Worked Examples

Example 1: Friction Force on a Sliding Box

A 20 kg box rests on a wooden floor. The coefficient of kinetic friction between the box and the floor is 0.35. What is the friction force acting on the box as it slides?

Given: mass (m) = 20 kg, μ_k = 0.35, g = 9.8 m/s²

Formula: F_f = μ × N, where N = mg

Step 1 — Find the normal force: N = 20 × 9.8 = 196 N

Step 2 — Find friction force: F_f = 0.35 × 196 = 68.6 N

The friction force opposing the box’s motion is 68.6 N.

Example 2: Gravitational Force Between Earth and the Moon

Calculate the gravitational force between Earth (mass = 5.97 × 10²⁴ kg) and an object of mass 80 kg standing on Earth’s surface (radius = 6.37 × 10⁶ m). Use G = 6.674 × 10⁻¹¹ N·m²/kg².

Given: M = 5.97 × 10²⁴ kg, m = 80 kg, r = 6.37 × 10⁶ m

Formula: F_g = G × M × m / r²

Step 1 — Square the radius: r² = (6.37 × 10⁶)² = 4.058 × 10¹³ m²

Step 2 — Multiply numerator: G × M × m = 6.674 × 10⁻¹¹ × 5.97 × 10²⁴ × 80 ≈ 3.189 × 10¹⁶

Step 3 — Divide: F_g = 3.189 × 10¹⁶ / 4.058 × 10¹³ ≈ 785 N

This matches the person’s weight (80 × 9.8 = 784 N), confirming the result is correct.

Example 3: Centripetal Force on a Car Rounding a Curve

A 1,200 kg car travels at 15 m/s around a circular highway curve of radius 50 m. What centripetal force is required to keep the car on the curve?

Given: m = 1,200 kg, v = 15 m/s, r = 50 m

Formula: F_c = m × v² / r

Step 1 — Square the speed: v² = 15² = 225 m²/s²

Step 2 — Multiply by mass: 1,200 × 225 = 270,000

Step 3 — Divide by radius: F_c = 270,000 / 50 = 5,400 N

The road must supply an inward (centripetal) force of 5,400 N through friction to keep the car on the curve.

Common Mistakes to Avoid

• Using weight instead of normal force in friction problems. The normal force equals mg only on a flat, horizontal surface. On a ramp or when an extra vertical force is applied, N changes. Always identify the actual normal force before substituting into F_f = μN.
• Forgetting to square the distance in the gravitational force formula. The formula is F_g = GMm / r², not GMm / r. Skipping the square dramatically overstates the force, especially over large distances.
• Treating centripetal force as a separate “extra” force. Centripetal force is not a new force that appears out of nowhere — it is the net inward force provided by friction, tension, gravity, or another existing force. Never add it as an extra term in a free-body diagram.
• Mixing up static and kinetic friction coefficients. μ_s (static) applies before an object starts moving; μ_k (kinetic) applies once it is sliding. Using the wrong value gives an incorrect force and the wrong answer on the ACT.
• Using diameter instead of radius in centripetal force problems. If a problem states a circular track’s diameter, you must halve it before substituting into F_c = mv² / r. Using the full diameter doubles r and cuts the calculated force in half.

Real-World Applications

• Car braking systems: Anti-lock brakes are engineered around static friction. Keeping tires rolling (static friction) rather than skidding (kinetic friction) maximizes stopping force because μ_s > μ_k for rubber on asphalt.
• Satellite orbits: Earth’s gravitational force on a satellite provides exactly the centripetal force needed to maintain a circular orbit. Engineers use F_g = GMm / r² to calculate the precise altitude and speed required for a stable orbit.
• Banked highway curves: Curved roads are tilted inward so that a component of the normal force contributes to centripetal acceleration, reducing reliance on friction alone and making turns safer at high speeds.
• Roller coaster loop design: At the top of a loop, gravity and the track’s normal force together supply the centripetal force. Designers use F_c = mv² / r to set the minimum speed that keeps riders safely pressed into their seats.

Frequently Asked Questions

What is the difference between static and kinetic friction?

Static friction acts on an object that is not yet moving and prevents it from starting to slide. Kinetic friction acts on an object already in motion. The coefficient of static friction (μ_s) is generally larger than the coefficient of kinetic friction (μ_k), which is why it takes more force to get an object moving than to keep it moving.

Does friction depend on the surface area of contact?

No — for most everyday surfaces, friction force does not depend on the contact area. It depends only on the normal force and the coefficient of friction. This is one of the classic counterintuitive results in introductory physics and is frequently tested on the ACT.

Why does gravitational force decrease with distance?

Gravitational force follows an inverse-square law: F_g = GMm / r². Doubling the distance between two objects reduces the gravitational force to one-quarter of its original value. This happens because gravity spreads outward in three dimensions, thinning over a larger spherical area as distance increases.

Is centripetal force the same as centrifugal force?

No. Centripetal force is a real, inward-directed force that causes circular motion. Centrifugal force is a fictitious force — it is the sensation of being pushed outward that you feel inside a rotating reference frame. In standard physics problems, always work with centripetal force, not centrifugal force.

What units should I use for the gravitational constant G?

G = 6.674 × 10⁻¹¹ N·m²/kg². To get the force in Newtons, masses must be in kilograms and distance must be in meters. Mixing units — for example, using kilometers for distance — is a common source of large errors in gravitational force calculations.

Can friction ever act in the direction of motion?

Yes. While friction often opposes sliding, static friction can act in the same direction as motion. For example, when you walk forward, static friction from the ground on your rear foot pushes you forward. The key is that friction opposes relative slipping between surfaces, not necessarily the overall direction of travel.