Question

2.5 x 103 kilogram truck with rubber tires moves through a 120 meter radius turn on a dry asphalt surface.

Answer 1

Answer:

a. The friction force acting on the truck's tires during the turn, is approximately 20,875 N

b. The maximum velocity with which the truck could make the bend is approximately 31.654 m/s

Explanation:

The given parameters of the truck are;

The mass of the truck, m = 2.5 × 10³ kg

The tire material = Rubber

The radius of the turn the through which the truck moves = 120 meters

The material and condition of the surface = Dry asphalt

a. The coefficient of friction of rubber tires on dry asphalt, μ ≈ 0.85 (researchgate (internet source))

The weight of the truck, W = m × g

Where;

m = The mass of the truck = 2.5 × 10³ kg

g = The acceleration due to gravity = 9.8 m/s²

∴ W = 2.5 × 10³ kg × 9.8 m/s² = 24,500 N

The force of friction acting on the tires, $F_f$ = W × μ

∴ $F_f$ = 24,500 N × 0.85 ≈ 20,875 N

The friction force acting on the truck's tires during the turn, $F_f$ ≈ 20,875 N

b. The centrifugal force the truck observes while turning through the bend, 'F', is given as follows;

F = m·v²/r

Therefore, the maximum velocity with which the truck could make the bend, 'v', is given when fiction force, '$F_f$', is equal to the centripetal force, F

When $F_f$ = F, we have;

W × μ = m·v²/r

∴ 20,875 N = 2.5 × 10³ kg × v²/(120 m)

∴ v² = 20,875 N × (120 m)/(2.5 × 10^3 kg) = 1002 m²/s²

v = √(1002 m²/s²) ≈ 31.654 m/s

The maximum velocity with which the truck could make the bend, v ≈ 31.654 m/s