SME
University Physics I: Mechanics · Problem type
An Atwood machine: two hanging blocks
Use this for “Blocks of 3 kg and 5 kg hang from the two ends of a string over a pulley. Find their acceleration and the string’s tension.”
Diagram
a
=
(m₂
−
m₁)
g
m₁ + m₂
a
=
(5
−
3)
×
9.81
3 + 5
=
2.453
m/
s
2
Block 1: m₁a = T − m₁g = 36.79 − 29.43 = 7.358 N
Block 2: m₂a = m₂g − T = 49.05 − 36.79 = 12.26 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)
+
×
9.81
Add the two blocks’ laws, m₂g − T = m₂a and T − m₁g = m₁a: the tension cancels, and dividing by m₁ + m₂ gives a.
T
Tension
worked out (not typed)
Formulas
a
=
(
m₂
−
m₁
)
×
9.81
m₁
+
m₂
2.4525
=
(5
−
3)
×
9.81
3 + 5
T
=
2
×
m₁
×
m₂
×
9.81
m₁
+
m₂
36.7875
=
2
×
3
×
5
×
9.81
3 + 5
Assumptions
•
A light string over a light, frictionless pulley, so the tension T is the same on both sides; g = 9.81 m/s².
•
The string doesn’t stretch, so the blocks move together: as m₂ goes down by some distance, m₁ goes up by the same.
•
a > 0 means the heavier m₂ goes down; with equal masses a = 0 and the blocks stay as they are.
Step-by-step
WE KNOW
m₁ = 3 kg, m₂ = 5 kg
FIND
acceleration (m₂ down) (a), tension (T)
Step 1 · Find acceleration (m₂ down): a
Use a = (m₂ − m₁) × 9.81 ÷ (m₁ + m₂)
Add
m₂g
−
T
=
m₂
a
and
T
−
m₁g
=
m₁
a
:
T
cancels.
The
difference
in
weight
moves
both
masses.
a
=
(5
−
3)
×
9.81
3 + 5
a
=
2
×
9.81
8
a
=
19.62
8
a
=
2.4525
m/
s
2
Step 2 · Find tension (T)
Use T = 2 × m₁ × m₂ × 9.81 ÷ (m₁ + m₂)
Put
a
into
T
−
m₁g
=
m₁
a
:
T
=
m₁
(g
+
a
),
which
simplifies
to
2m₁m₂g
over
m₁
+
m₂
.
T
=
2
×
3
×
5
×
9.81
3 + 5
T
=
2
×
3
×
5
×
9.81
8
T
=
6
×
5
×
9.81
8
T
=
30
×
9.81
8
T
=
294.3
8
T
=
36.7875
N
Check
Put the numbers back in. Both sides should match.
2.4525
=
(5
−
3)
×
9.81
3 + 5
✓
36.7875
=
2
×
3
×
5
×
9.81
3 + 5
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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