Statics · Problem type

Moment of a force about a point

Use this for “Find the moment of F = 30i + 40j N, acting at (2, 1) m, about the origin.”

Diagram

xyzr = 2.24 mF = 50 NMO = 50 N·m
r = ⟨2, 1, 0⟩, with length |r| = 2.24.
F = ⟨30, 40, 0⟩, with length |F| = 50.
MO = ⟨0, 0, 50⟩, square to both: (MO)·r = 0 and (MO)·F = 0.
Each quantity is drawn on its own scale (m, N, N·m): compare directions, not lengths.
This is an example. Type your own values to start a new problem.
MO (N·m)
=
x (m)
×
Fᵧ (N)
−
y (m)
×
Fₓ (N)
F
Force size
worked out (not typed)
d
Lever arm
worked out (not typed)

Formulas

MO
=
x
×
Fᵧ
−
y
×
Fₓ
+50 = 2 × 40 − 1 × 30
F
=
√
Fₓ² +
Fᵧ
2
50
=
√
30
2
+
40
2
d
=
|MO|
÷
F
1 = |50| ÷ 50

Assumptions

•
The force and its point of action lie in the xy-plane, so MO = r × F points along z only.
•
Counterclockwise moments are + and clockwise moments are −.
•
The lever arm d is the shortest distance from O to the force’s line of action, so |MO| = Fd.

Step-by-step

WE KNOW
x = 2 m, y = 1 m, Fₓ = 30 N, Fᵧ = 40 N
FIND
moment about O (MO), force size (F), lever arm (d)
Step 1 · Find moment about O (MO)
Use MO = x × Fᵧ − y × Fₓ
The
z-part
of
r
×
F
:
Fᵧ
turns
the
body
counterclockwise
about
O
with
arm
x
,
and
Fₓ
turns
it
clockwise
with
arm
y
.
MO = 2 × 40 − 1 × 30
MO = 80 − 30
MO = +50 N·m
Step 2 · Find force size (F)
Use F = √(Fₓ² + Fᵧ²)
The force’s parts are at right angles, so its size is the hypotenuse of their right triangle.
F
=
√
30
2
+
40
2
F
=
√
900 + 1,600
F
=
√
2
,500
F
=
50
N
Step 3 · Find lever arm (d)
Use d = |MO| ÷ F
The
size
of
the
moment
is
force
times
lever
arm,
|MO|
=
Fd,
so
divide
by
F
.
d
=
|50|
÷
50
d
=
50
50
d
=
1
m
Check
Put the numbers back in. Both sides should match.
+50 = 2 × 40 − 1 × 30 ✓
50
=
√
30
2
+
40
2
✓
1 = |50| ÷ 50 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.

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