Recent content by Jacky Lee

Hmm, okay. Thank you for all the help! :) I really appreciate it.

Should the friction force be negative? Even if I changed the sign on any of the numbers, my answer would still be incorrect. If I change the sign on either forces, I would get a smaller number and therefore a smaller velocity as an answer.

I made an error, it should be 18003.75N Maximum Frictional Force: ## 9.8 * 1500 * cos(15) = 14199.11 ## Force due to gravity: ## 9.8 * 1500 * sin(15) = 3804.64 ## Added together: ## 14199.11 + 3804.64 = 18003.75 N ## ## (m v^2) / r = 18003.75 ## ## (1500 v^2 / 70) = 18003.75## ## v = 29.0 m/s ##

Hmm, I still don't get the right answer. I tried adding the force of gravity in the direction of the incline to the frictional force (which would just double it), so I got: ## (m v^2) / r = 28398.22 ## ## (1500 v^2 / 70) = 28398.22 ## ## v = 36.4 ##

OH, I'm missing gravity, aren't I?

Forces on the tires: friction, normal, centripetal Am I missing anything?

I figured that when the centripetal force exceeds the maximum frictional force, the car would skid. Solving for the maximum frictional force only requires the normal force, so I didn't include any other forces. Where would I have to use the friction force?

Homework Statement A concrete highway curve of radius 70 m is banked at a 15° angle. What is the maximum speed with which a 1500 kg rubber tired car can take this curve without sliding? The static friction coefficient is 1. Homework Equations Centripetal Force = ## (m v^2) / r ## Maximum...