Hanging Mass and a Block on a Rough Table
Block 1, with mass 4.0 kg, rests on a horizontal table. The coefficients of static and kinetic friction between block 1 and the table are 0.45 and 0.30. Block 1 is connected by a light string over a frictionless pulley to hanging block 2. The string and pulley are ideal, and the table is long enough that block 1 never reaches the pulley. Take g=9.8 m/s². A student begins by assuming the system accelerates and computes the tension from Newton’s second law, but the result must be checked against the static-friction limit before it is accepted.
1. For block 2 mass 1.5 kg, determine whether the system moves. If it does not move, find the tension and the static friction force.
Read the worked answer
The normal force on block 1 is N = m1 g = (4.0)(9.8) = 39.2 N, so the maximum static friction is f_s,max = (0.45)(39.2) = 17.64 N.
The hanging weight is m2 g = (1.5)(9.8) = 14.7 N. This weight tries to pull block 1 to the right, so static friction can point left with magnitude 14.7 N. Since 14.7 N < 17.64 N, static friction can hold the system at rest.
Thus a = 0. The tension equals the hanging weight, T = 14.7 N, and the static friction on block 1 is 14.7 N to the left. Computing a nonzero acceleration from m2 g alone would be incorrect because it ignores the static-friction check.
2. Find the full range of block 2 masses for which the system can remain at rest. Explain any lower bound.
Read the worked answer
For equilibrium, the tension must equal m2 g. On block 1, static friction must balance that tension, so the required friction is f_s = m2 g. The static-friction limit requires m2 g ≤ f_s,max = 17.64 N, giving m2 ≤ 17.64/9.8 = 1.80 kg.
There is no positive lower bound beyond m2 ≥ 0: if m2 is very small, the required static friction is also very small, and static friction simply adjusts to keep block 1 at rest. The equilibrium range is 0 ≤ m2 ≤ 1.80 kg.
3. For block 2 mass 3.0 kg, find the acceleration of the system and the tension in the string.
Read the worked answer
Now the hanging weight is (3.0)(9.8) = 29.4 N, which exceeds the 17.64 N static-friction limit, so the system moves. Kinetic friction acts on block 1 with f_k = (0.30)(39.2) = 11.76 N.
For the two-block system along the direction of motion: 29.4 − 11.76 = (4.0 + 3.0)a 17.64 = 7.0a a = 2.52 m/s².
For block 2: m2 g − T = m2 a, so T = 3.0(9.8 − 2.52) = 21.84 N. Equivalently, for block 1: T − f_k = m1 a gives T = 11.76 + 4.0(2.52) = 21.84 N. The tension is less than the hanging weight because block 2 accelerates downward.