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.
angle = 15 degrees
F(friction) = ?
Coefficient of friction = ?
(I dont think rolling friction is a part of this problem as we have not been taught it yet, but I could be wrong)
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?