# Forces acting on a car on an incline

• TheBigDig
In summary: So, whether the handbrake is on or not, the car will move down the hill when the handbrake is released.In summary, a 1100 kg car parked on an incline of 1 in 30 will have a normal force acting on it of approximately 10774N. In the absence of friction, the car will slide down the hill with an acceleration of approximately 9.8sin(1.9) m/s^2. Whether the handbrake is on or not, the car will move down the hill when the handbrake is released.
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?

TheBigDig said:
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?

haruspex said:
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

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).

TheBigDig said:
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.

hutchphd

## 1. What is the force of gravity acting on a car on an incline?

The force of gravity acting on a car on an incline is equal to the car's mass multiplied by the acceleration due to gravity (9.8 m/s²).

## 2. How does the angle of the incline affect the forces acting on a car?

The steeper the incline, the greater the component of the car's weight that acts parallel to the incline. This results in a greater force pulling the car down the incline.

## 3. What other forces besides gravity act on a car on an incline?

Other forces that may act on a car on an incline include friction, air resistance, and the normal force from the surface the car is on. These forces may affect the car's motion and require additional calculations.

## 4. How does the weight of the car affect the forces on an incline?

The weight of the car directly affects the force of gravity acting on it. The heavier the car, the greater the force of gravity and the more force is required to move the car up the incline.

## 5. How can the forces acting on a car on an incline be calculated?

The forces acting on a car on an incline can be calculated using Newton's laws of motion and trigonometry. The weight of the car and the angle of the incline can be used to determine the force of gravity and the normal force, while the coefficient of friction and the angle of the incline can be used to calculate the force of friction. These forces can then be used to determine the net force on the car and its resulting motion.

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