Jean Blumer

2021-12-20

A550-N physics student stands on a bathroom scale in an elevator that is supported by a cable. The combined mass of student plus elevator is 850 kg. As the elevator starts moving, the scale reads 450 N.
(a) Find the acceleration of the elevator (magnitude and direction).
(b) What is the acceleration if the scale reads 670 N?
(c) If the scale reads zero, should the student worry? Explain.
(d) What is the tension in the cable in parts (a) and (с)?

### Answer & Explanation

stomachdm

Let the +y-axis be upward. The free-body diagrams for the elevator plus student and the student alone are shown in the figure below. The total mass of student plus elevator is $M=850kg$. The tension in the cable is denoted by T.
The normal force exerted by the scale on the student is denoted by n. By Newton's third law, this force is equal in magnitude to the downward pressure exerted by the student's feet on the scale (the scale reading), that is n represents the scale’s reading.
The mass of the student alone, whose actual weight is ${W}_{s}=550N,is{m}_{s}=\frac{{W}_{s}}{g}=550\frac{N}{9.80}\frac{m}{{s}^{2}}=56.1kg$
(a)Application of the y-equation of Newton's second law, $E{F}_{y}=m{a}_{y}$ gives
$n-{m}_{s}g={m}_{s}{a}_{y}.$
The scale’s reading corresponds to $n=450N$. Solving for dy, we find
${a}_{y}=\frac{n-{m}_{s}g}{{m}_{s}}=\frac{450N-550N}{56.1kg}=-1,78\frac{m}{{s}^{2}}$.
So the student, and thus the elevator, has an acceleration of magnitude 1.78 m/ s, which is directed downward.
(b) Repeating part (a) using the new scale reading, $n=670N$, we obtain
${a}_{y}=\frac{670N-550N}{56.1kg}=-2.14\frac{m}{{s}^{2}}$,
which represents an acceleration of magnitude 2.14 m/s” and upward direction
(c) In case the scale's reading was zero, that isn = 0, we have
${a}_{y}=\frac{0-{m}_{s}g}{{m}_{s}}=-g$.
The elevator is in free fall...Trouble!
(d) To find the tension T in the cable, we must apply Newton's second law to the student plus elevator this time.
Application of $EF,=M{a}_{y}$ gives
$T—Mg=M{a}_{y}$, NSH which leads to $T=M\left({a}_{y}+9\right)$.
In part (a), ${a}_{y}=—1.78\frac{m}{{s}^{2}}$. Hence,
$T=\left(850kg\right)\left(—1.78m/{s}^{2}+9.80m/{s}^{2}\right)=6817N.$
In part (c), ${a}_{y}=—g$. Hence,
$T=M\left(-g+9\right)=0$.
Yes, exactly! That is the trouble!

stomachdm

(a) $-1.78{m}^{2}$ (downward)
The equation of the forces acting on the student is:
$N—mg=ma\left(1\right)$
where
Nis the normal reaction of the scale on the student mg is the weight of the student
a is the acceleration of the student
The scale reads 450 N, so this is the normal reaction:
$N=450N$
Also, we know that the weight of the student is:
$mg=550N$
So we can find its mass:
$m=\frac{mg}{g}=\frac{550}{9.8}=56.1kg$
So now we can solve eq.(1) to find the acceleration:
$a=\frac{N-mg}{m}=\frac{450-550}{56.1}=1.78\frac{m}{{s}^{2}}$
where the negative sign means the acceleration is downward.
(b),upward
Again, the equation of the forces is
$N—mg=ma$
where this time, the reading of the scale (and so, the normal reaction) is
$N=670N$
Solving for the acceleration, we find
$a=\frac{N-mg}{m}=\frac{670-550}{56.1}=2.14\frac{m}{{s}^{2}}$
and the positive sign means the acceleration here is upward.
(c) Yes
Let's assume the scale is reading zero. In terms of forces, this means that the normal reaction on the student is zero:
N=0
So the equation of the forces simply becomes
mg =ma
Therefore the acceleration is
$a=g=-9.8\frac{m}{{s}^{2}}$
which means that the elevator is accelerating downward at 9.8m/s^{2} this means that the elevator is in free fall, so yes, the student should worry.
Let's now consider the equation of the forces on the elevator:
T-mg =ma(2)
where this time:
T is the tension in the cable
$mg=\left(850kg\right)\left(9.8\frac{m}{{s}^{2}}\right)=8330N$ is the weight of the elevator +student system
$4=-1.78\frac{m}{{s}^{2}}$ is the acceleration
Solving for T,
$T=mg+ma=8330+\left(850\right)\left(—1.78\right)=6817N$
(d) (b) 10149 N
Here we can use the same equation
$T—mg=ma$
where the only difference is that the acceleration is
$a=2.14\frac{m}{{s}^{2}}$
Solving the equation for T, we find
$T=mg+ma=8330+\left(850\right)\left(2.14\right)=10149$.
(d) (c) 0
Again, same equation
$T—mg=ma$
But this time, the acceleration is
$a=-9.8\frac{m}{{s}^{2}}$
So, we find:
$T=mg+ma=8330+\left(850\right)\left(-9.8\right)=0$
So, the tension in the cable is zero, since the elevator is in free fall.

