Physics Problem: Acceleration and Tension in a 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 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 its outer edge.
Question: What is the acceleration of the blocks and the tension in the rope on either side of the pulley?
🔹 Step 1: Calculate Forces Acting on Each Block
Normal force (R) on the 111 N block:
R = W = 111 N
Friction force (f):
f = μ × R = 0.3 × 111 = 33.30 N
Mass of 111 N block:
m₁ = 111 / 9.8 ≈ 11.33 kg
Mass of 258 N block:
m₂ = 258 / 9.8 ≈ 26.33 kg
🔹 Step 2: Write Equations of Motion
For the 111 N block (horizontal):
T₁ − f = m₁ × a
T₁ = 33.30 + 11.33a …(1)
For the 258 N block (vertical):
m₂g − T₂ = m₂ × a
T₂ = 258 − 26.33a …(2)
🔹 Step 3: Apply Rotational Dynamics to the Pulley
For a solid disk:
τ = Iα = (1/2)MR² × (a/R) = (1/2)Ma
Net torque due to tensions:
(T₂ − T₁)R = (1/2)MR × a
T₂ − T₁ = (1/2)M × a …(3)
Substitute T₁ and T₂ from (1) and (2) into (3):
(258 − 26.33a) − (33.30 + 11.33a) = (1/2)(1.25)a
Simplify:
258 − 26.33a − 33.30 − 11.33a = 0.625a
⇒ 224.7 − 37.66a = 0.625a
⇒ 224.7 = 38.285a
⇒ a ≈ 5.87 m/s²
🔹 Step 4: Find Tensions
Using (1):
T₁ = 33.30 + 11.33 × 5.87 = 99.81 N
Using (2):
T₂ = 258 − 26.33 × 5.87 = 103.44 N
- 🔸 Acceleration of the blocks: 5.87 m/s²
- 🔸 Tension on the left side (T₁): 99.81 N
- 🔸 Tension on the right side (T₂): 103.44 N