Chapter 4, Problem 174P

You are designing a high-speed elevator for a new skyscraper. The elevator will have a mass limit of 2400 kg (including passengers). For passenger comfort, you choose the maximum ascent speed to be 18 m/s, the maximum descent speed to be 10 m/s, and the maximum acceleration magnitude to be 1.2 m/s². Ignore friction. (a) What are the maximum and minimum upward forces that the supporting cables exert on the elevator car? (b) What is the minimum time it will take the elevator to ascend from the lobby to the observation deck, a vertical displacement of 640 m? (c) What are the maximum and minimum values of a 60 kg passenger’s apparent weight during the ascent? (d) What is the minimum time it will take the elevator to descend to the lobby from the observation deck?

To determine

Maximum and minimum upward force that the supporting cables exert on the elevator car.

Answer to Problem 174P

Maximum force is $\underset{\_}{26400\text{N}}$

Minimum force is $\underset{\_}{20640\text{N}}$

Explanation of Solution

Write the equation of force exerted by the supporting cables on the elevator car.

    $F=ma$        (I)

Here $m$ is the mass of the elevator and $a$ is the acceleration.

Write the equation of force due to gravity

    $F=mg$        (II)

Here $g$ is the acceleration due to gravity.

Then the maximum force would be

    $\begin{array}{c}{F}_{\max }=mg+ma\\ =m(g+a)\end{array}$        (III)

The minimum force would be

    $F_{\min }=m\left(g-a\right)$        (IV)

Conclusion

Substitute $2400\text{kg}$ for $m$, $1.2\text{m}/{\text{s}}^{\text{2}}$ for $a$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in (III)

    $\begin{array}{c}{F}_{\max }=2400\text{kg}\left(9.8\text{m}/{\text{s}}^{\text{2}}+1.2\text{m}/{\text{s}}^{\text{2}}\right)\\ =26400\text{N}\end{array}$

Substitute $2400\text{kg}$ for $m$, $1.2\text{m}/{\text{s}}^{\text{2}}$ for $a$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in (IV)

  $\begin{array}{c}{F}_{\min }=2400\text{kg}\left(9.8\text{m}/{\text{s}}^{\text{2}}-1.2\text{m}/{\text{s}}^{\text{2}}\right)\\ =20640\text{N}\end{array}$

Maximum force is $\underset{\_}{26400\text{N}}$.

Minimum force is $\underset{\_}{20640\text{N}}$.

To determine

Minimum time to ascent from the lobby to the observation deck.

Answer to Problem 174P

The time is $\underset{\_}{43\text{s}}$.

Explanation of Solution

Write the expression to find the time taken to reach the maximum ascent speed.

    $v=u+a{t}_1$        (I)

Here, $v$ is the maximum ascent speed, $u$ is the initial speed, $a$ is the acceleration, and $t_1$ is the time taken to reach the maximum ascent speed.

Write the expression to find the distance covered during acceleration.

    $s_1=ut+\frac{1}{2}a{t}_1^2$

Here, $s_1$ is the vertical displacement

Write the expression to find the distance covered.

    $s_2=v{t}_2$

Here, $s_2$ is the vertical displacement and $t_2$ is the time of displacement with uniform speed.

Conclusion:

Substitute $1.2\text{m}/{\text{s}}^{\text{2}}$ for $a$, $18\text{m/s}$ for $v$ and $0\text{m}/\text{s}$ for $u$ in (I)

    $\begin{array}{c}18\text{m/s}=0+\left(1.2\text{m}/{\text{s}}^{\text{2}}\right)t_1\\ t_1=\frac{18\text{m/s}}{1.2\text{m}/{\text{s}}^{\text{2}}}\\ =15\text{s}\end{array}$

Vertical distance covered during this time is,

Substitute $1.2\text{m}/{\text{s}}^{\text{2}}$ for $a$, $15\text{s}$ for $t_1$ and $0\text{m}/\text{s}$ for $u$ in (I)

    $\begin{array}{c}{s}_1=0\left(15\text{s}\right)+\frac{1}{2}\left(1.2\text{m}/{\text{s}}^{\text{2}}\right){\left(15\text{s}\right)}^2\\ =135\text{m}\end{array}$

Remaining vertical distance is,

    $\begin{array}{c}{s}_2=640\text{m}-135\text{m}\\ =505\text{m}\end{array}$

Elevator moves with uniform speed to cover this distance.

Expression to calculate the time taken to cover this distance,

    $t_2=\frac{s_2}{v}$

Substitute $505\text{m}$ and $18\text{m}/\text{s}$ for $v$.

