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Question

# When you weigh yourself on good old terra firma (solid ground), your weight is $142 \mathrm{lb}$. In an elevator your apparent weight is $121 \mathrm{lb}$. What are the direction and magnitude of the elevator's acceleration?

Solution

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We have to find the direction and magnitude of the elevator's acceleration.

$\textbf{Given values:}$

$W=142 \: \text{lb}$

$W_{a}=121 \: \text{lb}$

$a=9.81 \: \text{m}/\text{s}^2$

First, we have to find value for the net force :

\begin{align*} F&=W_{a}-W\\ F&=121 \: \text{lb}-142 \: \text{lb} \tag{Substitute values in equation.}\\ F&=-21 \: \text{lb}\\ \end{align*}

Now, from the next equation, we have to find value for $a$-acceleration :

\begin{align*} F&=m \cdot a \\ a&=\frac{F}{m} \tag{Where is W=m \cdot g \rightarrow m=\frac{W}{g}.}\\ a&=\frac{F}{\left(\dfrac{W}{g} \right)}\\ a&=\left( \frac{F}{W} \right) \cdot g\\ a&=\left( \frac{-21 \: \text{lb}}{142 \: \text{lb}} \right) \cdot (9.81 \: \text{m}/\text{s}^2) \tag{Substitute values in equation.}\\ a&=(-0.1478) \cdot (9.81 \: \text{m}/\text{s}^2) \\ a&=-1.449918 \: \text{m}/\text{s}^2\tag{Downward acceleration.}\\ \end{align*}

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