An Elevator's Motor

Q. An elevator can carry a maximum load of 1800 kg (elevator + passengers) and is moving up with a constant speed of 2 m/s. The frictional force opposing the motion is 4000 N. Taking g as 10 m/s², what is the minimum power delivered by the motor to the elevator, in watts? expects 4.4e+04 W

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Q. An elevator can carry a maximum load of 1800 kg $m = 1800$ kg (elevator + passengers) and is moving up $F_{m} = | F_{m} |$ with a constant speed $a = 0$ m/s² of 2 m/s $v_{0} = 2$ m/s. The frictional force opposing the motion $F_{s} = - | F_{s} |$ is 4000 N $| F_{s} | = 4000$ N. Taking g as $W = - | W |$ 10 m/s² $g = 10$ m/s², what is the minimum power delivered by the motor to the elevator, in watts?

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Inputs

mass

m

1800 kg

speed

v_{0}

2.0 m/s

friction

| F_{s} |

4000 N

gravity

g

10 m/s²

Outputs

power

P_{0,m}

The solution

Solved with Power

Given

$m = 1800$ kg
$v_{0} = 2$ m/s
$| F_{s} | = 4000$ N
$g = 10$ m/s²
$a = 0$ m/s² from “constant speed”
$W = - | W |$ from “Taking g as”
$F_{s} = - | F_{s} |$ from “opposing the motion”
$F_{m} = | F_{m} |$ from “moving up”

Find $P_{0,m}$ .

Working

second law(rising, y) $F_{net} = m \cdot a$
$F_{net} = 1800 \cdot 0$
$F_{net} = 0$ N
opposing the motion $F_{s} = - | F_{s} |$
$F_{s} = - 4000$
$F_{s} = −4000$ N
weight of a body(elevator) $| W | = m \cdot g$
$| W | = 1800 \cdot 10$
$| W | = 18000$ N
Taking g as $W = - | W |$
$W = - 18000$
$W = −18000$ N
net force is the resultant(elevator, rising, y) $F_{net} = F_{s} + F_{m} + W$
$0 = (−4000) + F_{m} + (−18000)$
$F_{m} = 22000$ N
power of a force(elevator, rising, 3) $P_{0,m} = F_{m} \cdot v_{0}$
$P_{0,m} = 22000 \cdot 2$
$P_{0,m} = 44000$ W

Conclusion

At a steady speed the motor only has to balance the weight and the friction; its power is that force times the speed it lifts at.

The Ruby

1require_relative "../physics"23question :an_elevator_motor do4  kind :numerical5  describes "An elevator can carry a maximum load of %{mass} (elevator + passengers) and is moving up with a " \6            "constant speed of %{speed}. The frictional force opposing the motion is %{friction}. Taking g as " \7            "%{gravity}, %{power}, in watts?"8  given mass: "1800 kg"9  given speed: "2 m/s"10  given friction: "4000 N"11  given gravity: "10 m/s2"12  asks power: "what is the minimum power delivered by the motor to the elevator"13end

1require_relative "../questions/an_elevator_motor"2require_relative "../science/work_energy_and_power/power"34solve :an_elevator_motor, using: Power do5  objects :elevator, :motor, :shaft6  axes :y7  elevator.goes(:rising)8  deduce_from "Taking g as", elevator.weight.direction => -y910  take mass: elevator.mass11  take speed: elevator.initial_velocity12  take friction: elevator.force(by: shaft).magnitude13  take gravity: g1415  deduce_from "constant speed", elevator.acceleration => "0 m/s2"16  deduce_from "opposing the motion", elevator.force(by: shaft).direction => -y17  deduce_from "moving up", elevator.force(by: motor).direction => y1819  power = elevator.power(by: motor, at: rising.start)20  answer power:, in: "W"2122  conclusion "At a steady speed the motor only has to balance the weight and the friction; its power " \23             "is that force times the speed it lifts at."24end

1require_relative "../solutions/an_elevator_motor"23play_with :an_elevator_motor do4  title "An Elevator's Motor"5end

