Find the force required to keep a 3000-lb car parked on a hill that makes an angle of 15° with the horizontal.

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Identify the forces acting on the car: the weight of the car acts vertically downward, and the force required to keep the car parked acts along the hill's surface, opposing the component of the weight pulling the car downhill.

Resolve the weight of the car into two components relative to the hill: one perpendicular to the hill's surface and one parallel to the hill's surface. The component parallel to the hill is responsible for the car's tendency to slide down.

Use the formula for the component of the weight parallel to the incline: \(F = W \times \sin(\theta)\), where \(W\) is the weight of the car (3000 lb) and \(\theta\) is the angle of the hill (15°).

Substitute the given values into the formula: \(F = 3000 \times \sin(15^\circ)\).

Calculate \(\sin(15^\circ)\) using a calculator or trigonometric table, then multiply by 3000 to find the force required to keep the car parked on the hill.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Resolving Forces on an Inclined Plane

When an object rests on a slope, its weight can be split into components parallel and perpendicular to the incline. The parallel component causes the object to slide down, calculated as weight times sine of the angle, while the perpendicular component presses into the surface.

Trigonometric Functions in Force Analysis

Sine and cosine functions relate the angle of the incline to the force components. Specifically, sine of the angle gives the ratio of the opposite side (parallel force) to the hypotenuse (weight), essential for determining the force needed to counteract sliding.

Equilibrium and Static Friction

To keep the car stationary, the applied force must balance the downhill component of weight, achieving equilibrium. Understanding static friction or the required force to prevent motion is key, as it counteracts the tendency to slide down the hill.

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