Incline-Pulley System That May Not Move
Block 1 of mass 4.0 kg rests on a fixed incline of 28° to the horizontal. It is connected by a light string over a frictionless pulley at the top of the incline to block 2 of mass 2.5 kg hanging vertically. The coefficients of friction between block 1 and the incline are μs = 0.30 and μk = 0.22. The system is released from rest and observed for several seconds; a student predicts that block 2 will descend because its weight is larger than the component of block 1’s weight along the incline. The instructor asks for a force analysis before any kinematic equation is used: decide whether the system moves, find the tension if it does not, and determine the smallest hanging mass that would just start block 1 moving up the incline. Use g = 9.8 m/s² and assume the pulley and string are ideal. A second part of the test uses a 3.5 kg hanging mass to check the kinetic case.
1. Decide whether the system moves when block 2 has mass 2.5 kg, and find the tension and the friction force if it is at rest.
Read the worked answer
Take the direction of impending motion as block 2 descending and block 1 moving up the incline. The driving force is
m₂g − m₁g sin 28° = (2.5)(9.8) − (4.0)(9.8)(0.4695) = 24.5 − 18.40 = 6.10 N.
The maximum static friction available is
f_s,max = μs m₁g cos 28° = (0.30)(4.0)(9.8)(0.8829) ≈ 10.4 N.
Because 6.10 N < 10.4 N, static friction can hold the system at rest. For block 2 in equilibrium, T = m₂g = 24.5 N. For block 1, T − m₁g sin 28° = 24.5 − 18.40 = 6.10 N up the incline, so static friction must act down the incline with magnitude 6.10 N. This is below the 10.4 N limit, so the assumption of rest is consistent.
2. Find the smallest hanging mass that would just start block 1 moving up the incline.
Read the worked answer
At the threshold of upward motion, the tension must balance both the down-slope weight component and the maximum static friction:
T = m₁g sin 28° + μs m₁g cos 28° = 18.40 + 10.38 ≈ 28.78 N.
For the hanging block at the threshold, T = m₂,min g, so
m₂,min = 28.78/9.8 ≈ 2.94 kg.
Any hanging mass greater than about 2.94 kg would cause block 1 to accelerate up the incline; at exactly this value the system is on the verge of motion. The student’s comparison of 24.5 N with 18.4 N ignored friction and therefore underestimated the hanging mass needed to move the system.
3. For a hanging mass of 3.5 kg, calculate the acceleration and the tension, stating any assumption about the direction of motion.
Read the worked answer
With m₂ = 3.5 kg, the driving force without friction is 34.3 − 18.40 = 15.9 N, which exceeds the 10.4 N static limit, so the system moves with block 1 up the incline. Kinetic friction acts down the incline:
f_k = μk m₁g cos 28° = (0.22)(4.0)(9.8)(0.8829) ≈ 7.61 N.
Net force on the system: 34.3 − 18.40 − 7.61 = 8.29 N, with total mass 7.5 kg. Thus a = 8.29/7.5 ≈ 1.11 m/s². For block 2, taking downward as positive: m₂g − T = m₂a, so T = 3.5(9.8 − 1.11) ≈ 30.4 N. Checking block 1: T − m₁g sin 28° − f_k = 30.4 − 18.40 − 7.61 ≈ 4.39 N, which is close to m₁a = 4.44 N; the small difference is rounding.