Block and Trolley System with Friction
This simulation demonstrates the motion of a block and trolley system with kinetic friction, based on Example 5.9 from the textbook.
Time
0.0 s
Position
0.00 m
Velocity
0.00 m/s
Acceleration
0.00 m/s²
Applied Force
Tension (T)
Friction Force (fk)
Weight
Normal Force
Physics Results:
Using the values from the textbook example:
For the 3 kg block:
ΣF = ma ⇒ 30 - T = 3a
For the 20 kg trolley:
ΣF = ma ⇒ T - fk = 20a
fk = μkN = 0.04 × (20×10) = 8 N
Solving these equations gives:
(30 - T) + (T - 8) = 3a + 20a ⇒ 22 = 23a
a = 22/23 ≈ 0.96 m/s²
T = 30 - 3a ≈ 27.1 N
ΣF = ma ⇒ 30 - T = 3a
For the 20 kg trolley:
ΣF = ma ⇒ T - fk = 20a
fk = μkN = 0.04 × (20×10) = 8 N
Solving these equations gives:
(30 - T) + (T - 8) = 3a + 20a ⇒ 22 = 23a
a = 22/23 ≈ 0.96 m/s²
T = 30 - 3a ≈ 27.1 N
Current simulation values will appear here when running.
