Connected Blocks with Friction and an Added Mass
In a laboratory experiment, block A of mass 4 kg rests on a rough horizontal bench and is connected by a light inextensible string over a smooth pulley at the bench edge to hanging block B of mass 3 kg. The coefficient of friction between A and the bench is 0.25. The system is released from rest. A student predicts the acceleration by using only the hanging weight and the total mass, ignoring friction, and obtains a value close to 4.2 m/s². The demonstrator asks the student to check whether the blocks actually move, and then to find the maximum additional mass that could be placed on top of A without the system moving. Use g = 9.8 m/s². The pulley is frictionless, the string is taut throughout, and the blocks are treated as particles. The additional mass would be fixed firmly to A so that it does not slide.
1. Show that the system moves when released, then calculate the acceleration and the tension in the string.
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The maximum friction available on A is Fmax = μ m_A g = 0.25(4)(9.8) = 9.8 N. The weight of B is m_B g = 3(9.8) = 29.4 N. Since 29.4 N > 9.8 N, the hanging weight exceeds limiting friction and the system moves. Let a be the acceleration and T the tension. For B, taking downward as positive: 3g − T = 3a. For A, taking motion toward the pulley as positive: T − F = 4a, and while moving F = 9.8 N. Adding the equations eliminates T: 3g − 9.8 = 7a, so 29.4 − 9.8 = 19.6 = 7a, giving a = 2.8 m/s². Then T = 3(g − a) = 3(9.8 − 2.8) = 21 N. Checking A: T − 9.8 = 21 − 9.8 = 11.2 = 4(2.8), as required.
2. Find the maximum additional mass M that can be added to A so that the system remains at rest on release. Explain why the tension must be compared with limiting friction, not with the actual friction needed for rest.
Read the worked answer
If the system remains at rest, the tension in the string equals the weight of B, so T = 3g = 29.4 N. The friction on A must balance this tension, so the required friction is also 29.4 N. The largest friction the bench can provide is μ(4 + M)g = 0.25(4 + M)(9.8). For rest to be possible, this limiting friction must be at least 29.4 N: 0.25(4 + M)(9.8) ≥ 3(9.8). Cancelling g gives 0.25(4 + M) ≥ 3, so 4 + M ≥ 12 and M ≥ 8? Wait, check the inequality direction: 0.25(4+M) ≥ 3 gives 4+M ≥ 12, so M ≥ 8. That would mean larger masses make rest possible, which is wrong because the normal reaction increases with added mass. Recheck: friction maximum increases with M, so adding mass helps rest. But the required friction is fixed at 29.4 N, so M can be as large as we like? Actually if M is very large, limiting friction is large, so rest is possible. The question should be the maximum additional mass that can be added while still moving, or the minimum mass to keep at rest. Let's correct: for rest, limiting friction ≥ required, so M ≥ 8 kg. But adding mass increases friction capacity, so the system stays at rest for M ≥ 8 kg, not M ≤ 8. The original question asks maximum mass without moving, which is unbounded. The intended answer is 8 kg as the threshold; for M > 8 the system remains at rest, for M < 8 it accelerates. I will state the threshold clearly.