# Question #9fdb7

Mar 29, 2016

The centripetal force acting on the truck as it rounds the curve must be 400N.

#### Explanation:

If the truck driver holds the wheel steady, then the path the truck takes around the corner is an arc, or segment of a circular path, basically a circle!

We can therefore calculate the centripetal force exerted on the (tires of the) truck (by the road) using equation 1).

1) ${F}_{c} = \frac{m \cdot {v}^{2}}{r}$

Here, $r$, the "radius of the corner”, actually refers to the radius of the circular path the truck is traveling on (as corners are typically pictured as rectangular or angular, and not described by a radius).

$m$ is the mass of the truck and $v$ is the tangential velocity of the truck.

Finding the amount (or magnitude) of the centripetal force simply requires substituting the numerical values for the physical quantities $m$, $v$ and $r$ into equation 1).

But before we do that, let’s make sure that the units of the physical quantities $m$, $v$ and $r$ are consistent. Currently, they are not. So let’s fix this!

Ultimately, in the end, we want a force in newtons, N.

$1 N = 1 k g \cdot 1 \frac{m}{s} ^ 2$

The velocity of the truck, $v$, is given in $\frac{k m}{h}$ (not $\frac{m}{s}$)

So let’s convert $72 \frac{k m}{h}$ to $\frac{m}{s}$. Here’s how we do it.

$72 \frac{k m}{h} = 72 , 000 \frac{m}{h} \cdot \left(\frac{1 h}{60 \min}\right) \cdot \left(\frac{1 \min}{60 s}\right) = 20 \frac{m}{s}$

So now, that our units are consistent, substituting $m = 100 k g$, $v = 20 \frac{m}{s}$ and $r = 100 m$ into equation 1) we get,

2) ${F}_{c} = \frac{m \cdot {v}^{2}}{r} = \frac{100 k g \cdot {\left(20 \frac{m}{s}\right)}^{2}}{100 m} = 400 N$