# What is the force exerted by the tire on the stone? [closed]

A stone with mass $m$ gets stuck in the tire of a car, whose radius is $r$. The car is moving at constant speed $v$.

The problem asks to determine the force $F$ exerted by the tire on the stone, when this is at maximum distance from the ground. Now, I know the centripetal force is $F_1=\dfrac{mv^2}{r}.$ Why is it that in the solution the weight of the stone is subtracted from $F_1$ to get $F$, rather than added? I mean, they're both downward.

## closed as off-topic by AccidentalFourierTransform, stafusa, Chris♦, glS, Jon CusterFeb 19 '18 at 16:09

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## 2 Answers

The centripetal force is the net force inwards.

Because gravity is pointing in the same direction as the tire and net inward force, then the force required by the tire on the stone is less.

Remember that $F_{net}$ is the sum of all the forces.

Here you should take care of centrifugal force not centripetal one. Centrifugal force is acting opposite to the centre thats why at the upper point of the tire mg is downward and mv2/r is upward so to calculate net force you have to subtract 1st on from 2nd one. For more detail you can go to the recommended link..http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html