centripetal force; that’s why we can write Tsin(23º) = mv²/r.
In fact, we are making an approximation. Because of the circular path, the speed is not 40mph everywhere inside the car. The speed is slightly less than 40mph on one side and slightly more than 40mph on the other side. But as long as car’s width is much smaller than the radius of the path, the effect is negligible. And the dice will be near the centre of the car anyway.
Hope that helps.
If the mass of the dice is m, then the weight is mg. Vertically there is no acceleration, so the vertical component of tension and the weight balance:
Tcos(23º) = mg
T = mg/cos(23º)
If the speed is v (converted from mph to metres/second) then the centripetal force acting on the dice is the horizontal component of T (which is Tsin(23º)).
Tsin(23º) = mv²/r
Substituting for T:
(mg/cos(23º))sin(23º) = mv²/r
Cancelling m and using sin/cos = tan gives:
gtan(23º) = v²/r
r = v²/(gtan(23º))