Vehicle resistance on grades — gradient (grade) resistance for a 980 kg car at 10° incline Compute the component of weight acting down the slope (gradient resistance) for a vehicle of mass 980 kg climbing a 10° incline. Use g = 9.81 m/s^2.

Introduction / Context:
When a vehicle climbs an incline, a component of its weight acts along the slope and opposes motion. This is called gradient (or grade) resistance and is essential for tractive effort and power calculations.

Given Data / Assumptions:

• Mass m = 980 kg.
• Incline angle θ = 10°.
• Acceleration due to gravity g = 9.81 m/s^2.
• Neglect rolling and aerodynamic resistances; compute only the gravitational component along the slope.

Concept / Approach:
The gravitational force is W = m * g. The component acting along the slope is W * sin(θ). This is the gradient resistance that the powertrain must overcome (in addition to other resistances) to maintain speed uphill.

Step-by-Step Solution:
W = m * g = 980 * 9.81 = 9603.8 Nsin(10°) ≈ 0.173648Gradient resistance = W * sin(θ)Gradient resistance = 9603.8 * 0.173648 ≈ 1669.4 N

Verification / Alternative check:
For small angles, sin(θ) ≈ θ in radians. 10° ≈ 0.1745 rad; 9603.8 * 0.1745 ≈ 1675 N, close to the precise 1669.4 N, confirming the calculation.

Why Other Options Are Wrong:

• 1.6694 N and 16.694 N: orders of magnitude too small.
• 166.94 N: off by a factor of 10.

Common Pitfalls:
Using cos(θ) instead of sin(θ) or forgetting to convert degrees to radians when using approximations. Also, mixing mass (kg) with weight (N) is a frequent mistake.

Final Answer:
1669.4 N