Initial Free Fall:
When a skydiver jumps from a plane, they initially accelerate downwards due to the force of gravity, Fg = mg, where m is the mass of the skydiver and g is the acceleration due to gravity (approximately 9.8 m/s²).
Increasing Air Resistance:
As the skydiver falls, their speed increases, which in turn increases the air resistance Fd acting against their fall.
Balancing Forces:
As the velocity increases, the drag force increases until it balances the gravitational force. At this point, the net force on the skydiver is zero, and they stop accelerating. The skydiver then continues to fall at a constant speed called the terminal velocity.
Mathematically, terminal velocity vt is reached when Fg = Fd.
Deploying the Parachute:
When the parachute is opened, the skydiver's cross-sectional area A dramatically increases, and so does the drag coefficient Cd.
Sudden Increase in Drag Force:
The increased area and drag coefficient lead to a significant increase in the drag force, even if the skydiver's speed v has not yet decreased much. This creates a large upward force opposing the skydiver's motion.
Rapid Deceleration:
This large drag force causes the skydiver to decelerate rapidly until the forces again balance. This new balance occurs at a much lower terminal velocity compared to free fall without the parachute.
New Terminal Velocity:
The new terminal velocity is much lower due to the increased drag. Mathematically, it can be expressed similarly:
$$mg = \dfrac{1}{2}\rho v_t^2C_d A$$
With the parachute open, both Cd and A are much larger, leading to a smaller vt.
In the following applet, the variation of velocity with time as a sky-diver falls through the air and opening the parachute in the middle of the dive is shown. An animated free-body diagram of sky-diver along the fall is displayed as the velocity changes. You can either begin the animation using the Start button or by dragging the scroll bar.
In this video, I took the elevator while turning on the accelerometer sensor in my phone using the PhyPhox app, which is free and downloadable on Android and iOS devices. The acceleration in the vertical direction is shown to first remain at zero with slight fluctuations. It then increases when the lift starts moving and goes back to zero as the lift reaches a constant speed. As the lift comes to a stop near the top floor, the acceleration becomes negative. The second half of the video shows me taking the elevator down.
The next video explains the observations using free-body diagrams.
To solve a two-body dynamics problem involving an external force \( F_{\text{ext}} \) pushing two boxes (Box 1 and Box 2), where the force acts directly on Box 1, and Box 1 exerts a pushing force on Box 2, we can apply Newton's Second Law in three different free-body diagrams as shown in the applet below:
Boxes: The masses of Box 1 and Box 2 are \( m_1 \) and \( m_2 \), respectively.
Forces:
For each box, apply Newton’s Second Law \( F = ma \).
The forces acting on Box 1 are:
The only force acting on Box 2 is the force \( F_{12} \) exerted by Box 1.
So, for Box 2:
\( F_{12} = m_2 a \)
The external force is exerted on the whole system which has the combined mass of \( (m_1 + m_2)\)
\( F_{\text{ext}} = (m_1 + m_2) a \)
By Newton's Third Law:
\( F_{12} = F_{21} \)
This implies that the force \( F_{12} \) exerted by Box 1 on Box 2 is equal in magnitude and opposite in direction to the force \( F_{21} \) exerted by Box 2 on Box 1.