Find Force of Friction for 990 kg Car on 4° Incline

[kimikims]

Any help?

A(n) 990 kg car is parked on a 4 degrees incline.
The acceleration of gravity is 9.8 m/s^2.
Find the force of friction keeping the car
from sliding down the incline. Answer in
units of N.

 

[quasar987]

Newton's second law of motion states that the force on an object is equal to its mass times its acceleration.

[tex]F=ma[/tex]

This means that if there is no (net!) force on an object (F=0), it has no acceleration either (a=0). This means that it's velocity doesn't change. This means that if the object is moving at a certain speed, it keeps doing so. And if the object is standing still, it keeps standing still.

With this in mind, if there is a force on our car (namely the force of gravity) and we want our car to stay still on the incline, this means we want the total force on it in the direction parallel to the incline to be 0.

You go from there.

(Hint: chose your system of coordinate parallel to the incline.)


P.S. It's A car

 

[wais]

To find the force of friction on the car, we can use the formula Ff = μmgcosθ, where μ is the coefficient of friction, m is the mass of the car, g is the acceleration of gravity, and θ is the angle of the incline. In this case, we have μ = 0.3 (assuming a typical value for the coefficient of friction between tires and pavement), m = 990 kg, g = 9.8 m/s^2, and θ = 4 degrees.

Plugging these values into the formula, we get:

Ff = (0.3)(990 kg)(9.8 m/s^2)(cos 4°)

= 2,886.4 N

Therefore, the force of friction keeping the car from sliding down the incline is approximately 2,886.4 N. It is important to note that this is the maximum possible friction force, as it assumes the car is on the brink of sliding down the incline. The actual force of friction may be slightly lower depending on the specific conditions of the car and the incline.