Posted by **Morgan** on Sunday, March 7, 2010 at 6:41pm.

Okay, I saw this one previously posted, but the answer it kept leading me to was wrong.

A sports car of mass 1400kg (including the driver) crosses the rounded top of a hill (radius 88m) at 23 m/s. Find the normal force exerted by road on the car.

I thought that the normal force was mass times acceleration, just what the road is doing to the car. So I multiplied 9.8 times 1400, but that was wrong. What am i doing wrong? The same applies to a driver (weighing 75kg in the car)

- Physics -
**bobpursley**, Sunday, March 7, 2010 at 6:42pm
normal force= apparent weight

= mg-mv^2/r

- Physics -
**Morgan**, Sunday, March 7, 2010 at 6:46pm
So, basically, the way I can look at problems like these is to subtract the two accelerations (normal's gravity and centripetal) then multiply them by the mass (F=ma)?

- Physics -
**bobpursley**, Sunday, March 7, 2010 at 6:47pm
yes.

- Physics -
**Alex**, Saturday, September 18, 2010 at 11:59pm
Why is the centripetal force subtracted from the force of gravity? Aren't both going in the same direction (since centripetal force is always in the direction of the center of the circle)

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