Chapter 4, Problem 112P

(a)

To determine

Acceleration of the elevator.

Answer to Problem 112P

Acceleration is $2.2\text{m}/{\text{s}}^{\text{2}}$ in upward direction.

Explanation of Solution

The reading of scale without box is $0.960\text{kN}$, mass of box is $20.0\text{kg}$, and the reading of scale with box is $1.200\text{kN}$.

Let the acceleration in upward direction is taken as positive and in downward direction is taken as negative.

Write the equation for the apparent weight of Luke in elevator moving upwards with constant acceleration without the box.

${W^{\prime }}_L=m_L\left(g+a_y\right)$ (I)

Here, the apparent weight of Luke in elevator is ${W^{\prime }}_L$, mass of Luke is $m_L$, gravitational acceleration is $g$, and the acceleration of elevator is $a_y$.

Write the equation for the apparent weight of Luke in elevator moving upwards with constant acceleration with the box.

${W^{\prime }}_{L+b}=\left(m_L+m_b\right)\left(g+a_y\right)$ (II)

Here, the apparent weight of Luke in elevator is ${W^{\prime }}_{L+b}$ and the mass of box is $m_b$.

Rewrite equation (I) in terms of $m_L$.

$m_L=\frac{{W^{\prime }}_L}{g+a_y}$

Rewrite the equation (II) by substituting the above relation for $m_L$.

$\begin{array}{c}{W^{\prime }}_{L+b}=\left(\frac{{W^{\prime }}_L}{g+a_y}+m_b\right)\left(g+a_y\right)\\ ={W^{\prime }}_L+\left(m_b\left(g+a_y\right)\right)\end{array}$

Rewrite the above relation in terms of $a_y$.

$a_y=\frac{{W^{\prime }}_{L+b}-{W^{\prime }}_L}{m_b}-g$

Conclusion:

Substitute $1.200\text{kN}$ for ${W^{\prime }}_{L+b}$, $0.960\text{kN}$ for $W^{\prime }$, $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$, and $20.0\text{kg}$ for $m_b$ in the above equation to find $a_y$.

$\begin{array}{c}{a}_y=\frac{\left(1.200\text{kN}-0.960\text{kN}\right)\left(\frac{{\text{10}}^{\text{3}}\text{N}}{\text{1}\text{kN}}\right)}{20.0\text{kg}}-9.80\text{m}/{\text{s}}^{\text{2}}\\ =+2.2\text{m}/{\text{s}}^{\text{2}}\end{array}$

Positive value indicates the acceleration is directed in upward direction.

Therefore, the acceleration is $2.2\text{m}/{\text{s}}^{\text{2}}$ in upward direction.

(b)

To determine

The weight of Luke.

Answer to Problem 112P

Luke’s weight is $784\text{N}$

Explanation of Solution

The reading of scale without box is $0.960\text{kN}$, mass of box is $20.0\text{kg}$, and the reading of scale with box is $1.200\text{kN}$.

Let the acceleration in upward direction is taken as positive and in downward direction is taken as negative.

Write the relation between ${W^{\prime }}_L$ and the weight of Luke.

${W^{\prime }}_L=W_L\left(1+\frac{a_y}{g}\right)$

Here, the weight of Luke is $W_L$.

Rewrite the above equation in terms of $W_L$.

$W_L=\frac{{W^{\prime }}_L}{\left(1+a_y/g\right)}$

Conclusion:

Substitute $0.960\text{kN}$ for $W^{\prime }$, $2.2\text{m}/{\text{s}}^{\text{2}}$ for $a_y$, and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the above equation to find $W_L$.

$\begin{array}{c}{W}_L=\frac{0.960\text{kN}}{\left(1+\left(\frac{2.2\text{m}/{\text{s}}^{\text{2}}}{9.8\text{m}/{\text{s}}^{\text{2}}}\right)\right)}\\ =\frac{0.960\text{kN}}{1.224}\\ =0.784\text{kN}\left(\frac{{\text{10}}^{\text{3}}\text{N}}{\text{1}\text{kN}}\right)\\ =784\text{N}\end{array}$

Therefore, the Luke’s weight is $784\text{N}$