2021-12-27

Step1
Given,
Weight of physics student = 550N
mass of elevator and student = 850kg
Step2
Step3
(a)
$N=W+ma$
$450=550+ma$
$ma=-100$
$a=\frac{-100}{550}\divideontimes 9.81$
$a=-1.78m/{s}^{2}$
(b)
$N=W+ma$
$670=550+ma$
$ma=120$
$a=\frac{120}{550}\divideontimes 9.81$
$a=2.14m/{s}^{2}$
(c)
if scale read zero then elevator is going down with acceleration of 9.81 $m/{s}^{2}$. he feel weightless
(d)
the temnsion in part (a)
$T=mg+ma$
$T=850\left(9.81—1.78\right)$
$T=6825.5N$
tension in part (c)
$T=mg+ma$
$T=mg—mg$
$T=0N$
Step 4
(a) magnitude of acceleration is 1.78 $m/{s}^{2}$ and direction is downward
(b)
magnitude of acceleration is 2.14 $m/{s}^{2}$ and direction is upward
(C)
if scale read zero then elevator is going down with acceleration of 9.81 $m/{s}^{2}$ . he feel weightless
(d)
for part (a) 6825.5 N
for part (c) ON

Jazz Frenia

Step 1:
(a) To find the acceleration of the elevator, we can analyze the forces acting on the system.
Let ${m}_{\text{student}}$ be the mass of the student and ${m}_{\text{elevator}}$ be the mass of the elevator. The combined mass of the student and elevator is given as ${m}_{\text{total}}={m}_{\text{student}}+{m}_{\text{elevator}}=850\phantom{\rule{0.167em}{0ex}}\text{kg}$.
The weight of the system (force due to gravity) is given by ${W}_{\text{total}}={m}_{\text{total}}·g$, where $g$ is the acceleration due to gravity.
When the elevator is at rest or moving at a constant velocity, the normal force exerted by the scale is equal in magnitude and opposite in direction to the weight of the system. Thus, we have $|{N}_{\text{scale}}|=|{W}_{\text{total}}|$.
In this case, when the scale reads 450 N, we have $|{N}_{\text{scale}}|=450\phantom{\rule{0.167em}{0ex}}\text{N}$, so $|{W}_{\text{total}}|=450\phantom{\rule{0.167em}{0ex}}\text{N}$.
Since $|{W}_{\text{total}}|={m}_{\text{total}}·g$, we can solve for $g$ as follows:
$g=\frac{|{W}_{\text{total}}|}{{m}_{\text{total}}}=\frac{450\phantom{\rule{0.167em}{0ex}}\text{N}}{850\phantom{\rule{0.167em}{0ex}}\text{kg}}\approx 0.529\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$
The acceleration of the elevator is equal to the acceleration due to gravity in magnitude but opposite in direction. Therefore, the acceleration of the elevator is ${a}_{\text{elevator}}=-0.529\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$.
Step 2:
(b) To find the acceleration when the scale reads 670 N, we can use the same approach. The magnitude of the normal force exerted by the scale is equal to the magnitude of the weight of the system. Thus, $|{W}_{\text{total}}|=670\phantom{\rule{0.167em}{0ex}}\text{N}$.
Using the formula $g=\frac{|{W}_{\text{total}}|}{{m}_{\text{total}}}$, we can calculate the acceleration as follows:
$g=\frac{670\phantom{\rule{0.167em}{0ex}}\text{N}}{850\phantom{\rule{0.167em}{0ex}}\text{kg}}\approx 0.788\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$
Again, the acceleration of the elevator is equal in magnitude to the acceleration due to gravity but opposite in direction. Therefore, the acceleration of the elevator is ${a}_{\text{elevator}}=-0.788\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$.
Step 3:
(c) If the scale reads zero, it means that the normal force exerted by the scale is zero. In this case, the weight of the system must be zero as well. However, the weight of the system is given by ${W}_{\text{total}}={m}_{\text{total}}·g$.
Since ${m}_{\text{total}}$ is positive and $g$ is the acceleration due to gravity (which is always positive), the weight of the system can never be zero unless ${m}_{\text{total}}$ is zero (which is not the case here).
Therefore, if the scale reads zero, it implies that there is no contact force between the student and the
scale, which means the scale is not functioning properly or is not in contact with the student. The student should worry about the malfunction of the scale rather than the situation of weightlessness.
Step 4:
(d) In part (a), the tension in the cable can be calculated using the following relationship: ${T}_{\text{cable}}={m}_{\text{total}}·g=850\phantom{\rule{0.167em}{0ex}}\text{kg}·9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}=8330\phantom{\rule{0.167em}{0ex}}\text{N}$.
Therefore, the tension in the cable in part (a) is ${T}_{\text{cable}}=8330\phantom{\rule{0.167em}{0ex}}\text{N}$.
In part (c), when the scale reads zero, it implies that there is no normal force exerted by the scale. However, the tension in the cable still needs to balance the weight of the system. Therefore, the tension in the cable is the same as in part (a), which is ${T}_{\text{cable}}=8330\phantom{\rule{0.167em}{0ex}}\text{N}$.

