Incline Friction Problem – Forces on Car Tires

Physics: Forces on a Car on an Inclined Road

Question:

See figure below. The coefficient of static friction between the car’s tires and the road is μ_s = 0.5, and the coefficient of kinetic friction is μ_k = 0.4. The weight of the car is 3000 lb, and the car has front-wheel drive.

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

Answer:

Part (a): Stationary Car on a 15° Slope

Given:

Weight (W) = 3000 lb
Angle α = 15°
μ_s = 0.5

1. Break Weight into Components

The car’s weight is resolved into components:

W_∥ = W × sin(α) = 3000 × sin(15°) ≈ 3000 × 0.2588 = 776.4 lb

W_⊥ = W × cos(α) = 3000 × cos(15°) ≈ 3000 × 0.9659 = 2897.7 lb

2. Static Friction

The friction force must oppose the component pulling the car downhill:

f_s(max) = μ_s × N = 0.5 × 2897.7 = 1448.9 lb

Since f_s needed = 776.4 lb < f_s(max), static friction is sufficient to hold the car.

3. Normal Force

N = W_⊥ = 2897.7 lb

✅ Final Answer for (a):

Normal Force (N): 2897.7 lb
Static Friction Force (uphill): 776.4 lb

Part (b): Maximum Climbable Angle at Constant Speed

Now: The car is moving up the slope at constant speed. That means we now use kinetic friction (μ_k) since there’s motion involved.

1. Force Balance for Constant Speed

To maintain constant speed, the driving force must overcome gravitational pull and kinetic friction:

Driving Force = W × sin(α) + f_k

At the limiting case (maximum angle), the friction is:

f_k = μ_k × N = μ_k × W × cos(α)

Equating driving force to resistive forces:

μ_k × cos(α) = sin(α) ⟹ tan(α) = μ_k

Now solve for α:

α = tan^-1(0.4) ≈ 21.8°

✅ Final Answer for (b):

The largest angle α the car can drive up at constant speed is 21.8°

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