Chapter 4, Problem 72P

(a)

To determine

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

Answer to Problem 72P

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 $\underset{\_}{20640\text{N}}$.

(b)

To determine

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

Answer to Problem 72P

The time is $\underset{\_}{102\text{s}}$.

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 $\underset{\_}{43\text{s}}$.

(c)

To determine

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

Answer to Problem 72P

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 $\underset{\_}{516\text{N}}$

(d)

To determine

Minimum time for the elevator to descend to the lobby.

Answer to Problem 72P

The time is $\underset{\_}{12.49\text{s}}$

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 $68\text{s}$.