Free body diagram, normal force, static friction

[cafe au lait]

Homework Statement

There is a roller coaster cart with passengers sitting on the top of a 23 degree incline. The ride is about to start. The combined mass of the cart and passengers is 363kg.

a) What would the free body diagram look like?
b) How do you solve for normal force?
c) How do you solve for maximum value of static friction AND coefficient of static friction?

m=363kg
g=9.81m/s²
theta=23 degrees

Homework Equations

F_normal=mg*cos(theta)
coefficient of static friction=tan(theta)
maximum value of static friction=coefficient of static friction*F_normal

The Attempt at a Solution

a) F_g arrow pointing straight down, F_normal arrow pointing perpendicular to the cart, F_friction arrow on right or left side of cart and F_pull arrow on the other.

b) F_normal=(363kg)(9.81m/s²)cos(23degrees)=3277.95N

c) coefficient of static friction=tan(23degrees)=0.42
maximum value of static friction=0.42(3277.95N)=1376.74N

I'm not very confident in my physics abilities so I was hoping someone could reassure me that I did this right. :) Thanks in advance!

 

[tiny-tim]

welcome to pf!

hi cafe au lait! welcome to pf!

(have a theta: θ and a mu: µ and a degree: ° )

a) F_g arrow pointing straight down, F_normal arrow pointing perpendicular to the cart, F_friction arrow on right or left side of cart and F_pull arrow on the other.

fine, except a roller-coaster doesn't have an F_pull … it coasts!

b) F_normal=(363kg)(9.81m/s²)cos(23degrees)=3277.95N

fine … there's no acceleration in the normal direction, so F_N = mg cos23°

c) coefficient of static friction=tan(23degrees)=0.42
maximum value of static friction=0.42(3277.95N)=1376.74N

i think you're not meant to do it that way, but by taking components along the slope …

use an equation like F_N = mg cos23° to get F_fr, and then calculate µ from F_fr and F_N

 

[cafe au lait]

Hi thanks for your help!

A) Okay, so how about I replace F<sub>pull with Fpush? My thinking behind that is a force (push) holds the cart at the top of the track until it's time to let go...

B) If there was acceleration, would that change the equation to FN=ma cos θ?

C) Ffr=(363kg)(9.81m/s<sup>2) cos23°=3277.95N

µ=Ffr/FN
µ=3277.95N/3277.95N=1

Uh that doesn't look right :(</sup></sub>

 

[tiny-tim]

hi cafe au lait!

A) Okay, so how about I replace F_{pull with Fpush? My thinking behind that is a force (push) holds the cart at the top of the track until it's time to let go...}

if the brake is on, then you're correct … there must be either a push or pull force

but i think that when the question says "The ride is about to start", it means that the brake has just been taken off

(otherwise it'd say something like "The ride has not started yet")

B) If there was acceleration, would that change the equation to F_{N=ma cos θ?}

that doesn't make sense … how can there be a normal acceleration if the rollercoaster stays on the track?

C) F_{fr=(363kg)(9.81m/s^{2) cos23°=3277.95N}}

_{^{i said an equation like that, not that actual equation!}}