Chapter 8, Problem 38P

(a)

To determine

The average power of the elevator motor.

Answer to Problem 38P

The average power of the elevator motor is $\underset{\_}{5.91\times {10}^3\text{ W}}$.

Explanation of Solution

Write the expression for average power

  $P_{av}=\frac{W}{\Delta t}$                                                                                                        (I)

Here, $P_{av}$ is the average power, $W$ is the work done and $\Delta t$ is duration of time interval for which work $W$ is done.

Consider the elevator as an isolated system.

Write the expression for work done by the motor on the elevator

  $W=\overrightarrow{F}\cdot \overrightarrow{d}$

Simplify the above expression.

    $W=Fd\cos \theta$                                                                                                 (II)

Here, $F$ is the total force acting on the elevator by the motor, $d$ is the displacement due to force, $W$ is the work done and $\theta$ is the angle between direction of force and displacement.

Since the direction of force and displacement is same, hence the angle $\theta$ is zero.

Substitute $0°$ for $\theta$ in equation (II).

  $W=Fd\cos 0°$

Simplify the above equation

  $W=Fd$

Substitute $Fd$ for $W$ in equation (I).

  $P_{av}=\frac{Fd}{\Delta t}$                                                                                                     (III)

Write the expression for total force when the system is moving with an acceleration.

  $\sum F=ma$

Here, $F$ is the total force acting on the system, $m$ is mass of the system, $a$ is acceleration of the system.

Write the expression for total force acting on the system

  $\sum F=F_{motor}+F_g$

Here, $\sum F$ is the total force, $F_{motor}$ is the force acting by motor and $F_g$ is the force due to gravitational force.

Substitute $ma$ for $\sum F$ in above equation.

  $F_{motor}+F_g=ma$                                                                                         (IV)

Consider the elevator as an isolated system and the motion of elevator is by force acting by motor, the elevator starts from rest when a force is applied by motor against the gravitational force of the elevator.

The gravitaional force is negative because it is in opposite to the direction of acceleration.

Write the expression for force due to gravity.

  $F_g=-mg$

Here, $g$ the the accleration due to gravity.

Substitute $-mg$ for $F_g$ in equation (IV).

  $F_{motor}+\left(-mg\right)=ma$

Simplify the above equation.

  $F_{motor}=m(a+g)$                                                                                         (V)

Consider the motion of elevator from rest to the speed of its cruising speed.

Write the equation of motion for velocity in terms of acceleration and time

  $v=u+at$

Here, $v$ is the final speed in the motion, $u$ is the initial speed in the motion, $a$ is the acceleration, and $t$ is the time period.

Since the initial speed is zero.

Substitute $0\text{m/s}$ for $u$ in above equation.

    $\begin{array}{l}v=0+at\\ =at\end{array}$

Simplify the above equation to find the expression for $a$

  $a=\frac{v}{t}$                                                                                                          (VI)

Write the equation of motion for distance travelled in terms of acceleration, time and initial speed.

  $d=ut+\frac{1}{2}a{t}^2$

Here, $d$ is the distance travelled by the elevator.

Since the initial speed is zero.

Substitute $0\text{m/s}$ for $u$ in above equation.

    $d=\frac{1}{2}a{t}^2$

Substitute $\frac{v}{t}$ for $a$ in above equation.

  $d=\frac{1}{2}\left(\frac{v}{t}\right)t^2$

Simplify the above equation.

  $d=\frac{1}{2}vt$                                                                                                    (VII)

Conclusion:

Substitute $1.75\text{ m/s}$ for $v$ and $3.00\text{ s}$ for $t$ in equation (VI).

  $\begin{array}{c}a=\frac{1.75\text{m/s}}{3\text{s}}\\ =\frac{175}{300}\frac{\text{m}}{s^2}\\ =\frac{7}{12}{\text{m/s}}^{\text{2}}\end{array}$

Substitute $\frac{7}{12}{\text{m/s}}^{\text{2}}$ for $a$, $650\text{ kg}$ for $m$ and $9.8{\text{ m/s}}^{\text{2}}$ for $g$ in equation (V).

