Chapter 12.1, Problem 12.22P

To unload a bound stack of plywood from a truck, the driver first tilts the bed of the truck and then accelerates from rest. Knowing that the coefficients of friction between the bottom sheet of plywood and the bed are μ_s = 0.40 and μ_k = 0.30, determine (a) the smallest acceleration of the truck which will cause the stack of plywood to slide, (b) the acceleration of the truck which causes corner A of the stack to reach the end of the bed in 0.9 s.

Fig. P12.22

To determine

Find the smallest acceleration of the truck which will cause the stack of plywood to slide.

Answer to Problem 12.22P

The smallest acceleration of the truck which will cause the stack of plywood to slide is $\underset{\_}{0.309{\text{m/s}}^2}$.

Explanation of Solution

Given information:

The coefficients of static friction $\left({\mu }_s\right)$ between the bottom sheet of plywood and the bed is 0.40.

The coefficients of static friction $\left({\mu }_k\right)$ between the bottom sheet of plywood and the bed is 0.30.

The tilting angle $\left(\theta \right)$ of bed of truck is $20°$.

The relative distance of plywood with respect to truck $\left(x_{P/T}\right)$ is 2 m.

Calculation:

Write the equation of Weight of plywood $\left(W_P\right)$.

$W_P=m_{P}g$

Here, $m_P$ is the mass of plywood and g is the acceleration of gravity.

Sketch the free body diagram and kinetic diagram of plywood as shown in Figure (1).

Refer Figure (1).

Apply the Newton’s law of equation along y-axis.

$\begin{array}{l}\Sigma F_y=-m_P{a}_T\sin 20°\\ N_1-W_P\cos 20°=-m_P{a}_T\sin 20°\end{array}$

Substitute $m_{P}g$ for $W_P$.

$\begin{array}{l}{N}_1-m_{P}g\cos 20°=-m_P{a}_T\sin 20°\\ N_1=m_P\left(g\cos 20°-a_T\sin 20°\right)\end{array}$

Write the equation of friction force $\left(F_1\right)$.

$F_1={\mu }_s{N}_1$

Here, $N_1$ is the normal force on the plywood.

Substitute 0.40 for ${\mu }_s$.

$F_1=0.40N_1$

Substitute $m_P\left(g\cos 20°-a_T\sin 20°\right)$ for $N_1$.

$F_1=0.40m_P\left(g\cos 20°-a_T\sin 20°\right)$ (1)

Apply the Newton’s law of equation along x-axis.

$\begin{array}{l}\Sigma F_x=m{a}_x\\ F_1-W_P\sin 20°=m_P{a}_T\cos 20°\end{array}$

Substitute $m_{P}g$ for $W_P$.

$\begin{array}{l}{F}_1-\left(m_{P}g\right)\sin 20°=m_P{a}_T\cos 20°\\ F_1=m_P\left(g\sin 20°+a_T\cos 20°\right)\end{array}$ (2)

Find the smallest acceleration of the truck which will cause the stack of plywood to slide.

Equate Equation (1) and (2).

$\begin{array}{l}0.40m_P\left(g\cos 20°-a_T\sin 20°\right)=m_P\left(g\sin 20°+a_T\cos 20°\right)\\ 0.40g\cos 20°-0.40a_T\sin 20°=g\sin 20°+a_T\cos 20°\\ 0.40g\cos 20°-g\sin 20°=a_T\cos 20°+0.40a_T\sin 20°\\ 0.0339g=1.0765a_T\end{array}$

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

$\begin{array}{l}0.0339\left(9.81\right)=1.0765a_T\\ a_T=0.309{\text{m/s}}^2\end{array}$

Thus, the smallest acceleration of the truck which will cause the stack of plywood to slide is $\underset{\_}{0.309{\text{m/s}}^2}$.

To determine

Find the acceleration of the truck which causes corner A of the stack to reach the end of the bed in 0.9 s.

