Chapter 8.1, Problem 8.37P

A 1.2-m plank with a mass of 3 kg rests on two joists. Knowing that the coefficient of static friction between the plank and the joists is 0.30, determine the magnitude of the horizontal force required to move the plank when (a) a = 750 mm, (b) a = 900 mm.

Fig. P8.37

To determine

Find the magnitude of the horizontal force required to move the plank.

Answer to Problem 8.37P

The magnitude of the horizontal force required to move the plank is $\underset{\_}{2.94\text{N}}$.

Explanation of Solution

Given information:

The length of the plank is $L=1.2\text{m}$.

The mass of each plank is $m=3\text{kg}$.

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

The distance between the points A and C in the plank is $a=750\text{mm}$.

Calculation:

Find the friction force (F) using the relation.

$F={\mu }_{s}N$

Show the free-body diagram of the member AB is vertical plane as in Figure 1.

Take moment about point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ N_C\left(a\right)-W\left(\frac{L}{2}\right)=0\\ N_C=\frac{WL}{2a}\end{array}$ (1)

Resolve the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ N_A+N_C-W=0\\ N_A+\frac{WL}{2a}-W=0\\ N_A=\frac{W}{2a}\left(2a-L\right)\end{array}$ (2)

Show the free-body diagram of the member AB is horizontal plane as in Figure 2.

Take moment about point A.

$\begin{array}{l}{\displaystyle \sum M_A=0}\\ F_C\left(a\right)-P\left(L\right)=0\\ F_C=\frac{PL}{a}\end{array}$ (3)

Resolve the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_z=0}\\ -F_A-P+F_C=0\\ -F_A-P+\frac{PL}{a}=0\\ F_A=\frac{P}{a}\left(L-a\right)\end{array}$ (4)

Find the weight of the plank (W) 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 3 kg for m and $9.81{\text{m/s}}^2$ for g.

$\begin{array}{c}W=3\times 9.81\\ =29.43\text{N}\end{array}$

Substitute 29.43 N for W, 1.2 m for L, and 750 mm for a in Equation (1).

$\begin{array}{c}{N}_C=\frac{29.43\times 1.2}{2\times 750\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}}\\ =23.544\text{N}\end{array}$

Substitute 29.43 N for W, 1.2 m for L, and 750 mm for a in Equation (2).

$\begin{array}{c}{N}_A=\frac{29.43}{2\times 750\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}}\left(2\times 750\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}-1.2\right)\\ =5.886\text{N}\end{array}$

Substitute 1.2 m for L, and 750 mm for a in Equation (3).

$\begin{array}{c}{F}_C=\frac{P\times 1.2}{750\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}}\\ =1.6P\end{array}$

Substitute 1.2 m for L, and 750 mm for a in Equation (4).

$\begin{array}{c}{F}_A=\frac{P}{750\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}}\left(1.2-750\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)\\ =0.6P\end{array}$

At point A, the plank to slip;

Find the horizontal force P using the relation.

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

Substitute 0.6P for $F_A$, 0.30 for ${\mu }_s$, and 5.886 N for $N_A$.

$\begin{array}{l}0.6P=0.30\times 5.886\\ P=2.94\text{N}\end{array}$

At point C, the plank to slip;

Find the horizontal force P using the relation.

$F_C={\mu }_s{N}_C$

Substitute 1.6P for $F_C$, 0.30 for ${\mu }_s$, and 23.544 N for $N_C$.

$\begin{array}{l}1.6P=0.30\times 23.544\\ P=4.41\text{N}\end{array}$

The smallest value of P will slip the plank. The plank will slip at A.

Therefore, the magnitude of the horizontal force required is $\underset{\_}{2.94\text{N}}$.

To determine

Find the magnitude of the horizontal force required to move the plank.

Answer to Problem 8.37P

The magnitude of the horizontal force required is $\underset{\_}{4.41\text{N}}$.

Explanation of Solution

Given information:

The length of the plank is $L=1.2\text{m}$.

The mass of each plank is $m=3\text{kg}$.

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

The distance between the points A and C in the plank is $a=900\text{mm}$.

Calculation:

Refer part (a) for calculation.

Substitute 29.43 N for W, 1.2 m for L, and 900 mm for a in Equation (1).

$\begin{array}{c}{N}_C=\frac{29.43\times 1.2}{2\times 900\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}}\\ =19.62\text{N}\end{array}$

Substitute 29.43 N for W, 1.2 m for L, and 900 mm for a in Equation (2).

$\begin{array}{c}{N}_A=\frac{29.43}{2\times 900\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}}\left(2\times 900\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}-1.2\right)\\ =9.81\text{N}\end{array}$

Substitute 1.2 m for L, and 900 mm for a in Equation (3).

$\begin{array}{c}{F}_C=\frac{P\times 1.2}{900\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}}\\ =1.3333P\end{array}$

Substitute 1.2 m for L, and 900 mm for a in Equation (4).

$\begin{array}{c}{F}_A=\frac{P}{900\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}}\left(1.2-900\text{mm}\times \frac{\text{1}\text{m}}{\text{1,000}\text{mm}}\right)\\ =0.3333P\end{array}$

At point A, the plank to slip;

Find the horizontal force P using the relation.

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

Substitute 0.3333P for $F_A$, 0.30 for ${\mu }_s$, and 9.81 N for $N_A$.

$\begin{array}{l}0.3333P=0.30\times 9.81\\ P=8.83\text{N}\end{array}$

At point C, the plank to slip;

Find the horizontal force P using the relation.

$F_C={\mu }_s{N}_C$

Substitute 1.3333P for $F_C$, 0.30 for ${\mu }_s$, and 19.62 N for $N_C$.

$\begin{array}{l}1.3333P=0.30\times 19.62\\ P=4.41\text{N}\end{array}$

The smallest value of P will slip the plank. The plank will slip at C.

Therefore, the magnitude of the horizontal force required is $\underset{\_}{4.41\text{N}}$.