AP Physics 1: Friction, Inclines, and Connected ObjectsAP Physics 1: Friction, Inclines, and Connected Objects

Newton's Laws Problem Solving: Friction, Inclines, and

Tension

Goal: Solve standard force problems: friction, inclined planes, elevators, and connected objects.

THE SHORT VERSION

Most Newton's-law problems follow the same recipe: draw an FBD for each object, pick axes, write ΣF = ma, and solve. This guide covers the four setups that appear most often.

1 Strategy

1. Sketch the situation and choose the object(s).

2. Draw an FBD for each object.

3. Choose axes (along the motion helps).

4. Write ΣF = ma on each axis, with signs.

5. Solve for the unknown, including connecting equations (same acceleration for connected objects; same tension for an ideal rope). 6. Check units and whether the answer makes sense.

2 Friction

Static friction Kinetic friction

When Surfaces not sliding Surfaces sliding

Size Adjusts up to a maximum: f_s ≤ μ_s F_N f_k = μ_k F_N

Coefficient μ_s (usually larger) μ_k (usually smaller)

Direction Opposes the tendency to slide Opposes the sliding

Friction depends on the normal force, not on the contact area.

On a horizontal surface with no vertical forces other than weight, F_N = mg, so f_k = μ_k mg.

If a box is pushed but doesn't move, static friction equals the push (net force zero).

3 Inclined Planes

Tilt your axes: x along the incline, y perpendicular. Angle θ is measured from the horizontal.

Weight components: mg sin θ parallel to the incline (down the slope), mg cos θ perpendicular to it.

Perpendicular balance: F_N = mg cos θ.

Frictionless: a = g sin θ (down the slope).

With kinetic friction (sliding down): a = g(sin θ − μ_k cos θ).

Example: a 5 kg block slides down a 30° incline with μ_k = 0.20. a = 9.8(0.500 − 0.20 × 0.866) = 9.8(0.327) ≈ 3.2 m/s² down the slope. Forces on a block on a 30 degree slope.

4 Elevators and Apparent Weight

Take up as positive: F_N − mg = ma. So F_N = m(g + a).

Accelerating up: you feel heavier (F_N > mg). Accelerating down: lighter (F_N < mg). Free fall: F_N = 0.

The scale reads F_N, not your true weight mg.

Constant velocity in an elevator: F_N = mg.

Example: a 60 kg person in an elevator accelerating upward at 2 m/s². F_N = 60(9.8 + 2) = 708 N.

5 Connected Objects

Same acceleration for objects tied together by a taut, inextensible rope.

Tension is the same on both ends of an ideal rope over an ideal pulley.

Treat the whole system for the acceleration, then isolate one object for the tension.

Two blocks on a frictionless table: a 20 N force pulls block A (3 kg), which is tied to block B (2 kg). System: a = 20/(3+2) = 4 m/s². Tension on B: T = m_B a = 2 × 4 = 8 N.

Atwood machine (two masses over a pulley, m₁ > m₂): a = (m₁ − m₂)g / (m₁ + m₂) and T = 2m₁m₂g / (m₁ + m₂). For m₁ = 3 kg, m₂ = 1 kg: a = 4.9 m/s² and T = 14.7 N.

Atwood machine: 3 kg and 1 kg over a pulley.

6 Words to know

Static friction

Friction when surfaces are not sliding; up to μ_s F_N.

Kinetic friction

Friction when sliding; μ_k F_N.

Coefficient of friction (μ)

Unitless number comparing friction to normal force.

Apparent weight

The normal force a scale reads.

Atwood machine

Two masses connected over a pulley.

Incline components

mg sin θ along the slope, mg cos θ perpendicular.

CHECK On an incline, if θ → 0 then a → 0 (frictionless), and if θ → 90° then a → g. This tests your formula. TRAP Normal force on an incline is mg cos θ, not mg.

AP Physics 1 Guide 5 · original study notes