Question

A ladder is leaning against a vertical wall, and both ends of the ladder are at the point of slipping. The coefficient of friction between the ladder and the horizontal surface is μ1 = 0.195 and the coefficient of friction between the ladder and the wall is μ2 = 0.253. Determine the maximum angle with the vertical the ladder can make without falling on the ground.

Answer 1

There are two answers that I have yet to confirm. Someone got 37.62 degrees and I got 22.305 degrees. What do you guys get? Oh the topic is static equilibrium. Thanks!

Answer 2

here we have to apply the cardinal equations of mechanics for equilibrium of a rigid body

Answer 3

your problem can be represented by this drawing:

where R_a and R_b are perpendicular to horizontal and vertical surfaces respectively, and F_mu_A, and F_mu_B are the friction forces which act on the ladder

Answer 4

here are the vector equations:
\[\Large \left\{ \begin{gathered}
{{\mathbf{F}}_{{\mathbf{\mu B}}}}{\mathbf{ + }}{{\mathbf{R}}_{\mathbf{B}}}{\mathbf{ + }}M{\mathbf{g + }}{{\mathbf{R}}_{\mathbf{A}}}{\mathbf{ + }}{{\mathbf{F}}_{{\mathbf{\mu A}}}}{\mathbf{ = 0}} \hfill \\
\\
{\mathbf{AG \times }}M{\mathbf{g + AB \times }}{{\mathbf{F}}_{{\mathbf{\mu B}}}}{\mathbf{ + AB \times }}{{\mathbf{R}}_{\mathbf{B}}}{\mathbf{ = 0}} \hfill \\
\end{gathered} \right.\]

Answer 5

please tell me when I may continue

Answer 6

Ok, so I see that the frictional forces are immediately equated but the sum of the torques are essential. Hmm. If you take a look at the answers I have posted, I'm not sure which one is right.

Answer 7

Solving that system above, I got this relationship:
\[\Large \tan \alpha = \frac{{{\mu _A}}}{{1 - {\mu _A}{\mu _B}}}\]
where:
\[\Large \begin{gathered}
{\mu _A} = 0.195 \hfill \\
{\mu _B} = 0.253 \hfill \\
\end{gathered} \]

Answer 8

Ok. So, I have the answer. My answer is correct. the answer is 22.305. The way to solve it was to find the minimum the bottom angle could be then that gives the maximum angle at the top. Thank you for trying anyway. The solution requires use of normal forces and relationships between the Fg and Fw and the muFg.

Answer 9

Here:
Sum Fx| Fgx=Fw
SumFy| Fgy=mg-mu2xFw
SumTorques | Fw(Lcos(theta))=mg(L/2)sin(theta) - Now,this leads to-->
Fw=Fg=(mg)/(2tan(theta))

For no slipping Fgx<=mu1(Fgy)

Just make the substitutions and it gets ugly but the final thing you get should be
theta= arctan[(1-mu1x mu2)/(2mu1)) , which equals 67.946 then subtract from 90 and viola! you get the answer of 22.305
I hope this helps in case you come across this question again.