Question

The 1500-kg truck reaches a speed of 50 km/h from rest in a distance of 60 m up the 10% incline with constant acceleration. Calculate the normal force under each pair of wheels and the effective coefficient of friction between the tires and the road during this motion.​

Answer 1

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Answer 2

Answer:

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Solving for Linear Acceleration:

V2f−V2o=2asVf2−Vo2=2as

since object starts from rest and distance traveled is 60m:

we get:

a=1.608m/s2a=1.608m/s2

Solving for Angle of Incline:

θ=5.71degθ=5.71deg

Solving for Equilibrium of Forces:

∑Fy=0∑Fy=0

N1+N2=(1500)(9.81)cos(θ)N1+N2=(1500)(9.81)cos⁡(θ)

∑Fx=ma∑Fx=ma

F−Wsin(θ)=maF−Wsin⁡(θ)=ma

F=3880NF=3880N

Establising Moment at Center of Gravity:

∑MG=0∑MG=0

1.5(N1)+3880(0.6)−1.5N2=01.5(N1)+3880(0.6)−1.5N2=0

Solving for N1 and N2

N1=6550NN1=6550N

N2=8100NN2=8100N

Solving for Coefficient of Friction:

SinceN2>N1SinceN2>N1

F=μkN2F=μkN2

μk=0.48μk=0.48