2.5 x 103 kilogram truck with rubber tires moves through a 120 meter radius turn on a dry asphalt surface.
1 answer:
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, = W × μ
∴ = 24,500 N × 0.85 ≈ 20,875 N
The friction force acting on the truck's tires during the turn, ≈ 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, '', is equal to the centripetal force, F
When = 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
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