acceleration braking-force

Question

3.1 The engine of a truck exerts a force of 70 kN on the truck as it travels up an incline of 1°. The truck experiences a resistance of 60 N per ton of the weight of the truck. The total mass of the truck and its engine is 240 ton. Calculate the following: 3.1.1 The acceleration of the truck 3.1.2 The braking force that would be required on the return journey to prevent the acceleration exceeding 0.02 m/s 2

Answer 1

3.1.1)

Exerted force F_n= 70 kN

Inclination angle θ = 1^o

experiencing resistance on the truck = 60 N per ton of the weight

The total mass m = 240 ton

1 ton = 1000 kg

m = 240000 kg

Friction coefficient ( μ ) = tanθ

μ = tan 1^o = 0.01745

Formula for force upwards in inclined friction plane :

F = W (sin θ + μ cos θ)

newton's first law F = ma

ma = mg (sin θ + μ cos θ)

a =  (9.8)( sin 1 + (0.01745)(cos1) )

a = 16.4928 m/s²

The acceleration of the truck is 16.4928 m/s²