Connected Particles, Pulleys, Inclined Planes with Friction & Moments

1

Resolving Forces on an Inclined Plane

For a particle of mass m on a plane inclined at angle θ to the horizontal, always resolve forces parallel and perpendicular to the slope (never horizontally and vertically, unless the particle is moving in a horizontal circle). Gravity mg resolves into two components: mg cos θ acting perpendicular to the slope (into the slope) and mg sin θ acting parallel to the slope (down the slope). If an additional external force P acts at an angle α to the slope (or horizontal), it also has a perpendicular component that changes the normal reaction R - so R is not automatically equal to mg cos θ!

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2

Static vs Dynamic Friction and Limiting Equilibrium (F_max = μR)

Friction F always acts parallel to the rough surface in the direction that opposes motion (or opposes the tendency to move). When a particle is in static equilibrium and not on the point of moving, friction is simply whatever value is needed to balance the parallel forces, satisfying the inequality F ≤ μR (where μ is the coefficient of friction). When the particle is in limiting equilibrium (on the point of slipping) OR is already moving/sliding, friction reaches its maximum value: F = F_max = μR.

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3

Connected Particles: Tow-Bars and Smooth Pulleys

When two particles are connected by a light inextensible string: 'light' means the string's mass is negligible so tension T is uniform along each straight section, and 'inextensible' means the string does not stretch so both particles share the exact same magnitude of velocity v and acceleration a. For tow-bars (moving in the same straight line), you can write F = ma for the whole system to find a immediately, then F = ma for the trailer alone to find tension T. For a pulley (where the particles move in different directions), never write a single 'whole-system' equation directly, write two separate F = ma equations (each in the direction of its own motion) and add them together to eliminate T!

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4

Two-stage pulley problems (hitting the ground & string going slack)

When the heavier mass B hits the floor after travelling distance h from rest: 1. Use SUVAT (v² = u² + 2ah) with the Stage 1 pulley acceleration a to find the speed v of both particles at the instant B hits the ground. 2. Immediately after B hits the ground, the string goes slack (Tension T drops to 0!). 3. In Stage 2, particle A continues moving for a further distance with initial speed u₂ = v, final speed v₂ = 0 (at its highest point), and a new deceleration a₂ determined solely by gravity (and friction if on a slope, with T = 0).

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5

Force on the Pulley

If a light inextensible string with tension T passes over a smooth fixed pulley, each side of the string pulls on the pulley with force T along the line of the string.1. If both sections of string hang vertically downwards, the total force exerted by the string on the pulley is simply 2T downwards. 2. If one section is vertical (or at angle θ₁) and the other lies along an inclined plane at angle θ to the horizontal, resolve the two tension forces of magnitude T horizontally and vertically (or use the cosine rule / angle bisector 2T cos(α/2), where α is the angle between the two string segments) to find the magnitude and direction of the resultant force on the pulley.

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6

Moments of Forces, Non-Uniform Rods, and Tilting vs Slipping

The moment of a force F about a pivot P is Moment = F × d_⊥ = F × d sin θ, where d is the distance from P to the point of application of F and θ is the angle between the force and the rod. A rigid body is in static equilibrium if and only if: 1. The resultant force in any two perpendicular directions is zero (ΣF_x = 0 and ΣF_y = 0); And 2. The sum of clockwise moments equals the sum of anticlockwise moments about any point. For a uniform rod of length L, its weight W acts at its midpoint (L / 2). For a non-uniform rod, its centre of mass is at some unknown distance x from one end. When a rod resting on two supports A and B is on the point of tilting (pivoting) about support B, the normal reaction at the other support A drops to zero: R_A = 0!

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1. Q1

   A particle A of mass 3 kg rests on a rough plane inclined at 30° to the horizontal, where the coefficient of friction between A and the plane is μ = 1 / (2√3). Particle A is connected by a light inextensible string passing over a smooth fixed pulley at the top of the plane to a particle B of mass 5 kg, which hangs freely. The system is released from rest with the string taut. Taking g = 9.8 m s⁻², find the acceleration of the system, the tension in the string, and the magnitude of the resultant force exerted by the string on the pulley. (8 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Resolve perpendicular to the slope for particle A (m_A = 3 kg): R = 3g cos 30° = 3(9.8)(√3 / 2) = 14.7√3 N [1 mark]. 2. Since the system moves, friction on A is limiting and acts down the slope: F_max = μR = [1 / (2√3)] × 14.7√3 = 7.35 N (or (3/4)g) [1 mark]. 3. Write Newton's Second Law for A up the slope: T - 3g sin 30° - F_max = 3a → T - 1.5g - 0.75g = 3a → T - 2.25g = 3a [1 mark]. 4. Write Newton's Second Law for B (m_B = 5 kg) vertically downwards: 5g - T = 5a [1 mark]. 5. Add the two equations to eliminate T: 5g - 2.25g = 8a → 2.75(9.8) = 8a → 26.95 = 8a → a = 3.37 m s⁻² (or (11/32)g) [1 mark]. 6. Substitute a back into 5g - T = 5a to find tension T: T = 5(9.8) - 5(3.36875) = 32.2 N (or (105/32)g) [1 mark]. 7. The two string segments at the pulley have angle α = 90° - 30° = 60° between them [1 mark]. 8. Magnitude of force on the pulley = 2T cos(60° / 2) = 2(32.16) cos 30° = 32.16√3 = 55.7 N (or by cosine rule √(T² + T² - 2T² cos 120°) = √3 T) [1 mark].

