Chapter 6, Problem 6.1CYU

Check Your Understanding Now calculate the scale reading when the elevator accelerates downward at a rate of $1.2{\text{0 m/s}}^{\text{2}}$ .

To determine

The reading on the scale when elevator accelerates downward with magnitude $1.20{\text{m/s}}^{\text{2}}$

Answer to Problem 6.1CYU

The reading on the scale in the elevator is $645\text{N}$

Explanation of Solution

Given info:

Mass of the person,

$m=75.0\text{kg}$

Acceleration of the elevator in downward direction,

$a=1.20{\text{m/s}}^{\text{2}}$

Formula used:

By 2^nd law of motion,

Net force on the object is defined as

$\begin{array}{l}F=ma\\ F=\text{net}\text{force}\text{on}\text{the}\text{object}\\ m=\text{mass}\text{of}\text{the}\text{object}\\ a=\text{acceleration}\text{of}\text{the}\text{obejct}\end{array}$

Calculation:

Here is the free-body diagram for that person who is going downward (negative direction of Y) with acceleration $a=1.20{\text{m/s}}^{\text{2}}$.

In the free-body diagram, $\overrightarrow{N}$ is normal force by the surface of the elevator shown in an upward direction. This is the reading of the scale in the elevator. $\overrightarrow{W}$ is the weight of the person shown in a downward direction.

Since the weight of the person = $\overrightarrow{W}=m\overrightarrow{g}$

Since the person’s acceleration is downward, so, the net force on the person would be downward.

The net force on the person,

${\overrightarrow{F}}_{net}=\overrightarrow{W}-\overrightarrow{N}$

Now, by the 2^nd law of motion,

$\begin{array}{l}{\overrightarrow{F}}_{net}=m\overrightarrow{a}\\ \overrightarrow{W}-\overrightarrow{N}=m\overrightarrow{a}\\ \overrightarrow{N}=\overrightarrow{W}-m\overrightarrow{a}\end{array}$

Since, all quantities are in the same line (along the y-axis), so, for instance, we ignore the vector notation, we just write its magnitude,

So, $N=W-ma$

$\begin{array}{l}N=mg-ma\\ N=m(g-a)\end{array}$

Substitute $75.0\text{kg}$ for m, $9.8{\text{m/s}}^{\text{2}}$ for g and $1.20{\text{m/s}}^{\text{2}}$ for a, we get,

$\begin{array}{l}N=75.0\text{kg}(9.8{\text{m/s}}^{\text{2}}-1.20{\text{m/s}}^{\text{2}})\\ N=645N\text{}\end{array}$

Conclusion:

Thus, the reading of scale on the elevator for the person is $645N\text{}$.