Answer to Problem 12.22P

The acceleration of the truck which causes corner A of the stack to reach the end of the bed in 0.9 s is $\underset{\_}{4.17{\text{m/s}}^2}$.

Explanation of Solution

Calculation:

The velocity of plywood relative to truck is zero.

Find the acceleration of the plywood relative to the truck $\left(a_{P/T}\right)$ using equation of kinematics.

$x_{P/T}=\left(v_{P/T}\right)t+\frac{1}{2}{a}_{P/T}{t}^2$

Here, $v_{P/T}$ is the velocity of plywood relative to truck.

Substitute 0 for $v_{P/T}$ 2 m for $x_{P/T}$, and 0.9 s for t.

$\begin{array}{l}2=0+\frac{1}{2}{a}_{P/T}{\left(0.9\right)}^2\\ a_{P/T}=4.94{\text{m/s}}^2\end{array}$

Sketch the free body diagram and kinetic diagram of truck as shown in Figure (2).

Refer Figure (2).

Apply the Newton’s law of equation along y-axis.

$\begin{array}{l}\Sigma F_y=-m_P{a}_y\\ N_2-W_P\cos 20°=-m_P{a}_T\sin 20°\end{array}$

Substitute $m_{P}g$ for $W_P$.

$\begin{array}{l}{N}_2-\left(m_{P}g\right)\cos 20°=-m_P{a}_T\sin 20°\\ N_2=m_P\left(g\cos 20°-a_T\sin 20°\right)\end{array}$ (3)

Here, $N_2$ is the normal force on truck.

Apply the Newton’s law of equation along x-axis.

$\begin{array}{l}\Sigma F_x=\Sigma m{a}_x\\ F_2-W_P\sin 20°=m_P{a}_T\cos 20°-m_P{a}_{P/T}\end{array}$

Substitute $m_{P}g$ for $W_P$.

$\begin{array}{l}{F}_2-\left(m_{P}g\right)\sin 20°=m_P{a}_T\cos 20°-m_P{a}_{P/T}\\ F_2=m_P\left(g\sin 20°+a_T\cos 20°-a_{P/T}\right)\end{array}$ (4)

Here, $F_2$ is the frictional force on truck.

Write the equation of frictional force on truck:

$F_2={\mu }_k{N}_2$

Substitute 0.30 for ${\mu }_k$ and $m_P\left(g\cos 20°-a_T\sin 20°\right)$ for $N_2$.

$F_2=0.30m_P\left(g\cos 20°-a_T\sin 20°\right)$ (5)

Find the acceleration of the truck $\left(a_T\right)$ which causes corner A of the stack to reach the end of the bed in 0.9 s

Equate Equation (4) and (5),

$\begin{array}{l}{m}_P\left(g\sin 20°+a_T\cos 20°-a_{P/T}\right)=0.30m_P\left(g\cos 20°-a_T\sin 20°\right)\\ a_T\cos 20°+0.30a_T\sin 20°=0.30g\cos 20°-g\sin 20°+a_{P/T}\\ a_T\left(\cos 20°+0.30\sin 20°\right)=0.30g\cos 20°-g\sin 20°+a_{P/T}\\ a_T=\frac{g\left(0.30\cos 20°-\sin 20°\right)+a_{P/T}}{\cos 20°+0.30\sin 20°}\end{array}$

Substitute $9.81{\text{m/s}}^2$ for g, $4.94{\text{m/s}}^2$ for $a_{P/T}$

$\begin{array}{c}{a}_T=\frac{9.81\left(0.30\cos 20°-\sin 20°\right)+4.94}{\cos 20°+0.30\sin 20°}\\ =\frac{4.35}{1.042}\\ =4.17{\text{m/s}}^2\end{array}$

Thus, the acceleration of the truck which causes corner A of the stack to reach the end of the bed in 0.9 s is $\underset{\_}{4.17{\text{m/s}}^2}$.