2. Q2

   A non-uniform rod AB of length 4 m and mass 12 kg rests horizontally on two smooth supports at C and D, where AC = 0.8 m and BD = 1.0 m. When a particle of mass M kg is placed on the rod at the end A, the rod is on the point of tilting about C. When the particle of mass M kg is instead placed at the end B, the rod is on the point of tilting about D. Find the distance of the centre of mass of the rod from A, and find the value of M. (6 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Let the centre of mass of the non-uniform rod be at a distance x metres from end A (where 0.8 < x < 3.0) [1 mark]. 2. When mass M is placed at A, the rod is on the point of tilting about C (0.8 m from A), so the normal reaction at D is zero (R_D = 0) [1 mark]. 3. Take moments about pivot C: anticlockwise moment = clockwise moment → Mg(0.8) = 12g(x - 0.8) → 0.8M = 12x - 9.6 (Equation 1) [1 mark]. 4. When mass M is placed at B, the rod is on the point of tilting about D (which is at 4.0 - 1.0 = 3.0 m from A), so R_C = 0; take moments about pivot D: 12g(3.0 - x) = Mg(1.0) → M = 36 - 12x (Equation 2) [1 mark]. 5. Substitute M = 36 - 12x into Equation 1: 0.8(36 - 12x) = 12x - 9.6 → 28.8 - 9.6x = 12x - 9.6 → 21.6x = 38.4 → x = 16/9 ≈ 1.78 m from A [1 mark]. 6. Substitute x = 16/9 into Equation 2: M = 36 - 12(16/9) = 36 - 64/3 = 44/3 ≈ 14.7 kg [1 mark].

3. Q3

   A uniform ladder AB of length 5 m and mass 20 kg rests with its top end A against a smooth vertical wall and its bottom end B on rough horizontal ground, inclined at an angle θ to the horizontal where tan θ = 4/3. A builder of mass 70 kg climbs the ladder. Given that the coefficient of friction between the ladder and the ground is μ = 0.45, find how far up the ladder from B the builder can climb before the ladder slips. (6 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. From tan θ = 4/3 (3-4-5 triangle), find sin θ = 4/5 = 0.8 and cos θ = 3/5 = 0.6 [1 mark]. 2. Resolve forces vertically for the whole system (ladder 20 kg + builder 70 kg): Ground normal reaction R_B = 20g + 70g = 90g N [1 mark]. 3. At the point of slipping, friction at the ground is limiting: F_max = μ R_B = 0.45 × 90g = 40.5g N (acting horizontally towards the wall) [1 mark]. 4. Since the vertical wall is smooth (no vertical friction at A), resolve horizontally: Wall normal reaction R_A = F_max = 40.5g N [1 mark]. 5. Let the builder be at distance x metres from B along the ladder, and take moments about the bottom B: Clockwise moment = R_A × 5 sin θ = 40.5g × 5(0.8) = 162g N m [1 mark]. 6. Equate to anticlockwise moments of the ladder's weight (at midpoint 2.5 m) and builder's weight (at x m): 20g(2.5 cos θ) + 70g(x cos θ) = 162g → 20(2.5)(0.6) + 70(0.6)x = 162 → 30 + 42x = 162 → 42x = 132 → x = 22/7 ≈ 3.14 m [1 mark].

4. Q4

   A box of mass 8 kg is held in limiting equilibrium on the point of slipping down a rough plane inclined at 25° to the horizontal by a horizontal force of magnitude P N pushing into the slope. Given that the coefficient of friction between the box and the plane is μ = 0.20 and g = 9.8 m s⁻², find the value of P. (5 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Draw a clear force diagram: since the box is on the point of slipping down the slope, limiting friction F_max = μR acts UP the slope [1 mark]. 2. Resolve forces perpendicular to the slope (noting that the horizontal force P has a component P sin 25° pushing into the slope): R = 8g cos 25° + P sin 25° [1 mark]. 3. Resolve forces parallel to the slope: P cos 25° + F_max = 8g sin 25° [1 mark]. 4. Substitute F_max = 0.20 R = 0.20(8g cos 25° + P sin 25°) into the parallel equation: P cos 25° + 0.20 P sin 25° = 8g sin 25° - 1.6g cos 25° [1 mark]. 5. Factor out P and solve: P(cos 25° + 0.20 sin 25°) = 9.8(8 sin 25° - 1.6 cos 25°) → P(0.9908) = 9.8(3.3809 - 1.4501) = 18.922 → P = 19.1 N [1 mark].

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