Question

(a) What is the acceleration of two falling sky divers (total mass = 132 kg including parachute) when the upward force of air resistance is equal to one-fourth of their weight? (b) After opening the parachute, the divers descend leisurely to the ground at constant speed. What now is the force of air resistance on the sky divers and their parachute?

See Fig. 4–44.

Short answer

The magnitude and direction of the acceleration of the elevator is $a=-2.45\mathrm{m}/{\mathrm{s}}^2$in the downward direction.

Step-by-step solution

Step 1. Determine the forces acting on skydivers

In this case, only two forces will be acting on the skydivers; weight (W) and the normal force (N). Consider the upward direction to be positive and apply Newton’s second law to calculate the skydivers’ acceleration.

Given data:

The total mass of the skydivers is $m=132\mathrm{kg}$.

The normal force is $N=\frac{W}{4}$.

The free body diagram of the skydivers is as follows:

The relation to calculate the vertical forces is given by:

$\begin{array}{c}\Sigma F_v=ma\\ N-W=ma\end{array}$

Here, a is the acceleration of the skydivers, W is the weight, and N is the normal force.

Step 2. Determine the acceleration of the skydivers

On plugging the values in the above relation, you get:

$\begin{array}{c}\frac{W}{4}-W=ma\\ \frac{mg}{4}-mg=ma\\ a=-0.75\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)\\ a=-7.35\mathrm{m}/{\mathrm{s}}^2\end{array}$

Here, g is the gravitational acceleration.

Thus, $a=-7.35\mathrm{m}/{\mathrm{s}}^2$is the acceleration of the skydivers. The negative sign indicates that the direction of acceleration is downward.

Step 3. Determine the force of the air resistance

When the skydivers are descending at a constant speed, the total force on them is zero.

The relation to calculate the force of the air resistance is given by:

$F=mg$

On plugging the values in the above relation, you get:

$\begin{array}{l}F=\left(132 \mathrm{kg}\right)\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)\\ F=1294.9\mathrm{N}\end{array}$

Thus, $F=1294.9\mathrm{N}$is the force of the air resistance.