Coefficient of Static Friction with Acceleration

[PeachBanana]

Homework Statement

A car can decelerate at -4.90 m/s^2 without skidding when coming to rest on a level road.
What would its deceleration be if the road were inclined at 11° uphill? Assume the same static friction coefficient.

Homework Equations

Force of Static Friction / Normal Force = coefficient of static friction
F (normal) - mg *sin 11° = 0

The Attempt at a Solution

(4.90m/s^2)/(9.8 m/s^2) = 0.5

I didn't know the mass so I just left it out.
F - mg *sin 11° = 0
F (normal) = mg * sin 11°
F = (9.8 m/s^2) * sin 11°
Normal force = 1.86 m/s^2

I know those are acceleration units so something more than likely went wrong there.

Force of static friction = (0.5)(1.86 m/s^2)
Force of Static Friction = 0.93 m/s^2

 

[Doc Al]

F (normal) - mg *sin 11° = 0

I don't understand this equation.

The Attempt at a Solution

(4.90m/s^2)/(9.8 m/s^2) = 0.5

Good.

To find the acceleration, use ƩF = ma. What's the net force when going uphill?

(Hint: What are the parallel and perpendicular components of the weight?)

 

[PeachBanana]

Doc Al - Sorry, the first one was a little unclear. I was attempting to find the normal force by saying the normal force equals the y component of the weight because there is no acceleration in the "y" direction. I think I should have used cosine instead of sine. Thank you for your advice.

 

[Doc Al]

I think I should have used cosine instead of sine.

Exactly.

 

[asteriatic]

The deceleration of the car on a level road is -4.90 m/s^2 and the normal force is 9.8 m/s^2. When the road is inclined at 11° uphill, the normal force decreases due to the component of the car's weight acting in the direction of the incline. Using the equation for the force of static friction (F=μN), we can calculate the coefficient of static friction (μ) as 0.93 m/s^2 / 9.8 m/s^2 = 0.095. This means that the car's deceleration on the inclined road would be -0.095 * 9.8 m/s^2 = -0.931 m/s^2. Therefore, the car would decelerate at a slower rate on the inclined road compared to the level road due to the decrease in normal force caused by the incline.