The laws

quantity	written	units	within	from
volume	$volume$	—	—	Vectors
mass	$m$	kg	0 … ∞	Vectors
speed	$speed$	—	—	Vectors
density	$density$	—	—	Vectors
number of moles	$n$	—	—	Vectors
angular frequency	${angular}_{frequency}$	—	—	Vectors
work	$W_{F}$	J	—	Vectors
current	$current$	—	—	Vectors
temperature	$temperature$	—	—	Vectors
pressure	$pressure$	—	—	Vectors
time	$t$	s	0 … 1e+05	Vectors
power	$P$	W	—	Vectors
total path length	$s$	—	—	Vectors
energy	$energy$	—	—	Vectors
gravitational potential	${gravitational}_{potential}$	—	—	Vectors
coefficient of friction	$μ$	—	—	Vectors
charge	$charge$	—	—	Vectors
acceleration	$a$	m/s²	—	Vectors
velocity	$v$	m/s	—	Vectors
displacement	$Δ x$	m	—	Vectors
angular velocity	${angular}_{velocity}$	—	—	Vectors
force	$F$	N	—	Vectors
angular momentum	${angular}_{momentum}$	—	—	Vectors
linear momentum	$p_{lin}$	—	—	Vectors
electric field	${electric}_{field}$	—	—	Vectors
average velocity	$v_{av}$	m/s	—	Vectors
magnetic moment	${magnetic}_{moment}$	—	—	Vectors
relative velocity	$v_{AB}$	—	—	Vectors
impulse	$impulse$	—	—	Vectors
first vector	$A$	—	0 … ∞	Vectors
second vector	$B$	—	0 … ∞	Vectors
angle between	$φ$	—	0 … 3.142	Vectors
resultant	$R$	—	0 … ∞	Vectors
resultant direction	$β$	—	0 … 3.142	Vectors
direction	$α$	—	-3.142 … 3.142	Vectors
x component	$A_{x}$	—	—	Vectors
y component	$A_{y}$	—	—	Vectors
position	$x$	m	—	Velocity
path length	$d$	m	—	Velocity
time taken	$Δ t$	s	0 … 1e+05	Velocity
average speed	$c_{av}$	m/s	—	Velocity
starting position	$x_{0}$	m	—	Velocity
final position	$x$	m	—	Velocity
change of velocity	$Δ v$	m/s	—	Acceleration
average acceleration	$a_{av}$	m/s²	—	Acceleration
starting velocity	$v_{0}$	m/s	—	Acceleration
final velocity	$v$	m/s	—	Acceleration
gravity	$g$	m/s²	—	UniformAcceleration
net force	$F_{net}$	N	—	NewtonsSecondLaw
weight	$W$	N	—	NewtonsSecondLaw
work by gravity	$W_{g}$	J	—	Work
net work	$W_{net}$	J	—	Work
displacement of elevator during rising along y	$Δ x$	m	—
position of elevator at end of rising along y	$x_{1}$	m	—
position of elevator at start of rising along y	$x_{0}$	m	—
average velocity of elevator during rising along y	$v_{av}$	m/s	—
time taken of rising	$Δ t$	s	0 … 1e+05
average speed of elevator during rising	$c_{av}$	m/s	—
path length of elevator during rising	$d$	m	—
change of velocity of elevator during rising along y	$Δ v$	m/s	—
velocity of elevator at end of rising from ground along y	$v_{1}$	m/s	—
velocity of elevator at start of rising from ground along y	$v_{0}$	m/s	—
average acceleration of elevator during rising along y	$a_{av}$	m/s²	—
acceleration of elevator during rising along y	$a$	m/s²	—
net force on elevator during rising along y	$F_{net}$	N	—
mass of elevator	$m$	kg	0 … ∞
mass of motor	$m_{m}$	kg	0 … ∞
mass of shaft	$m_{s}$	kg	0 … ∞
weight on ground	$\| W_{g} \|$	N	—
weight on elevator	$\| W \|$	N	—
weight on motor	$\| W_{m} \|$	N	—
weight on shaft	$\| W_{s} \|$	N	—
force on elevator by shaft during rising	$\| F_{s} \|$	N	—
force on elevator by motor during rising	$\| F_{m} \|$	N	—
weight on elevator along y	$W$	N	—
force on elevator by shaft during rising along y	$F_{s}$	N	—
force on elevator by motor during rising along y	$F_{m}$	N	—
work on elevator by shaft during rising	$W_{F,s}$	J	—
work on elevator by motor during rising	$W_{F,m}$	J	—
net work on elevator during rising	$W_{net}$	J	—
work by gravity on elevator during rising	$W_{g}$	J	—
power on elevator by shaft at start of rising	$P_{0,s}$	W	—
power on elevator by shaft at end of rising	$P_{1,s}$	W	—
power on elevator by motor at start of rising	$P_{0,m}$	W	—
power on elevator by motor at end of rising	$P_{1,m}$	W	—
displacement of elevator during rising	$\| Δ x \|$	m	—
position of elevator at end of rising	$\| x_{1} \|$	m	—
position of elevator at start of rising	$\| x_{0} \|$	m	—
average velocity of elevator during rising	$\| v_{av} \|$	m/s	—
change of velocity of elevator during rising	$\| Δ v \|$	m/s	—
velocity of elevator at end of rising from ground	$\| v_{1} \|$	m/s	—
velocity of elevator at start of rising from ground	$\| v_{0} \|$	m/s	—
average acceleration of elevator during rising	$\| a_{av} \|$	m/s²	—
acceleration of elevator during rising	$\| a \|$	m/s²	—
net force on elevator during rising	$\| F_{net} \|$	N	—

