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

a. Determine the force of friction acting on the truck's tires during the turn.


b. Calculate the maximum speed with which the truck could have made this turn.

Respuesta :

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, [tex]F_f[/tex] = W × μ

∴ [tex]F_f[/tex] = 24,500 N × 0.85 ≈ 20,875 N

The friction force acting on the truck's tires during the turn, [tex]F_f[/tex] ≈ 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, '[tex]F_f[/tex]', is equal to the centripetal force, F

When [tex]F_f[/tex] = 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

samueladesida43
  1. The force of friction acting on the truck's tires during the turn is 20,875 N
  2. The maximum velocity with which the truck could have made the turn is 31.654 m/s.

According to this question;

  1. The mass of the truck (m) = 2.5 × 10³ kg
  2. The radius of the turn = 120m
  3. The coefficient of friction of rubber tires on dry asphalt, μ ≈ 0.85

QUESTION 1:

  • Weight of the truck (W) = m × g

Where;

  • m = mass of the truck = 2.5 × 10³ kg
  • g = 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 force of friction acting on the truck's tires during the turn = 20,875 N

QUESTION 2:

  • Centrifugal force of the truck can be calculated as follows: F = mv²/r

  • However, force of friction (Ff) is equal to centripetal force (F). That is;

  • W × μ = m·v²/r

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

  • v² = 20,875 × 120/2.5 × 10³ = 1002 m²/s²

  • v = √1002 = 31.654m/s

  • The maximum velocity with which the truck could make the turn = 31.654 m/s.

Learn more about force of friction at: https://brainly.com/question/4230804?referrer=searchResults

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