Car on a hill - reasoning for static friction

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1. Apr 14, 2017

gelfand

• Member advised to use the homework template for posts in the homework sections of PF.
Problem :

A car of mass $m$ is parked on a slope.

What does static friction $\mu_s$ have to be for the car not to slide down if
the incline of the hill is $30$ degrees?

*******

For this problem I have the following forces, where the hill is inclined upwards
to the right:

Upwards : normal force of $mg \cos(\theta)$

Downwards : Force due to gravity of $mg \cos(\theta)$

Right (so up hill): Resistive Force due to friction $\mu_s mg \cos(\theta)$

Left (down hill): Force due to gravity : $mg \sin(\theta)$.

For the car to remain in place I need to have the frictional force to be greater
than the gravitational, if I set these up as

$$\mu_s mg \cos(\theta) = mg \sin(\theta)$$

Then reduce to
$$\mu_s = \tan(\theta)$$

Where $\theta = 30$ (degrees) , so $\mu_s = \frac{\sqrt{3}}{3}$, which is around
$0.58$. I *think* that this is a reasonable coefficient for friction? It's less
than 1, though I've not really got anything to go on there.

I'm not sure what else I could do / say about this problem either. I think that
I've considered all the forces, noted that the maximum that friction could be is
that of the equality $\mu_2 \cos(\theta) = \sin(\theta)$ (I've dropped $mg$
there...)

Static frictional force can change - whereas the kinetic friction is constant.
However - both coefficients $\mu_s$ and $\mu_k$ are themselves constants, so my
equation to find the coefficient $\mu_s$ is valid.

I've attached sketches of this problem as they were requested in the equation.

One other thing that I'm slightly unsure of is the expression for the force of
friction here.

I know that the force of friction is the normal multiplied by the frictional
coefficient. So in this case I have the normal as $mg \cos(\theta)$, giving
$\mu_s mg \cos(\theta)$ as the force of friction. I was slightly unsure because
it's actually $\sin(\theta)$ which is in the direction of the hill here.

Thanks.

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2. Apr 14, 2017

kuruman

Your analysis is correct. Note that there is no reason for μs to be less than unity.

On edit: Note that μs FN is the maximum force of static friction before the object starts sliding.

3. Apr 14, 2017

gelfand

Oh ok - i thought that the frictional coefficient was less than one for some reason.

When might I encounter something greater than one?

Thanks

4. Apr 14, 2017

kuruman

5. Apr 14, 2017

gelfand

6. Apr 14, 2017

kuruman

Yes, greater than 1 is not encountered often, but it is not an upper limit for some physical reason. Don't forget that if you place an object on an incline and you gradually increase the angle, sliding occurs at the critical angle θc such that μs = tan(θc). Having μs greater than 1 implies that the critical angle is greater than 45o. Everyday experience has us believe that this is not seen often.