Chapter 7, Problem 7.4P

The two homogeneous bars AB and BC are connected with a pin at B and placed between rough vertical walls. If the coefficient of static friction between each bar and the wall is 0.25, determine the largest angle 6 for which the assembly will remain at rest.

To determine

Largest angle θ for which the assembly will remain at rest

Answer to Problem 7.4P

The largest angle θ is equal to $6.75°$.

Explanation of Solution

Given information:

Co-efficient of static friction between each bar and wall is $0.25$.

For impending sliding, the friction force equals its limiting value;

  $F=F_{\max }={\mu }_{s}N$

  ${\mu }_s$ - Co-efficient of static friction

Steps to follow in the equilibrium analysis of a body are:

1. Draw the free body diagram.

2. Write the equilibrium equations.

3. Solve the equations for the unknowns.

Calculation:

FBD of assembly

Assume $N_A,N_C$ as the reaction force at point A and C.

Assume $F_A,F_C$ as the friction force at point A and C.

For the equilibrium of entire section, the bending moment about point A is equal to zero.

  ${\displaystyle \sum M_A=0}$

  $F_C\left(\left(48in\right)\cos \theta \right)-\left(8lb\right)\left(\left(12in\right)\cos \theta \right)-\left(10lb\right)\left(\left(36in\right)\cos \theta \right)=0$

Solve

  $\begin{array}{l}48F_C-96-360=0\\ F_C=9.5lb\end{array}$

Write equilibrium equation in vertical direction.

  $\uparrow {\displaystyle \sum F_y}=0$

  $\begin{array}{l}{F}_A+F_C-\left(8lb\right)-\left(10lb\right)=0\\ F_A=8.5lb\end{array}$

Write equilibrium equation in horizontal direction.

  $\to {\displaystyle \sum F_x}=0$

  $\begin{array}{l}{N}_A-N_C=0\\ N_A=N_C\end{array}$

The impending sliding will first occur at point C.

  $\frac{F_C}{N_C}>\frac{F_A}{N_A}$

The reaction force $N_C$ at point C

  $\begin{array}{l}{N}_C=\frac{F_C}{0.25}\\ =\frac{9.5lb}{0.25}\\ =38lb\end{array}$

FBD of bar BC

For the equilibrium of above section, the bending moment about point B is equal to zero.

  ${\displaystyle \sum M_B=0}$

  $\begin{array}{l}{F}_C\left(\left(24in\right)\cos \theta \right)-N_C\left(\left(24in\right)sin\theta \right)-\left(10lb\right)\left(\left(12in\right)\cos \theta \right)=0\\ 9.5\left(24\cos \theta \right)-38\left(24sin\theta \right)-10\left(12\cos \theta \right)=0\\ 108\cos \theta -912sin\theta =0\end{array}$

Therefore

  $\begin{array}{l}\theta ={\tan }^{-1}\left(\frac{108}{912}\right)\\ =6.75°\end{array}$

Conclusion:

The largest angle θ is equal to $6.75°$.