A car with front-wheel drive weighs 3000 lb. The coefficient of static friction is μₛ = 0.5 and the coefficient of kinetic friction is μₖ = 0.4 between the car’s tires and the road.

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Forces on a Car on an Inclined Slope | Physics Solution

Forces on a Car on an Inclined Slope

Question:
A car with front-wheel drive weighs 3000 lb. The coefficient of static friction is μₛ = 0.5 and the coefficient of kinetic friction is μₖ = 0.4 between the car’s tires and the road.

(a) If the car is stationary on a slope inclined at α = 15°, find all the forces acting on the tires by the road.
(b) What is the largest angle α the car can drive up at constant speed?

🧮 Part (a) – Forces on the Car at Rest

Given:
  • Weight of car, W = 3000 lb
  • Angle of incline, α = 15°
  • Static friction coefficient, μₛ = 0.5
1. Normal Force (N):
N = W × cos(α)
N = 3000 × cos(15°) ≈ 3000 × 0.9659 = 2897.7 lb
2. Component of Gravitational Force Along the Slope (F‖):
F‖ = W × sin(α)
F‖ = 3000 × sin(15°) ≈ 3000 × 0.2588 = 776.4 lb
3. Maximum Static Frictional Force (Fₛ):
Fₛ = μₛ × N = 0.5 × 2897.7 = 1448.85 lb

Since Fₛ > F‖, the car remains stationary. All forces are balanced.
Summary of Forces:
  • Gravitational force parallel to slope: 776.4 lb
  • Normal force: 2897.7 lb
  • Static frictional force: 1448.85 lb

🚗 Part (b) – Maximum Incline for Driving at Constant Speed

For driving at constant speed, the force pulling the car downhill (component of weight) must be balanced by the static friction force.
Let’s solve:
Fₛ = W × sin(α)
1448.85 = 3000 × sin(α)
sin(α) = 1448.85 / 3000 = 0.483
α = sin⁻¹(0.483) ≈ 28.8°
✅ Final Answers:

(a) Forces on the stationary car:
  • Gravitational force parallel to the slope: 776.4 lb
  • Normal force: 2897.7 lb
  • Static frictional force: 1448.85 lb
(b) Largest incline angle for constant-speed drive: 28.8°

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