    $\begin{array}{c}{t}_2=\frac{505\text{m}}{18\text{m}/\text{s}}\\ =28\text{s}\end{array}$

The minimum time taken by the elevator to ascend from the lobby to the observation desk is,

    $t=t_1+t_2$

Substitute $15\text{s}$ for $t_1$ and $28\text{s}$ for $t_2$ to find $t$.

    $\begin{array}{c}t=15\text{s}+\text{28}\text{s}\\ =\text{43}\text{s}\end{array}$

Therefore, the minimum time taken by the elevator to ascend from the lobby to the observation desk is $\underset{\_}{43\text{s}}$.

To determine

Maximum and minimum values of a passenger’s apparent weight during the ascend.

Answer to Problem 174P

The maximum weight during the ascent is $\underset{\_}{660\text{N}}$.

The minimum weight during the ascend is $\underset{\_}{516\text{N}}$

Explanation of Solution

The maximum weight would be

    $W=M\left(g+a\right)$        (IX)

Here $W$ is the weight and $M$ is the mass of the passenger

The minimum weight would be

    $W=M(g-a)$        (X)

Conclusion

Substitute $60\text{kg}$ for $M$, $1.2\text{m}/{\text{s}}^{\text{2}}$ for $a$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in (IX)

    $\begin{array}{c}{F}_{\max }=60\text{kg}\left(9.8\text{m}/{\text{s}}^{\text{2}}+1.2\text{m}/{\text{s}}^{\text{2}}\right)\\ =660\text{N}\end{array}$

Substitute $60\text{kg}$ for $M$, $1.2\text{m}/{\text{s}}^{\text{2}}$ for $a$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in (X)

    $\begin{array}{c}W=60\text{kg}\left(9.8\text{m}/{\text{s}}^{\text{2}}-1.2\text{m}/{\text{s}}^{\text{2}}\right)\\ =516\text{N}\end{array}$

The maximum weight during the ascent is $\underset{\_}{660\text{N}}$.

The minimum weight during the ascend is $\underset{\_}{516\text{N}}$

To determine

Minimum time for the elevator to descend to the lobby.

Answer to Problem 174P

The time is $\underset{\_}{68\text{s}}$.

Explanation of Solution

Write the expression to find the time taken to reach the maximum decent speed.

    $v=u+a{t}_1$        (II)

Here, $v$ is the maximum ascent speed, $u$ is the initial speed, $a$ is the acceleration, and $t_1$ is the time taken to reach the maximum descent speed.

Write the expression to find the distance covered during acceleration.

    $s_1=ut+\frac{1}{2}a{t}_1^2$

Here, $s_1$ is the vertical downward displacement

Write the expression to find the distance covered.

    $s_2=v{t}_2$

Here, $s_2$ is the vertical downward displacement and $t_2$ is the time of displacement with uniform speed.

Conclusion

Substitute $1.2\text{m}/{\text{s}}^{\text{2}}$ for $a$, $10\text{m/s}$ for $v$ and $0\text{m}/\text{s}$ for $u$ in (I)

    $\begin{array}{c}10\text{m/s}=0+\left(1.2\text{m}/{\text{s}}^{\text{2}}\right)t_1\\ t_1=\frac{10\text{m/s}}{1.2\text{m}/{\text{s}}^{\text{2}}}\\ =8.33\text{s}\end{array}$

Vertical distance covered during this time is,

Substitute $1.2\text{m}/{\text{s}}^{\text{2}}$ for $a$, $8.33\text{s}$ for $t_1$ and $0\text{m}/\text{s}$ for $u$ in (I)

    $\begin{array}{c}{s}_1=0\left(8.33\text{s}\right)+\frac{1}{2}\left(1.2\text{m}/{\text{s}}^{\text{2}}\right){\left(8.33\text{s}\right)}^2\\ =41.6\text{m}\end{array}$

Remaining vertical distance is,

    $\begin{array}{c}{s}_2=640\text{m}-41.6\text{m}\\ =598.4\text{m}\end{array}$

Elevator moves with uniform speed to cover this distance.

Expression to calculate the time taken to cover this distance,

    $t_2=\frac{s_2}{v}$

Substitute $598.4\text{m}$ and $10\text{m}/\text{s}$ for $v$.

    $\begin{array}{c}{t}_2=\frac{598.4\text{m}}{10\text{m}/\text{s}}\\ =59.84\text{s}\end{array}$

The minimum time taken by the elevator to ascend from the lobby to the observation desk is,

    $t=t_1+t_2$

Substitute $8.33\text{s}$ for $t_1$ and $59.84\text{s}$ for $t_2$ to find $t$.

    $\begin{array}{c}t=8.33\text{s}+59.84\text{s}\\ =68.2\text{s}\\ \approx 68\text{s}\end{array}$

Therefore, the minimum time taken by the elevator to ascend from the lobby to the observation desk is $\underset{\_}{68\text{s}}$.