Chapter 8.2, Problem 8.65P

A 15° wedge is forced under a 50-kg pipe as shown. Knowing that the coefficient of static friction at both surfaces of the wedge is 0.20, determine the largest coefficient of static friction between the pipe and the vertical wall for which slipping will occur at A.

Fig. P8.64 and P8.65

To determine

Find the largest coefficient of static friction between the pipe and the vertical wall for the condition.

Answer to Problem 8.65P

The largest coefficient of static friction between the pipe and the vertical wall is $\underset{\_}{0.442}$.

Explanation of Solution

Given information:

The mass of the pipe is $m=50\text{kg}$.

The value of angle $\theta$ is $15°$.

The coefficient of static friction at all surfaces is ${\left({\mu }_s\right)}_B=0.20$.

Calculation:

Find the weight (W) of the pipe using the relation.

$W=mg$

Here, the acceleration due to gravity is g.

Consider the acceleration due to gravity is $g=9.81{\text{m/s}}^2$.

Substitute 50 kg for m and $9.81{\text{m/s}}^2$ for g.

$\begin{array}{c}W=50\times 9.81\\ =490.5\text{N}\end{array}$

Show the free-body diagram of the pipe as in Figure 1.

Find the friction force at point B using the relation.

$F_B={\left({\mu }_s\right)}_B{N}_B$

Here, the normal force at point B is $N_B$.

Find the normal force at point B by taking moment about point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ N_B\left(r\cos \theta \right)-F_B\left(r\right)-F_B\sin \theta \left(r\right)-W\left(r\right)=0\\ N_{B}r\cos \theta -{\left({\mu }_s\right)}_B{N}_{B}r-{\left({\mu }_s\right)}_B{N}_{B}r\sin \theta -Wr=0\\ N_B\cos \theta -{\left({\mu }_s\right)}_B{N}_B-{\left({\mu }_s\right)}_B{N}_B\sin \theta -W=0\end{array}$

Substitute 490.5 N for W, 0.20 for ${\left({\mu }_s\right)}_B$ and $15°$ for $\theta$.

$\begin{array}{l}{N}_B\cos 15°-0.20N_B-0.20N_B\sin 15°-490.50=0\\ 0.966N_B-0.20N_B-0.052N_B-490.50=0\\ 0.714N_B=490.50\\ N_B=686.975\text{N}\end{array}$

Find the normal force at point A $\left(N_A\right)$ by resolving the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ N_A-N_B\sin \theta -F_B\cos \theta =0\\ N_A-N_B\sin \theta -{\left({\mu }_s\right)}_B{N}_B\cos \theta =0\end{array}$

Substitute 0.20 for ${\left({\mu }_s\right)}_B$, 686.975 N for $N_B$, and $15°$ for $\theta$.

$\begin{array}{l}{N}_A-686.975\sin 15°-0.20\times 686.975\times \cos 15°=0\\ N_A-177.802-132.713=0\\ N_A=310.515\text{N}\end{array}$

Find the friction force at point A $\left(F_A\right)$ by resolving the vertical component of forces.

$\begin{array}{l}+\uparrow {\displaystyle \sum F_y=0}\\ -F_A-W+N_B\cos \theta -F_B\sin \theta =0\\ -F_A-W+N_B\cos \theta -{\left({\mu }_s\right)}_B{N}_B\sin \theta =0\end{array}$

Substitute 490.5 N for W, 0.20 for ${\left({\mu }_s\right)}_B$, 686.975 N for $N_B$, and $15°$ for $\theta$.

$\begin{array}{l}-F_A-490.5+686.975\cos 15°-0.20\times 686.975\times \sin 15°=0\\ -F_A-490.5+663.567-35.560=0\\ F_A=137.507N\end{array}$

Find the coefficient of static friction at point A ${\left({\mu }_s\right)}_A$ using the relation.

$F_A={\left({\mu }_s\right)}_A{N}_A$

Substitute 137.507 N for $F_A$ and 310.515 N for $N_A$.

$\begin{array}{c}137.507={\left({\mu }_s\right)}_A\times 310.515\\ {\left({\mu }_s\right)}_A=0.442\end{array}$

Therefore, the largest coefficient of static friction between the pipe and the vertical wall is $\underset{\_}{0.442}$.