Calculating Coefficient of Friction for a Truck on an Incline

[menco]

Homework Statement

A truck weighing 15 tonnes is traveling down an decline of 15 degrees at 6.6m in 3s. He then stops and gets out. The truck then begins to roll down the decline as the brakes have failed.

1. What is the size of friction force acting on the rubber tyres of the charging vehicle
2. Coefficient of friction between the rubber tyres and the coal surface.

Knowns/Unknowns
m=15000kg
angle = 15 degrees
d=6.6m
t=3s
g=9.8m/s^2
F(friction) = ?
Coefficient of friction = ?

(I don't think rolling friction is a part of this problem as we have not been taught it yet, but I could be wrong)

Homework Equations

Fg = ma
Fg(parallel)= Fg sin (15)
Fg(perpendicular) = Fg cos (15)
Coefficient = F(friction)/ F(normal)

The Attempt at a Solution

1. Fg = 15000 x 9.8 = 147000 N
Fg(parallel) = 147000 sin (15) = 38046.4 N

Therefore F(friction) is 38046.4 N

2. Fg(perpendicular) = 147000 cos (15) = 141991 N = F(normal)
Coefficient = F(friction)/F(normal) = 38046.4/141991 = 0.27

Therefore the coefficient of friction is 0.27.

Am I at least on the right track if I got it wrong?

 

[tal444]

Looks right to me, although I could be wrong.

 

[menco]

Thanks for the response hopefully it is correct. I am just unsure if I am on the right track with Friction force

 

[menco]

solved thanks can someone please delete?

 

[tal444]

We don't delete finished threads. They're kept for anyone who might search up similar questions in the future.