Chapter 8.2, Problem 8.64P

A 15° wedge is forced under a 50-kg pipe as shown. The coefficient of static friction at all surfaces is 0.20. (a) Show that slipping will occur between the pipe and the vertical wall. (b) Determine the force P required to move the wedge.

Fig. P8.64 and P8.65

To determine

Show that the slipping will occur between the pipe and the vertical wall.

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 ${\mu }_s=0.20$.

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 A using the relation.

$F_A={\mu }_s{N}_A$

Here, the normal force at point A is $N_A$.

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

$\begin{array}{l}{\displaystyle \sum M_B=0}\\ W\left(r\sin \theta \right)+F_A\left(r+r\sin \theta \right)-N_A\left(r\cos \theta \right)=0\\ Wr\sin \theta +{\mu }_s{N}_{A}r\left(1+\sin \theta \right)-N_{A}r\cos \theta =0\\ W\sin \theta +{\mu }_s{N}_A\left(1+\sin \theta \right)-N_A\cos \theta =0\end{array}$

Substitute 490.5 N for W, 0.20 for ${\mu }_s$ and $15°$ for $\theta$.

$\begin{array}{l}490.5\sin 15°+0.20N_A\left(1+\sin 15°\right)-N_A\cos 15°=0\\ 126.951+0.252N_A-0.966N_A=0\\ 0.714N_A=126.951\\ N_A=177.803\text{N}\end{array}$

Find the friction force at point B $\left(F_B\right)$ by resolving the horizontal component of forces.

$\begin{array}{l}+\nearrow {\displaystyle \sum F_x=0}\\ -F_B-W\sin \theta -F_A\sin \theta +N_A\cos \theta =0\\ -F_B-W\sin \theta -{\mu }_s{N}_A\sin \theta +N_A\cos \theta =0\end{array}$

Substitute 490.5 N for W, 0.20 for ${\mu }_s$, 177.803 N for $N_A$, and $15°$ for $\theta$.

$\begin{array}{l}-F_B-490.5\sin 15°-0.20\times 177.803\times \sin 15°+177.803\cos 15°=0\\ -F_B-126.951-9.204+171.745=0\\ F_B=35.59\text{N}\end{array}$

Find the normal force at point B $\left(N_B\right)$ by resolving the vertical component of forces.

$\begin{array}{l}+\nwarrow {\displaystyle \sum F_y=0}\\ N_B-W\cos \theta -F_A\cos \theta -N_A\sin \theta =0\\ N_B-W\cos \theta -{\mu }_s{N}_A\cos \theta -N_A\sin \theta =0\end{array}$

Substitute 490.5 N for W, 0.20 for ${\mu }_s$, 177.803 N for $N_A$, and $15°$ for $\theta$.

$\begin{array}{l}{N}_B-490.5\cos 15°-0.20\times 177.803\times \cos 15°-177.803\times \sin 15°=0\\ N_B-473.787-34.349-46.019=0\\ N_B=554.155\text{N}\end{array}$

Find the maximum friction force at point B using the relation.

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

Substitute 0.20 for ${\mu }_s$ and 554.155 N for $N_B$.

$\begin{array}{c}{\left(F_B\right)}_m=0.20\times 554.155\\ =110.831\text{N}\end{array}$

The friction force at point B is less than the maximum friction force at point B.

$F_B=35.59\text{N}<{\left(F_B\right)}_m=110.831\text{N}$

Therefore, the slipping will not occur at point B and the slipping will occur between the pipe and the vertical wall.

To determine

Find the force P required to move the wedge.

Answer to Problem 8.64P

The force P required to move the wedge is $\underset{\_}{283\text{N}(\leftarrow )}$.

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 ${\mu }_s=0.20$.

Calculation:

Show the free-body diagram of the wedge as in Figure 2.

Find the normal force $\left(N_1\right)$ by resolving the vertical component of forces.

$\begin{array}{l}\uparrow {\displaystyle \sum F_y=0}\\ N_1-N_B\cos \theta +F_B\sin \theta =0\end{array}$

Substitute 554.155 N for $N_B$, $15°$ for $\theta$, and 35.59 N for $F_B$.

$\begin{array}{l}{N}_1-554.155\cos 15°+35.59\sin 15°=0\\ N_1-535.273+9.211=0\\ N_1=526.062\text{N}\end{array}$

Find the force P by resolving the horizontal component of forces.

$\begin{array}{l}+\to {\displaystyle \sum F_x=0}\\ F_B\cos \theta +N_B\sin \theta +F_1-P=0\\ F_B\cos \theta +N_B\sin \theta +{\mu }_s{N}_1-P=0\end{array}$

Substitute 554.155 N for $N_B$, $15°$ for $\theta$, 526.062 N for $N_1$, 0.20 for ${\mu }_s$, and 35.59 N for $F_B$.

$\begin{array}{l}35.59\cos 15°+554.155\sin 15+0.20\times 526.062-P=0\\ 34.377+143.426+105.212-P=0\\ P=283\text{N}(\leftarrow )\end{array}$

Therefore, the force P required to move the wedge is $\underset{\_}{283\text{N}(\leftarrow )}$.