Engineering Mechanics: Statics and Dynamics, 14th Edition
Authors: Russell C. Hibbeler
ISBN-13: 978-0133915426
See our solution for Question 17P from Chapter 8 from Hibbeler's Engineering Mechanics.
Problem 17P
Step-by-Step Solution
We have given the following values:
The weight of the man is $W = 180\;{\rm{lb}}$.
The angle of inclination of the ladder on the verge of slipping is $\theta = 60^\circ $.
We are asked to calculate the coefficient of static friction ${\mu _s}$ between the friction pad at A and the ground 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 at A is given by:
\[{F_A} = {\mu _s}{N_A}\]Substitute the value of and ${N_A}$ in the above equation:
\[{F_A} = {\mu _s} \times \left( {180\;{\rm{lb}}} \right)\]Step 5
Apply equilibrium equation for the moment about point B:
\[\begin{array}{c} \sum {{M_B}} = 0\\ {N_A} \times \left( {10\;{\rm{ft}} \times \cos 60^\circ } \right) - {F_A} \times \left( {10\;{\rm{ft}} \times \sin 60^\circ } \right) - W \times \left( {3\;{\rm{ft}}} \right) = 0 \end{array}\]Substitute the value of ${N_A}$ and ${F_A}$ in the above equation:
\[\begin{array}{c} \left( {180\;{\rm{lb}}} \right) \times 10\;{\rm{ft}} \times \cos 60^\circ - {\mu _s} \times \left( {180\;{\rm{lb}}} \right) \times 10\;{\rm{ft}} \times \sin 60^\circ - \left( {180\;{\rm{lb}}} \right) \times 3\;{\rm{ft}} = 0\\ {\mu _s} \times \left( {180\;{\rm{lb}}} \right) \times 10\;{\rm{ft}} \times \sin 60^\circ = 360\;{\rm{lb}} \cdot {\rm{ft}}\\ {\mu _s} = 0.231 \end{array}\]