Need help find the centripetal forces

Question

A banked curve makes it safer to drive on curved roads even if it seems more dangerous. The following is a 1200kg car on a banked curve at 20 degrees from the horizontal. However, the car is driving at 25 m/s, which is actually higher than the recommended speed limit for the banked curve. This means friction is required to provide the remaining force needed to complete the turn. The radius is 80m.

a) Find the centripetal force needed to keep this car at 25 m/s, in N.

b) Find the amount of centripetal force provided only by the normal force, in N.

c) Find the remaining amount of centripetal force needed from friction to keep the car moving at 25 m/s, in N.

Answer 1

A great figure / breakdown for the forces (minus friction) in this scenario is here: http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/carbank.html

1) The centripetal force is defined as F = ma = m v² / r. Given the values in this problem, you should get F = 9,375 N

2) Ignoring friction and using a standard horizontal/vertical x/y coordinate system, we have the following following forces. This figure may help: https://image3.slideserve.com/6740164/banked-curves35-l.jpg

∑F_x = N sin θ = Centripetal Force

ΣF_y = N cos θ - mg = 0

Therefore, N = mg / cos θ, and the Centripetal Force = N sin θ = (mg / cos θ) ⋅ sin θ

Centripetal Force = mg tan θ = 4,280 N

3) The answer from (1) gave us the total centripetal force, and the answer from (2) gave us the contribution from the normal force. We just need to take their difference to get the force of friction that contributes to the centripetal force: 5095 N