Maximum incline angle for parked car with 0.8 friction coefficient

[Medgirl314]

Homework Statement

On how steep an incline(in degrees) can a car be parked without sliding down if the coefficient of static friction between the tires and the road is 0.8?

Homework Equations

F=ma

The Attempt at a Solution

My physics teacher showed how to find the acceleration for an object on an incline. He got the equation a=gsin(theta)-(mu)gcos(theta)

(Sorry, I can't seem to find the sybmols for mu and theta on LaTex.


I understand how he got the equation. It makes perfect sense. However, I'm having a hard time transforming the equation to get the angle of inclination. Adding mgcos(theta) to both sides yields (mu)gcos(theta)+a=gsin(theta). I would think that the acceleration is negligible since it's not moving, and I think the g's cancel out, correct? But this equation just gives 0.8cos(theta)=sin(theta). But where do I go from here? I don't know cos(theta), sin(theta), or theta. I've been looking at the theta as part of separate terms, but if they canceled out, then it would just be 0.8cos=sin. I tried the inverse of this on a hunch and got sin=36.9 degrees. This seems pretty reasonable, is this my answer? If so, why is it my answer?

Thanks so much!

 

[Rococo]

Homework Statement

On how steep an incline(in degrees) can a car be parked without sliding down if the coefficient of static friction between the tires and the road is 0.8?

Homework Equations

F=ma

The Attempt at a Solution

My physics teacher showed how to find the acceleration for an object on an incline. He got the equation a=gsin(theta)-(mu)gcos(theta)

(Sorry, I can't seem to find the sybmols for mu and theta on LaTex.


I understand how he got the equation. It makes perfect sense. However, I'm having a hard time transforming the equation to get the angle of inclination. Adding mgcos(theta) to both sides yields (mu)gcos(theta)+a=gsin(theta). I would think that the acceleration is negligible since it's not moving, and I think the g's cancel out, correct? But this equation just gives 0.8cos(theta)=sin(theta). But where do I go from here? I don't know cos(theta), sin(theta), or theta. I've been looking at the theta as part of separate terms, but if they canceled out, then it would just be 0.8cos=sin. I tried the inverse of this on a hunch and got sin=36.9 degrees. This seems pretty reasonable, is this my answer? If so, why is it my answer?

Thanks so much!

You have the equation:

$0.8cosθ=sinθ$

There a simple trigonometric identity relating sinθ and cosθ. You can use it to turn this into an equation involving only one term which has θ in it.

 

[Medgirl314]

Thanks! I thought so! But I've researched it, and looked back in my videos, and plugged cosines and sines of angles into my calculator, and I can't find it anywhere.

 

[Rococo]

Thanks! I thought so! But I've researched it, and looked back in my videos, and plugged cosines and sines of angles into my calculator, and I can't find it anywhere.

The identity you need to use is:

$tanθ=\frac{sinθ}{cosθ}$

 

[Medgirl314]

Okay, thank you! Looking at the diagram, I see how this equation makes sense. However, I still don't see how I have enough information. It looks like the hypotenuse length and the opposite length may be the same, does this mean that the tan and cos functions cancel out, leaving sin(0.8)? I don't think so, because that number seems far too small.

 

[Rococo]

Okay, thank you! Looking at the diagram, I see how this equation makes sense. However, I still don't see how I have enough information. It looks like the hypotenuse length and the opposite length may be the same, does this mean that the tan and cos functions cancel out, leaving sin(0.8)? I don't think so, because that number seems far too small.

You are trying to find the angle of inclination, which is θ.

You have the equation $0.8cosθ=sinθ$

If you divide both sides of this equation by $cosθ$, then you get:

$0.8=\frac{sinθ}{cosθ}$

Now, we know that $\frac{sinθ}{cosθ}=tanθ$

So, we now have the following equation:

$tanθ=0.8$

You are trying to find the angle of inclination, which is θ. You know that $tanθ=0.8$. Can you see how you can now find what θ is?

 

[Medgirl314]

I understand how you got tan(theta)=0.8. I still can't seem to find a trigonometric relationship to get to the answer. If we multiply by the reciprocal we get 0.8(cos)/tan, but that doesn't make sense. I really do want to find the answer myself, so is there a way you can help show the relationship without giving away the answer?

Edit: Is there a inverse relationship I can use? 0.8(inverse tangent)=38.66 degrees.

Thanks so much!

 

[Rococo]

I understand how you got tan(theta)=0.8. I still can't seem to find a trigonometric relationship to get to the answer. If we multiply by the reciprocal we get 0.8(cos)/tan, but that doesn't make sense. I really do want to find the answer myself, so is there a way you can help show the relationship without giving away the answer?

Edit: Is there a inverse relationship I can use? 0.8(inverse tangent)=38.66 degrees.

Thanks so much!

Yes, you know that $tanθ=0.8$. Therefore:

$θ=tan^{-1}(0.8)$

Which gives a value of $θ=38.66°$.

 

[Medgirl314]

Ah, right. I forgot the "why" of the inverse function for a moment, and just went off a hunch. Thank you! I get it now.