Andre BalkonE

(a) To find the acceleration of the elevator when the scale reads 450 N, we can use Newton's second law of motion. The net force acting on the elevator is the difference between the tension in the cable and the weight of the elevator and student. Mathematically, we can express this as:
${F}_{\text{net}}=T-mg$
where ${F}_{\text{net}}$ is the net force, $T$ is the tension in the cable, $m$ is the mass of the elevator and student, and $g$ is the acceleration due to gravity.
We also know that the net force is equal to the mass of the system multiplied by the acceleration of the elevator. Mathematically, this can be written as:
${F}_{\text{net}}=ma$
Setting these two equations equal to each other, we have:
$T-mg=ma$
Substituting the given values, we have:
$450\phantom{\rule{0.167em}{0ex}}\text{N}-\left(850\phantom{\rule{0.167em}{0ex}}\text{kg}×9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}\right)=850\phantom{\rule{0.167em}{0ex}}\text{kg}×a$
Simplifying the equation, we can solve for the acceleration $a$:
$a=\frac{450\phantom{\rule{0.167em}{0ex}}\text{N}-850\phantom{\rule{0.167em}{0ex}}\text{kg}×9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}}{850\phantom{\rule{0.167em}{0ex}}\text{kg}}$
Calculating the value, we find:
$a\approx \frac{450\phantom{\rule{0.167em}{0ex}}\text{N}-8330\phantom{\rule{0.167em}{0ex}}\text{N}}{850\phantom{\rule{0.167em}{0ex}}\text{kg}}\approx -9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$
Therefore, the acceleration of the elevator is approximately $-9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$, in the downward direction.
(b) To find the acceleration when the scale reads 670 N, we can use the same equation as in part (a):
$T-mg=ma$
Substituting the given values, we have:
$670\phantom{\rule{0.167em}{0ex}}\text{N}-\left(850\phantom{\rule{0.167em}{0ex}}\text{kg}×9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}\right)=850\phantom{\rule{0.167em}{0ex}}\text{kg}×a$
Simplifying the equation, we can solve for the acceleration $a$:
$a=\frac{670\phantom{\rule{0.167em}{0ex}}\text{N}-8330\phantom{\rule{0.167em}{0ex}}\text{N}}{850\phantom{\rule{0.167em}{0ex}}\text{kg}}$
Calculating the value, we find:
$a\approx \frac{670\phantom{\rule{0.167em}{0ex}}\text{N}-8330\phantom{\rule{0.167em}{0ex}}\text{N}}{850\phantom{\rule{0.167em}{0ex}}\text{kg}}\approx -7.5\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$
Therefore, the acceleration of the elevator is approximately $-7.5\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$, in the downward direction.
(c) If the scale reads zero, the student should not worry. When the scale reads zero, it means that the tension in the cable is equal to the weight of the elevator and student:
$T-mg=0$
Simplifying the equation, we can solve for the tension $T$:
$T=mg$
Since the tension is equal to the weight of the system, there is no net force acting on the elevator. In this case, the elevator and the student are in free fall and experience weightlessness. Therefore, the student should not worry.
(d) In part
(a), the tension in the cable can be calculated by substituting the known values into the equation:
$T=mg$
$T=850\phantom{\rule{0.167em}{0ex}}\text{kg}×9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$
Calculating the value, we find:
$T\approx 8330\phantom{\rule{0.167em}{0ex}}\text{N}$
Therefore, the tension in the cable in part (a) is approximately 8330 N.
In part (c), when the scale reads zero, it means that the tension in the cable is also zero. In this case, the tension is no longer supporting the elevator and the student. Therefore, the tension in the cable in part (c) is zero.

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