Physics Problem: Block and Pulley System
Question:
A 111 N block sits on a table; the coefficient of kinetic friction between the block and the table is 0.300. This block is attached to a 258 N block by a rope that passes over a pulley; the second block hangs down below the pulley. The pulley is a solid uniform disk with a mass of 1.25 kg and an unknown radius. The rope passes over the pulley on the outer edge.
What is the acceleration of the blocks and the tension in the rope on either side of the pulley?
Solution:
Step 1: Given Data
- Weight of block A: 111 N ⇒ mA = 111 / 9.8 = 11.33 kg
- Weight of block B: 258 N ⇒ mB = 258 / 9.8 = 26.33 kg
- μk = 0.300
- Mass of pulley: mp = 1.25 kg
Pulley is a solid disk, so moment of inertia: I = ½ mp R²
Step 2: Equations of Motion
Block A (on the table):
- Kinetic friction: fk = μk × N = μk × mAg = 0.3 × 11.33 × 9.8 = 33.33 N
- Net force: T1 − fk = mAa
- T1 = mAa + 33.33 … (1)
Block B (hanging):
- Net force: mBg − T2 = mBa
- T2 = mBg − mBa = 258 − 26.33a … (2)
Pulley (torque):
Using τ = Iα and α = a / R
(T2 − T1)R = ½ mp R² × (a / R)
T2 − T1 = ½ mp a = 0.625a
… (3)
Step 3: Solving the Equations
Substitute Eq. (1) and Eq. (2) into Eq. (3):
(258 − 26.33a) − (11.33a + 33.33) = 0.625a
224.67 − 37.66a = 0.625a
224.67 = 38.285a ⇒ a = 224.67 / 38.285 ≈ 5.87 m/s²
Step 4: Calculate Tensions
Using Eq. (1):
T1 = 11.33 × 5.87 + 33.33 = 66.49 + 33.33 = 99.8 N
Using Eq. (2):
T2 = 258 − 26.33 × 5.87 = 258 − 154.5 = 103.5 N
Final Answers:
- Acceleration of the blocks: 5.87 m/s²
- Tension on the table side (T₁): 99.8 N
- Tension on the hanging side (T₂): 103.5 N