Determine acceleration on slope using coefficient of static friction

In summary, the maximum acceleration of the sports car on the hill is 11.6 m/s^2, considering the coefficient of static friction and the angle of the incline. Mass is not necessary for this calculation.
  • #1
linzeluv
2
0
THE PROBLEM:
A sports car is accelerating up a hill that rises 21.2 ° above the horizontal. The coefficient of static friction between the wheels and the road is μs = 0.880. It is the static frictional force that propels the car forward. (a) What is the magnitude of the maximum acceleration that the car can have? (b) What is the magnitude of the maximum acceleration if the car is being driven down the hill?


I set up my fbd with the static frictional force acting in the -x direction, the normal force acting perpedicular to the inclined plane, and the weight acting in the -y direction. Without knowing the weight of the car or if it is at a constant velocity, I'm not sure how to go about setting up this problem.
 
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  • #2
Welcome to PF linzeluv,

I think you'll find the in the end you don't need the mass of the car, it'll just cancel out in your calculations.

Secondly, you already know that it isn't traveling at a constant velocity because it's accelerating!
 
  • #3
Setting up the equation

I thought this would be on the right track, but my answer for part A still isn't right:

Maximum Acceleration = Net Force in X-direction/ mass = -mg sin 21.2deg + maximum static frictional force/ mass

Maximum static frictional force = coefficient of static friction * Normal Force

Net Force in y-direction = ma = 0

Net Force in y-direction = Normal Force - mg cos 21.2deg = 0
Normal Force = mg cos 21.2deg

Final Equation

Maximum acceleration = -mg sin 21.2deg + .880 (mg cos 21.2deg)/ mass

Mass cancels out and answer is 11.6 m/s^2

Not sure where I'm going wrong.
 
  • #4
linzeluv said:
I thought this would be on the right track, but my answer for part A still isn't right:

Maximum Acceleration = Net Force in X-direction/ mass = -mg sin 21.2deg + maximum static frictional force/ mass

Maximum static frictional force = coefficient of static friction * Normal Force

Net Force in y-direction = ma = 0

Net Force in y-direction = Normal Force - mg cos 21.2deg = 0
Normal Force = mg cos 21.2deg

Final Equation

Maximum acceleration = -mg sin 21.2deg + .880 (mg cos 21.2deg)/ mass

Mass cancels out and answer is 11.6 m/s^2

Not sure where I'm going wrong.
All your working and your final equation is correct, but your final answer is wrong. You must have punched the numbers into your calculator incorrectly.
 
  • #5
make sure your calculator is in degrees mode not radians, I fell foul of that several times :-)
 

What is the coefficient of static friction?

The coefficient of static friction is a dimensionless quantity that measures the maximum amount of friction that can exist between two surfaces in contact before they start to move relative to each other.

How is the coefficient of static friction related to acceleration on a slope?

The coefficient of static friction is directly related to the acceleration on a slope. It is a factor in the calculation of the maximum force that can be exerted on an object before it starts to slide down the slope.

How do you determine the coefficient of static friction?

The coefficient of static friction can be determined experimentally by measuring the force required to start an object moving on a surface and dividing it by the normal force (the force perpendicular to the surface) acting on the object.

What factors affect the coefficient of static friction?

The coefficient of static friction can be affected by the nature of the surfaces in contact, the roughness of the surfaces, and the presence of any lubricants or contaminants.

Is the coefficient of static friction the same for all surfaces?

No, the coefficient of static friction can vary depending on the materials and surfaces in contact. For example, the coefficient of static friction between rubber and glass will be different than the coefficient of static friction between rubber and concrete.

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