Decelerating Car

[Mivz18]

I'm having trouble with the third part to this question:

On a level road with its brakes on, the shortest distance in which a car traveling with 93 km/hr can stop is 91 m. This shortest distance occurs when the driver uses anti-lock brakes which means that that the car brakes without skidding.

The first part asks what the magnitude of the acceleration of the car is and I found that to be the absolute value of 3.67.

The second part asks what the coefficient between the tires and the pavement is and I found that to be 0.374 .

The third part then asks How steep a hill can the car park on, with the angle to the horizontal. I have no clue how to go about this. The only hint or help given on the online program is "Since the car stopped without skidding, the coefficient of friction you found in part (b) is actually the static coefficient of friction, which is the proper coefficient of friction for this part as well. "

How do I go about figuring this out with the information I have?

[NateTG]

Can you draw an FBD for a block (car) resting on an inclined plane?

[Mivz18]

ok, I get three forces. I don't know how to demonstrate that here, but there is a normal force, static friction going up and mg force angled down in which you find by means of vector addition. So where would I go from there?

[NateTG]

Well, if you say that the angle of the slope is \theta then you should be able to calculate the component of gravity that acts along the slope which must be counteracted by, and the component which acts perpendicular to the slope which must be counteracted by the normal force. If you recall that F_{friction}=\mu_s N you should be able to set up an equation with \theta as the unknown.

[Mivz18]

I'm still kind of lost on this problem. When I try to look for Fnet, I get :

Fnet = W + N + Fs or
Fnet = mg + mgy + (mu)mg or
Fnet = mg + mgcosx + (mu)mg

Am I doing this correct so far??

[Doc Al]

You are correct that there are three forces acting on the car:
gravity, acting down
friction, acting up the incline
normal force, acting perpendicular to the incline

Now analyze the components of those forces parallel and perpendicular to the plane. (Since the car isn't moving, it's in equilibrium: the net force is zero in any direction.) Get two equations, one for each of those directions. Combine them to solve for the angle of the incline.

[Mivz18]

I'm in a bind. I figured out the Fnet for each direction:

Fnet of x = 0 = N - mg of y = mg - mgcosx
Fnet of y = 0 = Fs - mg of x = (mu)mg - mgsinx

Then, when I try to solve for theta (x), I get mg(1 - cosx) = mg( (mu) - sinx)
Simplified, 1-cosx = (mu) - sinx or 0.626 = cosx - sinx

Where do I go from here??? Is this a trig equation trick that I'm overlooking? Because it isn't a half angle or anything like that? Or am I going in the wrong direction with this?

[Doc Al]

I'm in a bind. I figured out the Fnet for each direction:

Fnet of x = 0 = N - mg of y = mg - mgcosx
Fnet of y = 0 = Fs - mg of x = (mu)mg - mgsinx

Your force equations have an error: the normal force is not mg. I'll rewrite them:
(1) N = mg cos\theta
(2) \mu N = mg sin\theta

Now see if you can solve for \theta.