Block on a Rough Incline: Checking the Static Case Before Using F = ma
Block A, of mass 2.0 kg, rests on a rough plane inclined at 30° to the horizontal; the coefficients of friction between A and the plane are μs = 0.40 and μk = 0.25. A light inextensible string runs from A up the slope, parallel to the incline, over a frictionless pulley fixed at the top, and hangs vertically to block B of mass 1.5 kg. The string is taut throughout, and the bench is long enough that neither block reaches the pulley or the floor. The system is released from rest; take g = 9.8 m s⁻². A student argues that B must accelerate downwards because B's weight, 14.7 N, exceeds the component of A's weight along the slope, 9.8 N. The lab partner disagrees, noting that the static case must be checked before any acceleration is calculated and that friction cannot be assumed to act at its limiting value.
1. Decide whether the system moves when released. Support your decision with a calculation, and if it does not move, state the magnitude and direction of the friction force acting on A.
Read the worked answer
Forces along the slope, friction neglected first. B's weight pulls A up the slope with T ≈ m_B g = 1.5 × 9.8 = 14.7 N. A's weight component down the slope is m_A g sin30° = 2.0 × 9.8 × 0.500 = 9.8 N. Resultant without friction = 14.7 − 9.8 = 4.9 N up the slope, so the impending motion of A is up the slope and friction would act down the slope.
Can static friction supply 4.9 N? f_max = μs m_A g cos30° = 0.40 × 2.0 × 9.8 × 0.866 = 6.8 N. Since 4.9 N < 6.8 N, static friction can balance the resultant exactly. The system therefore remains at rest; the student's force values are right but the conclusion ignores friction.
Values at rest. For B: m_B g − T = 0, so T = 14.7 N. For A: T − m_A g sin30° − f = 0, so f = 14.7 − 9.8 = 4.9 N, directed down the slope (towards the bottom of the incline). This is below the 6.8 N limit, so the assumption of no sliding is self-consistent.
2. B is replaced by heavier blocks until A just begins to slide. Find the minimum mass of B for which motion starts, and state the direction of motion.
Read the worked answer
At the threshold, A is about to move up the slope, so friction is at its limiting value and acts down the slope. Resolving along the slope for the whole system at the point of slipping (a = 0): m_B g = m_A g sin30° + μs m_A g cos30° Dividing by g (so the answer is independent of g): m_B = m_A (sin30° + μs cos30°) = 2.0 × (0.500 + 0.40 × 0.866) = 2.0 × 0.846 = 1.69 kg. So B must exceed about 1.7 kg for A to slide, and when it does, A moves up the slope while B descends. For any B between 1.5 kg and 1.69 kg the system stays at rest with static friction adjusting below its maximum. Note that this threshold is not the same as the 1.5 kg value obtained by ignoring friction - friction raises the mass needed.
3. With B of mass 2.5 kg, find the acceleration of the blocks and the tension in the string.
Read the worked answer
Motion occurs, so kinetic friction acts down the slope on A. Take the direction of motion (B down, A up the slope) as positive.
Net driving force: F = m_B g − m_A g sin30° − μk m_A g cos30° = 24.5 − 9.8 − (0.25 × 2.0 × 9.8 × 0.866) = 24.5 − 9.8 − 4.24 = 10.46 N. The mass being accelerated is 2.0 + 2.5 = 4.5 kg, so a = 10.46 / 4.5 = 2.3 m s⁻².
Tension, from B (which accelerates downwards): m_B g − T = m_B a → T = 2.5 × (9.8 − 2.32) = 18.7 N ≈ 19 N.
Check on A: T = m_A g sin30° + f_k + m_A a = 9.8 + 4.24 + (2.0 × 2.32) = 18.7 N ✓. The tension is less than B's weight (24.5 N), which is consistent with B accelerating downwards.