SME
University Physics I: Mechanics · Problem type
A ladder against a smooth wall: will it slip?
Use this for “A 300 N ladder leans on a smooth wall at 65° to the floor. How hard does the wall push, and how rough must the floor be to hold it?”
Diagram
Up and down: Nf = W = 200 N
Torques
about
the
foot:
Nw
×
L
sin
θ
=
W
×
1
2
L
cos
θ,
so
Nw
=
W
2 tan θ
=
200
2 × tan 60°
=
57.74
N
Level:
f
=
Nw
=
57.74
N,
so
it
holds
when
μₛ
≥
f
Nf
=
0.289
This is an example. Type your own values to start a new problem.
Nw (N)
=
W (N)
2
×
tan(
°)
Torques about the foot: Nw × L sin θ = W × (L ÷ 2) cos θ. Divide by L sin θ to get Nw.
Nf
Normal force from the floor
worked out (not typed)
f
Friction at the foot
worked out (not typed)
μₛ
Least static friction coefficient
worked out (not typed)
Formulas
Nf
=
W
200 = 200
Nw
=
W
2 × tan(
θ
)
57.735
=
200
2 × tan(60°)
f
=
Nw
57.735 = 57.735
μₛ
=
f
Nf
0.2887
=
57.735
200
Assumptions
•
The ladder is uniform, so its weight W acts at its middle.
•
The wall is smooth: it only pushes level, Nw. The rough floor pushes up, Nf, and its friction f points toward the wall.
•
At rest the forces balance both ways, and the torques about the foot cancel, so Nf and f drop out of that balance.
•
The ladder holds while μₛNf ≥ f, so the least μₛ is f ÷ Nf; the ladder’s length cancels.
Step-by-step
WE KNOW
W = 200 N, θ = 60°
FIND
normal force from the floor (Nf), normal force from the wall (Nw), friction at the foot (f), least static friction coefficient (μₛ)
Step 1 · Find normal force from the floor (Nf)
Use Nf = W
Up and down: the smooth wall pushes only level, so the floor alone holds the weight.
Nf = 200 N
Step 2 · Find normal force from the wall (Nw)
Use Nw = W ÷ (2 × tan(θ))
Torques
about
the
foot:
Nw
×
L
sin
θ
=
W
×
1
2
L
cos
θ
,
and
the
length
L
cancels.
Nw
=
200
2 × tan(60°)
Nw
=
200
2 × 1.732
Nw
=
200
3.464
Nw = 57.735 N
Step 3 · Find friction at the foot (f)
Use f = Nw
Level forces: the friction at the foot balances the wall’s push.
f
=
57.735
N
Step 4 · Find least static friction coefficient (μₛ)
Use μₛ = f ÷ Nf
The
floor
holds
while
μₛNf
≥
f
,
so
the
least
μₛ
is
f
divided
by
Nf.
μₛ
=
57.735
200
μₛ = 0.2887
Check
Put the numbers back in. Both sides should match.
200 = 200 ✓
57.735
=
200
2 × tan(60°)
✓
57.735 = 57.735 ✓
0.2887
=
57.735
200
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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