Calculating Stopping Distance on 10° Slope w/ Forces of Friction

[Chapin]

A Corvette can brake to a stop from 60 mi/hr (26.82 m/s) in 123 ft (37.49 m) on a flat surface. What is his stopping distance on a roadway sloping downward 10 degrees?

--This question is in our Forces of Friction section, and we can find the acceleration and the coefficient of friction. What equations do we use to get the stopping distance on the 10 deg slope?

 

[Doc Al]

Originally posted by Chapin
--This question is in our Forces of Friction section, and we can find the acceleration and the coefficient of friction. What equations do we use to get the stopping distance on the 10 deg slope?

First find the coefficient of friction using the data for a flat surface. Then, for the sloping case, consider all the forces acting on the car when the brakes are applied. Find the net force, and thus the acceleration. Then you can calculate the new stopping distance.

 

[Chapin]

Reply

OK, I have the deceleration to be -9.59 m/s^2 and the coeffiecent of friction to be .978 on the flat surface.

I am having a problem finding the net force on the slope because mass is not given.

I guess I need to derive an equation to do this but it is kicking my butt.

Does this look right?
\sum F_x=mgsin(10)-.978\\\sum F_y=mgcos(10)

 

[Doc Al]

Originally posted by Chapin
I am having a problem finding the net force on the slope because mass is not given.

Just call it "m" for now; it will drop out.

Does this look right?
\sum\F_x=mgsin(10)-.978

Partly. There are two forces acting on the car along the plane:
- the weight, which is mgsinθ (acting down)
- the friction, which is μN (acting up; where N is the normal force)
The normal force, N, equals mgcosθ so, F_friction = μ mgcosθ.

Thus, F_net= mgsinθ - μ mgcosθ = ma

Solve for a.

 

[Chapin]

Thank you

Thank you, thank you and thank you.