Chapter 8, Problem 8.136RP

A 120-lb cabinet is mounted on casters that can be locked to prevent their rotation. The coefficient of static friction between the floor and each caster is 0.30. If h = 32 in., determine the magnitude of the force P required to move the cabinet to the right (a) if all casters are locked, (b) if the casters at B are locked and the casters at A are free to rotate, (c) if the casters at A are locked and the casters at B are free to rotate.

Fig. P8.136

To determine

Find the magnitude of the force P requires to move the cabinet to the right if all casters are locked.

Answer to Problem 8.136RP

The magnitude of the force P required is $\underset{\_}{36\text{lb}(\to )}$.

Explanation of Solution

Given information:

The weight of the cabinet is $W=120\text{lb}$.

The coefficient of static friction between the floor and the caster is ${\mu }_s=0.30$.

The height at which the force P is acting from the bottom is $h=32\text{in}\text{.}$.

Calculation:

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

For tipping; $N_A=F_A=0$.

Find the tipping force P by taking moment about point B.

$\begin{array}{l}{\displaystyle \sum M_B=0}\\ 120\left(12\right)-P_{tip}\left(32\right)=0\\ P_{tip}=45\text{lb}\end{array}$

Find the friction force $\left(F_A\right)$  at point A using the relation.

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

Find the friction force $\left(F_B\right)$ at point B using the relation.

$F_B={\mu }_s{N}_B$

Resolve the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ N_A+N_B-120=0\\ N_A+N_B=120\end{array}$

Resolve the horizontal component of forces.

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ P-F_A-F_B=0\\ P-{\mu }_s{N}_A-{\mu }_s{N}_B=0\\ P-{\mu }_s\left(N_A+N_B\right)=0\end{array}$

Substitute 0.30 for ${\mu }_s$ and 120 lb for $\left(N_A+N_B\right)$.

$\begin{array}{l}P-0.30\times 120=0\\ P=36\text{lb}(\to )\end{array}$

Here,

$P=36\text{lb}<P_{tip}=45\text{lb}$

Therefore, the magnitude of the force P required is $\underset{\_}{36\text{lb}(\to )}$.

To determine

Find the magnitude of the force P requires to move the cabinet to the right if the casters at B are locked and at A is free.

Answer to Problem 8.136RP

The magnitude of the required force P is $\underset{\_}{30\text{lb}(\to )}$.

Explanation of Solution

Given information:

The weight of the cabinet is $W=120\text{lb}$.

The coefficient of static friction between the floor and the caster is ${\mu }_s=0.30$.

The height at which the force P is acting from the bottom is $h=32\text{in}\text{.}$.

Calculation:

Refer to Figure 1;

Find the friction force $\left(F_A\right)$  at point A using the relation.

$F_A=0$

Find the friction force $\left(F_B\right)$ at point B using the relation.

$F_B={\mu }_s{N}_B$

Resolve the horizontal component of forces.

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ P-0-F_B=0\\ P-{\mu }_s{N}_B=0\\ N_B=\frac{P}{{\mu }_s}\end{array}$

Find the magnitude of the force P by taking moment about point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ -120\left(12\right)-P\left(32\right)+N_B\left(24\right)=0\\ -1,440-32P+24\frac{P}{{\mu }_s}=0\end{array}$

Substitute 0.30 for ${\mu }_s$.

$\begin{array}{l}-1,440-32P+24\times \frac{P}{0.30}=0\\ -32P+80P=1,440\\ P=30\text{lb}\end{array}$

Here,

$P=30\text{lb}<P_{tip}=45\text{lb}$

Therefore, the magnitude of the force P required is $\underset{\_}{30\text{lb}(\to )}$.

To determine

Find the magnitude of the force P requires to move the cabinet to the right if the casters at A are locked and at B is free.

Answer to Problem 8.136RP

The magnitude of the force P required is $\underset{\_}{12.86\text{lb}(\to )}$.

Explanation of Solution

Given information:

The weight of the cabinet is $W=120\text{lb}$.

The coefficient of static friction between the floor and the caster is ${\mu }_s=0.30$.

The height at which the force P is acting from the bottom is $h=32\text{in}\text{.}$.

Calculation:

Refer to Figure 1;

Find the friction force $\left(F_A\right)$  at point A using the relation.

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

Find the friction force $\left(F_B\right)$ at point B using the relation.

$F_B=0$

Resolve the horizontal component of forces.

$\begin{array}{l}{\displaystyle \sum F_x=0}\\ P-0-F_A=0\\ P-{\mu }_s{N}_A=0\\ N_A=\frac{P}{{\mu }_s}\end{array}$

Find the magnitude of the force P by taking moment about point A.

$\begin{array}{l}{\displaystyle \sum M_B=0}\\ 120\left(12\right)-P\left(32\right)-N_A\left(24\right)=0\\ 1,440-32P-24\frac{P}{{\mu }_s}=0\end{array}$

Substitute 0.30 for ${\mu }_s$.

$\begin{array}{l}1,440-32P-24\times \frac{P}{0.30}=0\\ 32P+80P=1,440\\ P=12.86\text{lb}\end{array}$

Here,

$P=12.86\text{lb}<P_{tip}=45\text{lb}$

Therefore, the magnitude of the force P required is $\underset{\_}{12.86\text{lb}(\to )}$.