Forces acting on a car on an incline

[TheBigDig]

Homework Statement

A 1100 kg car is parked on an incline of 1 in 30. What is the normal force acting on the car? In the absence of friction what would be the acceleration if the handbrake were released?

Relevant Equations

$\vec{F} = \mu_k \vec{N}$
$F = mgsin\alpha$
$N = mgcos\alpha$

For part 1, I got $tan \alpha = 1/30$
$\alpha = 1.9^{\circ}$
$mgcos(1.9) = 10774N$

I'm a little thrown off by the second part. Are we supposed to assume that in the absence of friction, F = N and then substitute F = ma to solve for this?

 

[haruspex]

Homework Statement:: A 1100 kg car is parked on an incline of 1 in 30. What is the normal force acting on the car? In the absence of friction what would be the acceleration if the handbrake were released?
Relevant Equations:: $\vec{F} = \mu_k \vec{N}$
$F = mgsin\alpha$
$F = mgcos\alpha$

F = N

It would help if you were to distinguish forces in different directions instead of writing "F" in every case.
Which direction is the normal force? In which direction will the car move if the brake is released?

 

[TheBigDig]

It would help if you were to distinguish forces in different directions instead of writing "F" in every case.
Which direction is the normal force? In which direction will the car move if the brake is released?

Sorry yes of course. The normal force is perpendicular to the surface and the car will roll backward. I believe the model should correspond to this

 

[James2020]

The force acting parallel to the slope causes the car’s acceleration so, since F=ma, you have mgsin(θ)=ma so a=gsin(θ).

 

[PeroK]

Homework Statement:: A 1100 kg car is parked on an incline of 1 in 30. What is the normal force acting on the car? In the absence of friction what would be the acceleration if the handbrake were released?

If there is no friction, then whether the handbrake is on or not is irrelevant. The car will slide down the hill in any case.

If there is enough friction, then the handbrake will hold the car, but the car will roll down the hill if the handbrake is released.