High School

A tow truck tows a 1550 kg car along a level road. The cable attaching the car to the tow truck makes an angle of 37° above the horizontal. The coefficient of friction between the car and the road is [tex]$u=0.198$[/tex], and the tension in the cable is [tex]$6510 \, N$[/tex]. Determine the magnitude of the acceleration of the car.

Answer :

The car's acceleration is determined by calculating the net force and using Newton's second law. The acceleration is approximately 2.30 m/s².

To determine the acceleration of a 1550 kg car towed by a truck, we need to consider the forces involved:

  1. Calculate the force due to friction.
  2. Resolve the tension in the cable into horizontal and vertical components.
  3. Calculate the net force and use it to determine the acceleration.

Step-by-Step Solution:

  • Force of Friction: The force of friction [tex]F_{friction}[/tex] is calculated using the formula

[tex]F_{friction[/tex] = μ × N

where μ is the coefficient of friction (0.198) and N is the normal force.

  • Normal Force Calculation: The normal force is influenced by the vertical component of the tension, [tex]T_{vertical[/tex] and the gravitational force (mg).

N = mg - [tex]T_{vertical[/tex] .

  • Given, T = 6510 N and the angle θ = 37°.

[tex]T_{vertical[/tex] = T × sin(37°)

[tex]T_{vertical[/tex] = 6510 N × 0.6018 ≈ 3917 N.

  • N = mg - [tex]T_{vertical[/tex]

N = (1550 kg × 9.8 m/s²) - 3917 N ≈ 12183 N - 3917 N = 8266 N.

N ≈ 12183 N - 3917 N

N ≈ 8266 N.

  • Friction Force:

[tex]F_{friction[/tex] = μ × N = 0.198 * 8266 N ≈ 1636 N.

  • Resolve Tension into Horizontal Component:

[tex]T_{horizontal[/tex] = T * cos(37°)

[tex]T_{horizontal[/tex] = 6510 N * 0.7986 ≈ 5200 N.

  • Net Force:

F = [tex]T_{horizontal[/tex] - [tex]F_{friction[/tex] = 5200 N - 1636 N = 3564 N.

  • Acceleration: Using Newton's second law, F = ma, we can solve for the acceleration a.

a = F / m

a = 3564 N / 1550 kg ≈ 2.30 m/s².

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