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 part (a) and (c)?

Short answer

1. Acceleration of the elevator is, -1.78$m/s^2$ .
2. Acceleration if the scale reads 670 N is, 2.14$m/s^2$ .
3. If the scale reads zero, the student should worry.
4. Tension in part (a) is, 6817 N and in part (c) is zero.

Step-by-step solution

Identification of given data

The given data can be listed below as follows,

• The actual weight on the bathroom scale elevator is,$W_s=550N$ .

• The total mass of the student and elevator is,$M=850kg$ .

• The reading of scale as the elevator start moving is,$n=650N$ .

Significance of Newton’s second law

Newton’s second law gives the relationship between the force and the acceleration. This law allows us to calculate the acceleration of a given mass and known forces of an object.

Determination of acceleration of the elevator

(a)

The mass of the student alone whose weight is$W_s=550N$ is expressed as,

$m_s=\frac{W_s}{g}$

Here g is the acceleration due to gravity

Substitute$9.80m/s^2$ for g and 550 N for$W_s$ in the above equation, and we get,

$$\begin{gathered} m_s=\frac{550N}{9.80m/s^2} \\ {m}_s=56.1kg \end{gathered}$$

The relation of acceleration in terms of Newton’s second law is expressed as,

$F_y=m{a}_y$

which gives,

$$\begin{gathered} n-W_s=m_s{a}_y \\ {a}_y=\frac{n-W_s}{m_s} \end{gathered}$$

Here$a_y$ is the acceleration of the elevator,$W_s$ is the actual weight, n is the normal force exerted by the scale,$m_s$ is the mass of the student alone.

Substitute 450 N for n , 550 N for$W_s$ , 56.1 kg for$m_s$ in the above equation, and we get,

$$\begin{gathered} a_y=\frac{450N-550N}{56.1kg} \\ =-1.78m/s^2 \end{gathered}$$

Hence, the elevator has an acceleration of$1.78m/s^2$ , which is directed downward.

Determination of acceleration of scale reads .b)

The relation of acceleration is expressed as,

$a_y=\frac{n-W_s}{m_s}$

Substitute 670 N for n , 550 N for $W_s$, and 56.1 kg for $m_s$in the above equation, and we get,

$$\begin{gathered} a_y=\frac{670N-550N}{56.1kg} \\ =2.14m/s^2 \end{gathered}$$

Hence, the acceleration is 2.14 $m/s^2$which is in an upward direction.

If the scale reads zeroc)

The relation of acceleration is expressed as,

$a_y=\frac{n-W_s}{m_s}$

Substitute n = 0 because scale reading is zero, 550 N for $W_s$, and 56.1 kg for $m_s$in the above equation, and we get,

$$\begin{gathered} a_y=\frac{0-550N}{56.1kg} \\ =-9.80m/s^2 \\ =-g \end{gathered}$$

Hence, the value of acceleration is equal to the negative of acceleration due to gravity which means the elevator is in free fall, and it is a troubled student should worry about this.

Determination of tension in the cabled)

Newton’s second law is expressed as follows,

$F=m{a}_y$

which gives,

$$\begin{gathered} T-Mg=M{a}_y \\ T=M\left(g+a_y\right) \end{gathered}$$

In part (a)

Substitute -1.78$m/s^2$ for $a_y$, 850 kg for M ,$9.8m/s^2$ for g in the above equation, and we get,

$$\begin{gathered} T=\left(850kg\right)\times \left(-1.78m/s^2+9.80m/s^2\right) \\ =6817N \end{gathered}$$

Hence, the tension in part (a) is 6817 N .

In part (c)

Substitute$-9.8m/s^2$ for g in the above equation , and we get,

$$\begin{gathered} T=\left(850N\right)\times \left(-9.80m/s^2+9.80m/s^2\right) \\ =0 \end{gathered}$$

Hence, the tension in part (c) is zero.