SME
Dynamics
Kinetics of particles
Use this for “A 10 kg crate on a bench (μₖ = 0.2) is tied by a rope over a pulley to a 4 kg block hanging at the edge. Find the acceleration and the rope’s tension.”
Diagram
a
=
(m₂
−
μₖm₁)
g
m₁ + m₂
a
=
(4
−
0.2
×
10)
×
9.81
10 + 4
a
=
1.401
m/
s
2
Block 1: m₁a = T − μₖm₁g = 33.63 − 19.62 = 14.01 N
Block 2: m₂a = m₂g − T = 39.24 − 33.63 = 5.606 N
A light pulley: the tension is the same on both sides.
This is an example. Type your own values to start a new problem.
a (m/s²)
=
(
m₂ (kg)
−
×
m₁ (kg)
)g
m₁ (kg)
+
m₂ (kg)
T (N)
=
m₂ (kg)
(g
−
a (m/s²)
)
Hanging block: m₂g − T = m₂a. Crate: T − μₖm₁g = m₁a. Adding the two gives a, and the first one gives T.
Formulas
a
=
(
m₂
−
μₖ
×
m₁
)
×
9.81
m₁
+
m₂
1.4014
=
(4
−
0.2
×
10)
×
9.81
10 + 4
T
=
m₂
×
(9.81
−
a
)
33.6343 = 4 × (9.81 − 1.4014)
Assumptions
•
A light rope over a light, frictionless pulley: the tension T is the same on both sides; g = 9.81 m/s².
•
The rope doesn’t stretch, so the crate and the hanging block share one acceleration a.
•
The crate is already sliding, so kinetic friction μₖm₁g acts against its motion.
•
If m₂ ≤ μₖm₁, the blocks can’t speed up: once at rest, they don’t start.
Step-by-step
WE KNOW
m₁ = 10 kg, m₂ = 4 kg, μₖ = 0.2
FIND
acceleration (a), rope tension (T)
Step 1 · Find acceleration (a)
Use a = (m₂ − μₖ × m₁) × 9.81 ÷ (m₁ + m₂)
Add
m₂g
−
T
=
m₂
a
(hanging
block)
and
T
−
μₖm₁g
=
m₁
a
(crate):
T
cancels,
leaving
one
equation
for
a
.
a
=
(4
−
0.2
×
10)
×
9.81
10 + 4
a
=
(4
−
2)
×
9.81
14
a
=
2
×
9.81
14
a
=
19.62
14
a
=
1.40143
m/
s
2
Step 2 · Find rope tension (T)
Use T = m₂ × (9.81 − a)
Newton’s
second
law
on
the
hanging
block:
m₂g
−
T
=
m₂
a
,
so
T
is
m₂g
less
m₂
a
.
T
=
4
×
(9.81
−
1.40143)
T
=
4
×
8.409
T
=
33.6343
N
Check
Put the numbers back in. Both sides should match.
1.40143
=
(4
−
0.2
×
10)
×
9.81
10 + 4
✓
33.6343 = 4 × (9.81 − 1.40143) ✓
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Relative velocity of two moving particles
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