  $\begin{array}{c}{F}_{motor}=\left(650\text{Kg}\right)\left(\frac{7}{12}{\text{m/s}}^{\text{2}}+9.8{\text{m/s}}^{\text{2}}\right)\\ =6749.167\text{N}\end{array}$

Substitute $0$ for $u$, $1.75\text{ m/s}$ for $v$ and $3\text{ s}$ for $t$ in equation (VI).

  $\begin{array}{c}d=\frac{1}{2}\left(1.75\text{m/s}\right)\left(3\text{s}\right)\\ =2.625\text{m}\end{array}$

Substitute $6749.167\text{ N}$ for $F$, $2.625\text{ m}$ for $d$ and $3\text{ s}$ for $\Delta t$ in equation (I) to find $P_{av}$

  $\begin{array}{c}{P}_{av}=\frac{\left(6749.167\text{N}\right)\left(2.625\text{m}\right)}{3\text{s}}\\ =5905.52\text{W}\\ =5.91\times {10}^3\text{ W}\end{array}$

Thus, the average power of the elevator motor is $5.91\times {10}^3\text{ W}$.

(b)

To determine

The comparison between average power and the power when elevator moves with cruising speed.

Answer to Problem 38P

The instantaneous power required is $\underset{\_}{11.15\times {10}^3\text{W}}$ is $\underset{\_}{5.24\times {10}^3\text{ W}}$ more than average power calculated in part (a).

Explanation of Solution

Write the expression for instaneous power when the elevator is moving at a constant cruising speed

  $P_{inst}=\overrightarrow{F}\cdot \overrightarrow{v}$

Simplify the above expression

    $P_{inst}=Fv\cos \theta$

Here, $P_{inst}$ is the power at any instance, $F$ is the force acting, $v$ is the velocity in effect of acted force on system and $\theta$ is the angle between force and velocity vector.

Since the force and velocity are in same direction, hence the angle between $\overrightarrow{F}$ and $\overrightarrow{v}$ is zero.

Substitute $0°$ for $\theta$ in above equation.

    $P_{inst}=Fv\cos 0°$

Simplify the above equation.

  $P_{inst}=Fv$                                                                                                  (VIII)

Here, $F$ is the total force acting, and $v$ is the speed of system.

Since the elevator moves at constant speed when it reaches cruising speed, hence the acceleration will be zero.

Write the expression for comparison in both powers

  $\Delta P=P_{inst}-P_{av}$                                                                                         (VIII)

Here, $\Delta P$ is the difference in powers in two situations, $P_{isnt}$ is the power when the elevator is moving with constant speed, and $P_{av}$ is the power required by elevator to raise its speed from rest to cruising speed.

Since the elevator is moving at constant speed hence the acceleration is zero.

Substitute $0$ for $a$ in equation (V).

    $F_{motor}=m(0+g)$

Simplify the above equation

    $F_{motor}=mg$                                                                                                 (IX)

Conclusion:

Substitute $650\text{ kg}$ for $m$, and $9.8{\text{ m/s}}^{\text{2}}$ for $g$ in equation (IX).

  $\begin{array}{c}{F}_{motor}=\left(650\text{kg}\right)(9.8{\text{m/s}}^{\text{2}})\\ =6370\text{N}\end{array}$

Substitiute $6370\text{ N}$ for $F$, and $1.75\text{ m/s}$ for $v$ in equation (VIII).

  $\begin{array}{c}{P}_{inst}=(6370\text{N})(1.75\text{m/s})\\ =11147.50\text{W}\\ \text{=11}\text{.15}\times {\text{10}}^3\text{ W}\end{array}$

Substitute $\text{11}\text{.15}\times {\text{10}}^3\text{ W}$ for $P_{inst}$ and $\text{5}\text{.91}\times {\text{10}}^3\text{ W}$ for $P_{av}$ in equation (VIII).

  $\begin{array}{l}\Delta P=\text{11}\text{.15}\times {\text{10}}^3\text{ W}-5.\text{91}\times {\text{10}}^3\text{ W}\\ =5.\text{24}\times {\text{10}}^3\text{ W}\end{array}$

Thus, the instantaneous power required is $\underset{\_}{11.15\times {10}^3\text{W}}$ is $\underset{\_}{5.24\times {10}^3\text{ W}}$ more than average power calculated in part (a).