Chapter 4, Problem 17P

(a)

To determine

The acceleration of two divers falling freely in high upward air resistance.

Answer to Problem 17P

Solution:

Divers are falling with the acceleration of $-7.35{\text{ m/s}}^{\text{2}}$

Explanation of Solution

Given:

Total mass of the two divers, $m=132\text{kg}$

The upward for due to air drag, ${\displaystyle \sum F_a}=\frac{1}{4}{F}_g=0.25F_g$

where, $F_g$ is the weight of divers

Formula Used:

From Newton’s second law of motion:

  ${\displaystyle \sum F}=ma$

Here, m is the mass and a is the acceleration.

Calculation:

Free body diagram:

  

  $\begin{array}{l}{\displaystyle \sum F}=ma\\ F_g-F_a=ma\\ F_g-0.25F_g=(132)a\\ mg–0.25mg=\left(132\right)a\\ \left(132\right)\left(9.8\right)–0.25\left(132\right)\left(9.8\right)=132a\\ 7.35=a\\ a=7.35{\text{ m/s}}^{\text{2}}\end{array}$

Due to upward air resistance, both thedivers feel less acceleration compared to free fall acceleration when air resistance is negligible.

Conclusion:

The acceleration felt by the divers is $7.35{\text{ m/s}}^{\text{2}}$ downward.

(b)

To determine

The force due air resistance when divers fall with constant speed.

Answer to Problem 17P

Solution:

The upward force due to air is $1293.6\text{ N}$ .

Explanation of Solution

Given:

Total mass of the two divers, $m=132\text{kg}$

As speed is constant, $a=0{\text{ m/s}}^{\text{2}}$

Formula Used:

From Newton’s second law of motion:

  ${\displaystyle \sum F}=ma$

Here, m is the mass and a is the acceleration.

Calculation:

  

  $\begin{array}{l}{\displaystyle \sum F_a}=ma\\ F_a-F_g=ma\\ F_a–\left(132\right)\left(9.8\right)=0\\ F_a=1293.6\text{ N}\end{array}$

When divers fall with constant speed, Net force on divers is 0. This condition is attained when upward force becomes equal to weight of divers which is $1293.6\text{ N}$ .