Q 20E

Question

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)?

Step-by-Step Solution

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Answer

a) Acceleration of the elevator is, $- 1.78 {m/s}^{2}$ .

b) Acceleration if the scale reads 670N is, $2.14 {m/s}^{2}$ .

c) If the scale reads zero, the student should worry.

d) Tension in part (a) is, 6817N and in part (c) is zero.

1Step 1: Identification of given data

The given data can be listed below as follows,

The actual weight on the bathroom scale elevator is, $W_{s} = 550 N$ .

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

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

2Step 2: 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.

3Step 3: Determination of acceleration of the elevator

(a)

The mass of the student alone whose weight is $W_{s} = 550 N$ is expressed as,

$m_{s} = \frac{W_{s}}{g}$

Here g is the acceleration due to gravity

Substitute $9.80 {m/s}^{2}$ for g and 5N for $W_{s}$ in the above equation, and we get,

$m_{s} = \frac{550 N}{9.80 {m/s}^{2}} m_{s} = 56.1 kg$

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

$F_{y} = m a_{y}$

which gives,

$n - W_{s} = m_{s} a_{y} a_{y} = \frac{n - W_{s}}{m_{s}}$

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 450N for n, 550N for $W_{s}$ , 56.1kg for $m_{s}$ in the above equation, and we get,

$\begin{array}{rcl} a_{y} & = & \frac{450 N - 550 N}{56.1 kg} \\ = & - 1.78 {m/s}^{2} \end{array}$

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

4Step 4: Determination of acceleration of scale reads 670N. b)

The relation of acceleration is expressed as,

$a_{y} = \frac{n - W_{s}}{m_{s}}$

Substitute 670N for n, 550N for $W_{s}$ , and 56.1kg for $m_{s}$ in the above equation, and we get,

$\begin{array}{rcl} a_{y} & = & \frac{670 N - 550 N}{56.1 k g} \\ = & 2.14 {m/s}^{2} \end{array}$

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

5Step 5: If the scale reads zero c)

The relation of acceleration is expressed as,

$a_{y} = \frac{n - W_{s}}{m_{s}}$

Substitute n = 0 because scale reading is zero, 550N for $W_{s}$ , and 56.1kg for $m_{s}$ in the above equation, and we get,

$\begin{array}{rcl} a_{y} & = & \frac{0 - 550 N}{56.1 kg} \\ = & - 9.80 {m/s}^{2} \\ = & - g \end{array}$

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.

6Step 6: Determination of tension in the cable d)

Newton’s second law is expressed as follows,

$F = m a_{y}$

which gives,

$T - M g_{} = M a_{y} T = M (g + a_{y})$