Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 16P from Chapter 8 from Hibbeler's Engineering Mechanics.
Problem 16P
Step-by-Step Solution
We have given the following values:
The weight of the man is $W = 180\;{\rm{lb}}$.
The coefficient of static friction between the friction pad A and the ground is ${\mu _s} = 0.4$.
We are asked to calculate the inclination angle $\theta $ of the ladder when the ladder is on the verge of slipping.
Step 2
Draw a labeled free-body diagram of the ladder.
Here, ${F_A}$ is the friction force at A, ${N_A}$ is the normal reaction force at point A, and ${N_B}$ is the normal reaction force at point B.
Step 3
Apply equilibrium equation of forces along the vertical direction:
\[\begin{array}{c} \sum {{F_y}} = 0\\ {N_A} - W = 0\\ {N_A} = W \end{array}\]Substitute the value of W in the above equation:
\[{N_A} = 180\;{\rm{lb}}\]Step 4
The friction force is given by:
\[{F_A} = {\mu _s}{N_A}\]Substitute the value of ${\mu _s}$ and ${N_A}$ in the above equation:
\[\begin{array}{c} {F_A} = \left( {0.4} \right) \times \left( {180\;{\rm{lb}}} \right)\\ {F_A} = 72\;{\rm{lb}} \end{array}\]Step 5
Apply equilibrium equation for the moment about point B:
\[\begin{array}{c} \sum {{M_B}} = 0\\ {N_A} \times 10\;{\rm{ft}} \times \cos \theta - {F_A} \times 10\;{\rm{ft}} \times \sin \theta - W \times 3\;{\rm{ft}} = 0 \end{array}\]Substitute the value of ${N_A}$ and ${F_A}$ in the above equation:
\[\begin{array}{c} \left( {180\;{\rm{lb}}} \right) \times 10\;{\rm{ft}} \times \cos \theta - \left( {72\;{\rm{lb}}} \right) \times 10\;{\rm{ft}} \times \sin \theta - \left( {180\;{\rm{lb}}} \right) \times 3\;{\rm{ft}} = 0\\ \cos \theta - 0.4\sin \theta = 0.3 \end{array}\]Step 6
On applying hit and trial method to solve the above equation, we get:
\[\theta = 52.0^\circ \]