SME
Grade 11 Science · Problem type
Balancing a seesaw
Use this for “A 300 N child sits 2 m from the pivot. Where must a 400 N child sit to balance, and how hard does the pivot push up?”
Diagram
Balanced: F₁d₁ = F₂d₂: 300 × 2 = 400 × 1.5 = 600 N·m
Fₚ = F₁ + F₂ = 300 + 400 = 700 N
The heavier weight sits closer to the pivot; the pivot holds up both.
This is an example. Type your own values to start a new problem.
F₁ (N)
×
d₁ (m)
=
F₂ (N)
×
d₂ (m)
τ
Torque on each side
calculated · N·m
Fₚ
Pivot force
calculated
Formulas
τ
=
F₁
×
d₁
600 = 300 × 2
τ
=
F₂
×
d₂
600 = 400 × 1.5
Fₚ
=
F₁
+
F₂
700 = 300 + 400
Assumptions
•
Balanced means no net torque and no net force: the two torques match, and the pivot holds up both weights.
•
Each torque is a weight times its distance from the pivot.
•
The plank is light and pivoted at its middle, so its own weight is left out.
Step-by-step
WE KNOW
F₁ = 300 N, d₁ = 2 m, F₂ = 400 N
FIND
torque on each side (τ), second distance (d₂), pivot force (Fₚ)
Step 1 · Find torque on each side (τ)
Use τ = F₁ × d₁
The first child’s torque: weight times distance from the pivot.
τ
=
300
×
2
τ
=
600
N·m
Step 2 · Find second distance (d₂)
Use τ = F₂ × d₂
Divide the torque by the weight: the heavier child sits closer.
d₂
=
τ
F₂
d₂
=
600
400
d₂
=
1.5
m
Step 3 · Find pivot force (Fₚ)
Use Fₚ = F₁ + F₂
No net force: the pivot pushes up as hard as both weights push down.
Fₚ = 300 + 400
Fₚ = 700 N
Check
Put the numbers back in. Both sides should match.
600 = 300 × 2 ✓
600 = 400 × 1.5 ✓
700 = 300 + 400 ✓
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Torque, rotation and static equilibrium
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Angle in radians and arc length
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