Chapter 4, Problem 4.50P

To determine

(a)

The equation of motion for the system.

Answer to Problem 4.50P

The equation of motion for the system is $8.33\dot{v}+v=34.51$.

Explanation of Solution

Given information:

The mass of the block on an inclined plane is $50\text{kg}$, the angle of inclination is $25°$, the friction coefficient between the block and the plane is $6\text{N}\cdot \text{s}/\text{m}$.

Figure-(1) shows the mechanical system diagram of the block on the inclined plane condition.

Figure-(1)

Here, the component of force acting in the horizontal direction along the inclined plane is $mg\sin \theta$, the component of force acting in vertical direction and opposite to reaction force is $mg\cos \theta$.

Write the expression for equation of motion for the system along the inclined plane.

$m\dot{v}=mg\sin \theta -cv$ ..... (I)

Here, the force acting on the block is $m\dot{v}$, the velocity of the block is $v$ ,the acceleration of the block is $\dot{v}$, the mass of the block is $m$, the angle of inclination is $\theta$, the coefficient of friction is $c$, acceleration due to gravity is $g$.

Calculation:

Substitute $50\text{}\text{Kg}$ for $m$, $25°$ for $\theta$, $9.81\text{m}/{\text{sec}}^{\text{2}}$ for $g$, and $6\text{N}\cdot \text{s}/\text{m}$ for $c$ in Equation (I)

$\begin{array}{l}\left(50\text{kg}\right)\dot{v}=\left(50\text{kg}\right)\left(9.81\text{m}/{\text{sec}}^{\text{2}}\right)\sin \left(25°\right)-\left(6\text{N}\cdot \text{s}/\text{m}\right)v\\ \left(50\text{kg}\right)\dot{v}+\left(6\text{N}\cdot \text{s}/\text{m}\right)v=\left(490\text{kg}\cdot \text{m}/{\text{sec}}^{\text{2}}\right)\sin \left(25°\right)\\ \left(50\text{kg}\right)\dot{v}+\left(6\text{N}\cdot \text{s}/\text{m}\right)v=207.08\text{kg}\cdot \text{m}/{\text{sec}}^{\text{2}}\\ 8.33\dot{v}+v=34.51\end{array}$.

Conclusion:

The equation of motion for the system is $8.33\dot{v}+v=34.51$.

To determine

(b)

The equation of motion for speed $V\left(t\right)$ if the initial speed of the block is $5\text{m}/\text{s}$.

Answer to Problem 4.50P

The equation of motion for speed $V\left(t\right)$ is $V\left(t\right)=34.548-34.548e^{-\frac{t}{8.33}}$.

Explanation of Solution

Given information:

The initial speed of the block is $5\text{m}/\text{s}$.

Write the expression for equation of motion for the system.

$8.33\dot{v}+v=34.548$ ..... (IV)

Apply Laplace transform for Equation (IV)

$\begin{array}{l}8.33sV\left(s\right)+V\left(s\right)=\frac{34.548}{s}\\ \left(8.33s+1\right)V\left(s\right)=\frac{34.548}{s}\\ s\left(8.33s+1\right)V\left(s\right)=34.548\\ V\left(s\right)=\frac{34.548}{s\left(8.33s+1\right)}\end{array}$ ...... (V)

Apply partial fraction technique for Equation (V).

$\begin{array}{l}\frac{A}{s}+\frac{B}{8.33s+1}=\frac{34.548}{s\left(8.33s+1\right)}\\ \frac{A\left(8.33s+1\right)+Bs}{s\left(8.33s+1\right)}=\frac{34.548}{s\left(8.33s+1\right)}\\ A\left(8.33s+1\right)+Bs=34.548\end{array}$ ..... (VI)

Substitute $0$ for $s$ in equation (VI)

$\begin{array}{l}A\left(8.33\left(0\right)+1\right)+B\left(0\right)=34.548\\ A+B\left(0\right)=34.548\\ A=34.548\end{array}$

Equate coefficient on both sides of Equation

$0=8.33A+B$ ...... (VII)

Substitute $34.548$ for $A$ in Equation (VII)

$\begin{array}{l}0=8.33\left(34.548\right)+B\\ B=-287.784\end{array}$.

Calculation:

Substitute $34.548$ for $A$ and $-287.784$ for $B$ in Equation (VI)

$\begin{array}{l}\frac{34.548\left(8.33s+1\right)+\left(-287.784\right)s}{s\left(8.33s+1\right)}=\frac{34.548}{s\left(8.33s+1\right)}\\ \frac{34.548\left(8.33s+1\right)}{s\left(8.33s+1\right)}-\frac{\left(287.784\right)s}{s\left(8.33s+1\right)}=\frac{34.548}{s\left(8.33s+1\right)}\end{array}$

Substitute $\frac{34.548\left(8.33s+1\right)}{s\left(8.33s+1\right)}-\frac{\left(287.784\right)s}{s\left(8.33s+1\right)}$ for $\frac{34.548}{s\left(8.33s+1\right)}$ in Equation (V)

$\begin{array}{l}V\left(s\right)=\frac{34.548\left(8.33s+1\right)}{s\left(8.33s+1\right)}-\frac{\left(287.784\right)s}{s\left(8.33s+1\right)}\\ V\left(s\right)=\frac{34.548}{s}-\frac{34.548}{s+\frac{1}{8.33}}\end{array}$

Take inverse Laplace transform.

$V\left(t\right)=34.548-34.548e^{-\frac{t}{8.33}}$.

Conclusion:

The equation of motion for speed $V\left(t\right)$ is $V\left(t\right)=34.548-34.548e^{-\frac{t}{8.33}}$.

To determine

(c)

The steady-state speed of the block

The time required to reach the speed.

Answer to Problem 4.50P

The steady-state speed of the block is $34.548\text{m}/\text{s}$

The time required to reach the speed $33.32\text{s}$.

Explanation of Solution

Write the expression for first order differential equation along with dynamic value.

$\tau \dot{v}+v=Ku$ ...... (VIII)

Here, the time constant is $\tau$, the steady-state speed is $K$, the velocity of the block is $v$.

Write the expression for time required to reach the steady-state speed

$t=4\tau$ ...... (IX)

Here, the time required to reach the steady-state is $t$.

Calculation:

Compare Equation (IV) and Equation (VIII).

$\tau =8.33$

Substitute $0$ for $\dot{v}$ in Equation (IV)

$\begin{array}{c}8.33\left(0\right)+v=34.548\\ v=34.548\end{array}$

Substitute $8.33$ for $\tau$ in Equation (IX)

$\begin{array}{c}t=4\left(8.33\right)\\ =33.32\text{s}\end{array}$.

Conclusion:

The steady-state speed of the block is $34.548\text{m}/\text{s}$

The time required to reach the speed $33.32\text{s}$.

To determine

(d)

The effect of initial velocity on the steady-state speed.

Explanation of Solution

Write the expression for steady-state speed.

$\dot{v}=0$

Since the steady-state speed is zero therefore, the initial velocity is independent of the steady-state speed and hence, the initial velocity does not affect the steady-state speed value.

Conclusion:

The initial velocity does not affect the steady-state speed value.