SME
Grade 11 Science · Physics
Torque, rotation and static equilibrium
Use this for “An 80 N force pushes a 0.25 m wrench at 30° to the handle. What torque does it make?”
Diagram
F⊥ = F sin θ = 80 × sin 30° = 40 N
τ = rF⊥ = 0.25 × 40 = 10 N·m
Only the part of F across the arm turns it; the part along the arm pulls on the pivot.
This is an example. Type your own values to start a new problem.
τ (N·m)
=
r (m)
×
F (N)
×
sin(
θ
°)
F⊥
Force across the arm
calculated
Formulas
F
⊥
=
F
×
sin(
θ
)
40 = 80 × sin(30°)
τ
=
r
×
F
⊥
10 = 0.25 × 40
Assumptions
•
The lever arm r runs from the pivot to where the force pushes.
•
Only the part of F at right angles to the arm turns it: all of F at 90°, none at 0°.
•
The same force turns harder on a longer arm: τ = rF sin θ.
Step-by-step
WE KNOW
r = 0.25 m, F = 80 N, θ = 30°
FIND
force across the arm (F⊥), torque (τ)
Step 1 · Find force across the arm (F⊥)
Use F⊥ = F × sin(θ)
The
part
of
the
force
across
the
arm
is
F
times
sin
θ
.
F
⊥
=
80
×
sin(30°)
F
⊥
=
80
×
0.5
F
⊥
=
40
N
Step 2 · Find torque (τ)
Use τ = r × F⊥
Torque is the lever arm times the force across it.
τ
=
0.25
×
40
τ
=
10
N·m
Check
Put the numbers back in. Both sides should match.
40 = 80 × sin(30°) ✓
10 = 0.25 × 40 ✓
Topic check
See what you know: 5 questions on this topic
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