x_{0} \equiv x_{i} (start of motion)

starting position

x \equiv x_{i} (end of motion)

final position

v_{0} \equiv v_{i} (start of motion)

starting velocity

v \equiv v_{i} (end of motion)

final velocity

t \equiv Δ t_{m}

time

A = \sqrt{{A_{x}}^{2} + {A_{y}}^{2}}

magnitude from components

d_{i} (whole) = d_{i} (part 1) + d_{i} (part 2)

path in parts

Δ t_{w} = Δ t_{p} + Δ t_{p}

time in parts

a_{i} (part 1) = a_{i} (whole)

steady through its parts

m_{b} \neq 0

has mass?

R^{2} = A^{2} + B^{2} + 2 \cdot A \cdot B \cdot \cos φ

law of cosines

R \cdot \sin β = B \cdot \sin φ

law of sines

R \cdot \cos β = A + B \cdot \cos φ

along the first

A_{x} = A \cdot \cos α

resolved along x

A_{y} = A \cdot \sin α

resolved along y

Δ x = x_{1} - x_{0}

change of position(rising, y)

v_{av} = \frac{Δ x}{Δ t}

displacement a time(rising, y)

c_{av} = \frac{d}{Δ t}

path a time(rising, y)

Δ v = v_{1} - v_{0}

velocity gained(rising, y)

a_{av} = \frac{Δ v}{Δ t}

velocity gained a time(rising, y)

v_{1} = v_{0} + a \cdot Δ t

velocity after a time(rising, y)

x_{1} = x_{0} + v_{0} \cdot Δ t + \frac{a \cdot {Δ t}^{2}}{2}

position after a time(rising, y)

{v_{1}}^{2} = {v_{0}}^{2} + 2 \cdot a \cdot (x_{1} - x_{0})

velocity after a distance(rising, y)

x_{1} - x_{0} = \frac{(v_{1} + v_{0}) \cdot Δ t}{2}

mean of the velocities(rising, y)

F_{net} = m \cdot a

second law(rising, y)

| W | = m \cdot g

weight of a body(elevator)

F_{net} = F_{s} + F_{m} + W

net force is the resultant(elevator, rising, y)

W_{F,s} = F_{s} \cdot Δ x

work of a steady force(elevator, rising, 1)

W_{F,m} = F_{m} \cdot Δ x

work of a steady force(elevator, rising, 2)

W_{net} = F_{net} \cdot Δ x

work of the net force(elevator, rising)

W_{g} = W \cdot Δ x

work of the weight(elevator, rising)

P_{0,s} = F_{s} \cdot v_{0}

power of a force(elevator, rising, 1)

P_{1,s} = F_{s} \cdot v_{1}

power of a force(elevator, rising, 2)

P_{0,m} = F_{m} \cdot v_{0}

power of a force(elevator, rising, 3)

P_{1,m} = F_{m} \cdot v_{1}

power of a force(elevator, rising, 4)

| Δ x | = \sqrt{{Δ x}^{2}}

magnitude from components(displacement of elevator during rising)

| x_{1} | = \sqrt{{x_{1}}^{2}}

magnitude from components(position of elevator at end of rising)

| x_{0} | = \sqrt{{x_{0}}^{2}}

magnitude from components(position of elevator at start of rising)

| v_{av} | = \sqrt{{v_{av}}^{2}}

magnitude from components(average velocity of elevator during rising)

| Δ v | = \sqrt{{Δ v}^{2}}

magnitude from components(change of velocity of elevator during rising)

| v_{1} | = \sqrt{{v_{1}}^{2}}

magnitude from components(velocity of elevator at end of rising from ground)

| v_{0} | = \sqrt{{v_{0}}^{2}}

magnitude from components(velocity of elevator at start of rising from ground)

| a_{av} | = \sqrt{{a_{av}}^{2}}

magnitude from components(average acceleration of elevator during rising)

| a | = \sqrt{a^{2}}

magnitude from components(acceleration of elevator during rising)

| F_{net} | = \sqrt{{F_{net}}^{2}}

magnitude from components(net force on elevator during rising)

| W | = \sqrt{W^{2}}

magnitude from components(weight on elevator)

| F_{s} | = \sqrt{{F_{s}}^{2}}

magnitude from components(force on elevator by shaft during rising)

| F_{m} | = \sqrt{{F_{m}}^{2}}

magnitude from components(force on elevator by motor during rising)

W = - | W |

Taking g as

F_{s} = - | F_{s} |

opposing the motion

F_{m} = | F_{m} |

moving up

m \neq 0

has mass?(elevator)

m_{m} \neq 0

has mass?(motor)

m_{s} \neq 0

has mass?(shaft)

F_{net} = 0

free of forces?(elevator, rising)