A ladder leaning against a wall can slip if the angle with the floor is too small. The bottom of the ladder tends
to slide outward, while the top tends to slide downward. Static friction at the floor and, if present, static
friction at the wall help prevent this motion.
1. Forces on the ladder
For a uniform ladder of mass \(m\) and length \(L\), the weight \(mg\) acts at the midpoint. The contact forces are:
- Floor normal force \(N_f\), upward.
- Floor friction \(f_f\), horizontally toward the wall.
- Wall normal reaction \(N_w\), horizontally away from the wall.
- Wall friction \(f_w\), upward if the top of the ladder tends to slide downward.
The ladder is at the limiting angle when the required friction reaches its maximum available value.
2. Smooth wall, rough floor
If the wall is smooth, then \(f_w=0\). Force balance gives
\[
f_f=N_w,
\qquad
N_f=mg.
\]
At impending slip,
\[
f_f=\mu_fN_f=\mu_fmg.
\]
Taking torques about the bottom contact point,
\[
N_wL\sin\theta
=
mg\frac{L}{2}\cos\theta.
\]
Since \(N_w=\mu_fmg\),
\[
\mu_fmgL\sin\theta
=
mg\frac{L}{2}\cos\theta.
\]
Cancel \(mgL\):
\[
\mu_f\sin\theta=\frac12\cos\theta.
\]
Therefore,
\[
\tan\theta_{\min}=\frac{1}{2\mu_f}.
\]
\[
\theta_{\min}=\tan^{-1}\!\left(\frac{1}{2\mu_f}\right).
\]
3. Rough wall and rough floor
If the wall has friction coefficient \(\mu_w\), then wall friction can help support the ladder. At the limiting
condition:
\[
f_f=\mu_fN_f,
\qquad
f_w=\mu_wN_w.
\]
Force balance gives
\[
f_f=N_w,
\]
\[
N_f+f_w=mg.
\]
Substitute \(f_w=\mu_wN_w\):
\[
N_f+\mu_wN_w=mg.
\]
Torque balance about the bottom contact point gives
\[
N_wL\sin\theta+f_wL\cos\theta
=
mg\frac{L}{2}\cos\theta.
\]
Substitute \(f_w=\mu_wN_w\):
\[
N_wL\sin\theta+\mu_wN_wL\cos\theta
=
mg\frac{L}{2}\cos\theta.
\]
Divide by \(L\cos\theta\):
\[
N_w(\tan\theta+\mu_w)=\frac{mg}{2}.
\]
Combining this with the limiting friction condition gives
\[
\tan\theta_{\min}
=
\frac{1-\mu_f\mu_w}{2\mu_f}.
\]
Hence,
\[
\theta_{\min}
=
\tan^{-1}\!\left(\frac{1-\mu_f\mu_w}{2\mu_f}\right).
\]
If \(\mu_f\mu_w\ge1\), the ideal model predicts no positive minimum angle. In a real ladder problem, geometry,
deformation, and contact limits would still matter.
4. Force values at the limiting angle
Once the limiting angle is known, the wall reaction can be found from the torque equation:
\[
N_w=\frac{mg}{2(\tan\theta_{\min}+\mu_w)}.
\]
The wall friction is
\[
f_w=\mu_wN_w.
\]
The floor normal force is
\[
N_f=mg-f_w.
\]
The floor friction force is
\[
f_f=N_w.
\]
5. Slip/no-slip check for a chosen angle
An entered angle \(\theta\) is safe in this ideal model when
\[
\theta\ge\theta_{\min}.
\]
If
\[
\theta<\theta_{\min},
\]
the ladder requires more friction than is available and would begin to slip.
6. Why mass and length cancel
The ladder’s mass and length affect the actual force values, but they cancel in the limiting angle formula. This
happens because all torques scale with \(L\), and all contact forces at the limiting condition scale with \(mg\).
Therefore, for the ideal uniform-ladder model, \(\theta_{\min}\) depends only on the friction coefficients.
7. Worked example: smooth wall
Suppose
\[
\mu_f=0.4,
\qquad
\mu_w=0.
\]
Then
\[
\theta_{\min}
=
\tan^{-1}\!\left(\frac{1}{2(0.4)}\right).
\]
\[
\theta_{\min}
=
\tan^{-1}(1.25).
\]
\[
\theta_{\min}\approx51.3^\circ.
\]
8. Worked example: rough wall
If the wall also has friction, for example
\[
\mu_f=0.4,
\qquad
\mu_w=0.2,
\]
then
\[
\theta_{\min}
=
\tan^{-1}\!\left(\frac{1-(0.4)(0.2)}{2(0.4)}\right).
\]
\[
\theta_{\min}
=
\tan^{-1}(1.15).
\]
\[
\theta_{\min}\approx49.0^\circ.
\]
Wall friction lowers the minimum angle because it helps support the top of the ladder.
Key idea: the minimum angle is found by combining force balance, torque balance, and the maximum static friction
condition.
