Question

A 550-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 (c)?

Answer 1

(a) $-1.78 m/s^2$ (downward)

The equation of the forces acting on the student is:

$N-mg = ma$ (1)

where

N is 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 = 450 N

Also, we know that the weight of the student is:

$mg = 550 N$

So we can find its mass:

$m=\frac{mg}{g}=\frac{550}{9.8}=56.1 kg$

So now we can solve eq.(1) to find the acceleration:

$a=\frac{N-mg}{m}=\frac{450-550}{56.1}=-1.78 m/s^2$

where the negative sign means the acceleration is downward.

(b) $2.14 m/s^2$, 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 = 670 N

Solving for the acceleration, we find

$a=\frac{N-mg}{m}=\frac{670-550}{56.1}=2.14 m/s^2$

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

$a=g=-9.8 m/s^2$

which means that the elevator is accelerating downward at $9.8 m/s^2$: this means that the elevator is in free fall, so yes, the student should worry.

(d) (a) 6817 N

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

$mg=(850 kg)(9.8 m/s^2)=8330 N$ is the weight of the elevator+student system

$a=-1.78 m/s^2$ is the acceleration

Solving for T,

$T=mg+ma=8330+(850)(-1.78)=6817 N$

(d) (b) 10149 N

Here we can use the same equation

$T-mg = ma$

where the only difference is that the acceleration is

$a=2.14 m/s^2$

Solving the equation for T, we find

$T=mg+ma=8330+(850)(2.14)=10149 N$

(d) (c) 0

Again, same equation

$T-mg = ma$

But this time, the acceleration is

$a=-9.8 m/s^2$

So, we find:

$T=mg+ma=8330+(850)(-9.8)=0$

So, the tension in the cable is zero, since the elevator is in free fall.