Chapter 8.1, Problem 8.42P

(a) Show that the beam of Prob. 8.41 cannot be moved if the top surface of the dolly is slightly lower than the platform. (b) Show that the beam can be moved if two 175-lb workers stand on the beam at B, and determine how far to the left the beam can be moved.

8.41 A 10-ft beam, weighing 1200 lb, is to be moved to the left onto the platform as shown. A horizontal force P is applied to the dolly, which is mounted on frictionless wheels. The coefficients of friction between all surfaces are μ_s = 0.30 and μ_s = 0.25, and initially, χ = 2 ft. Knowing that the top surface of the dolly is slightly higher than the platform, determine the force P required to start moving the beam. (Hint: The beam is supported at A and D.)

Fig. P8.41

To determine

Show that the beam cannot be moved if the top surface of the dolly is slightly lower than the platform.

Explanation of Solution

Given information:

The length of the beam is 10 ft.

The weight of the beam is $W=1,200\text{lb}$.

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

The coefficient of kinetic friction between the surfaces is ${\mu }_k=0.25$.

Calculation:

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

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

$\begin{array}{l}{\displaystyle \sum M_C=0}\\ N_B\left(8\right)-1,200\left(3\right)=0\\ N_B=450\text{lb}(\uparrow )\end{array}$

Find the normal force at point C by resolving the vertical component of forces.

$\begin{array}{l}{\displaystyle \sum F_y=0}\\ N_C-1,200+N_B=0\\ N_C-1,200+450=0\\ N_C=750\text{lb}(\uparrow )\end{array}$

Find the maximum friction force at point C ${\left(F_C\right)}_m$ using the relation.

${\left(F_C\right)}_m={\mu }_s{N}_C$

Substitute 0.30 for ${\mu }_s$ and 750 lb for $N_C$.

$\begin{array}{c}{\left(F_C\right)}_m=0.30\times 750\\ =225\text{lb}\end{array}$

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

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

Substitute 0.30 for ${\mu }_s$ and 450 lb for $N_B$.

$\begin{array}{c}{\left(F_B\right)}_m=0.30\times 450\\ =135\text{lb}\end{array}$

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

${\left(F_B\right)}_m=135\text{lb}<{\left(F_C\right)}_m=225\text{lb}$

The sliding is about to happen at point B.

Therefore, the beam $\underset{\_}{cannot}$ be moved.

To determine

Show that the beam can be moved if two 175-lb workers stand on the beam at B.

Find the distance the beam moves to the left.

Answer to Problem 8.42P

The distance the beam moves to the left is $\underset{\_}{x=2.90\text{ft}}$.

Explanation of Solution

Given information:

The length of the beam is 10 ft.

The weight of the beam is $W=1,200\text{lb}$.

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

The coefficient of kinetic friction between the surfaces is ${\mu }_k=0.25$.

Calculation:

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

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

$\begin{array}{l}{\displaystyle \sum M_C=0}\\ N_B\left(10-x\right)-1,200\left(5-x\right)-350\left(10-x\right)=0\\ N_B\left(10-x\right)-6,000+1,200x-3,500+350x=0\\ N_B=\frac{9,500-1,550x}{10-x}\end{array}$ (1)

Find the normal reaction at point C by taking moment about point B.

$\begin{array}{l}{\displaystyle \sum M_B=0}\\ 1,200\left(5\right)-N_C\left(10-x\right)=0\\ 6,000-N_C\left(10-x\right)=0\\ N_C=\frac{6,000}{10-x}\end{array}$ (2)

When two 175 lb workers stand on the end B: $x=2\text{ft}$.

Substitute 2 ft for x in Equation (1).

$\begin{array}{c}{N}_B=\frac{9,500-1,550\left(2\right)}{10-2}\\ =\frac{6,400}{8}\\ =800\text{lb}\end{array}$

Substitute 2 ft for x in Equation (2).

$\begin{array}{c}{N}_C=\frac{6,000}{10-2}\\ =750\text{lb}\end{array}$

Find the maximum friction force at point C ${\left(F_C\right)}_m$ using the relation.

${\left(F_C\right)}_m={\mu }_s{N}_C$

Substitute 0.30 for ${\mu }_s$ and 750 lb for $N_C$.

$\begin{array}{c}{\left(F_C\right)}_m=0.30\times 750\\ =225\text{lb}\end{array}$

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

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

Substitute 0.30 for ${\mu }_s$ and 800 lb for $N_B$.

$\begin{array}{c}{\left(F_B\right)}_m=0.30\times 800\\ =240\text{lb}\end{array}$

The maximum friction force at point B is greater than the maximum friction force at point C.

${\left(F_B\right)}_m=240\text{lb}>{\left(F_C\right)}_m=225\text{lb}$

The sliding is about to happen at point C.

Therefore, the beam $\underset{\_}{can}$ be moved.

The beam will stop moving when the friction force at point C is equal to the maximum friction force at point B.

$F_C={\left(F_B\right)}_m$ (3)

Find the friction force at point C $\left(F_C\right)$ using the relation.

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

Substitute 0.25 for ${\mu }_k$ and $\frac{6,000}{10-x}$ for $N_C$.

$\begin{array}{l}{F}_C=0.25\left(\frac{6,000}{10-x}\right)\\ F_C\text{=}\frac{1,500}{10-x}\end{array}$

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

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

Substitute 0.30 for ${\mu }_s$ and $\frac{9,500-1,550x}{10-x}$ for $N_B$.

$\begin{array}{c}{\left(F_B\right)}_m=0.30\left(\frac{9,500-1,550x}{10-x}\right)\\ =\frac{2,850-465x}{10-x}\end{array}$

Substitute $\frac{1,500}{10-x}$ for $F_C$ and $\frac{2,850-465x}{10-x}$ for ${\left(F_B\right)}_m$ in Equation (3).

$\begin{array}{l}\frac{1,500}{10-x}=\frac{2,850-465x}{10-x}\\ 1,500=2,850-465x\\ x=2.90\text{ft}\end{array}$

Therefore, the distance the beam moves to the left is $\underset{\_}{x=2.90\text{ft}}$.