3. Acceleration, friction, drag effects and terminal velocity

Doc Brown's Physics Revision Notes

Suitable for GCSE/IGCSE Physics/Science courses or their equivalent

 This page will help answer questions such as ...

  What causes the drag effect in fluids?

 What is air resistance? What has it got to do with friction?

  What do we mean by 'terminal velocity?

  Describe an experiment to illustrate terminal velocity?

  How does a parachute work?

Friction and motion  (solid surface contact and objects moving through a fluid)

You first appreciate friction when two solid surfaces rub together. When you pull a heavy object across the floor you experience the resistive force between the two surfaces. This resistive force is called friction and is caused by the atoms of the two surfaces bumping in to each other from opposite directions. The moment you stop applying a force, the dragged object ceases moving immediately.

The force of friction always acts in the opposite direction to an object's movement and it can involve ANY type of surface contact effect.

Therefore when a solid object moves through a fluid - gas or liquid, there are collisions between the fluid particles and the surface of the object which creates the resistive force of friction.

In fluids this resistive force is called the drag effect. A boat ploughing through water experience the drag effect of the water brushing over the surface of the hull. An aircraft of skydiver moving through the atmosphere experience a drag effect, which in this case is called air resistance.

If an object is moving at a steady speed the thrust or driving force (engine  or gravity etc.) is being balanced by the opposing force of friction.

e.g. a car moving at a constant velocity means the thrust force from the engine is balanced by the friction of moving parts in the engine, wheel contact with the road surface and the air resistance (the latter being the drag effect).

If you increase the power output of the car by pressing harder on the accelerator pedal you overcome more of the resistive force and move faster, but the friction e.g. the air resistance increases with speed (see next section), so eventually all the opposing forces balance and you attain a higher steady speed.

BUT, if you want a faster constant speed, the engine must work harder and burn more fuel to overcome the increased drag effect as air resistance increases with speed.

If you take your foot of the accelerator and apply the brake, you increase the force of friction so the car slows down. Theoretically the drag force is reduced with decreased speed, but its a very small effect here compared to the friction between the brake pads and brake discs on the car wheels.

If a moving object has no thrust force acting on it, then it will always slow down and come to a halt e.g. on a level road, if you take your foot of the accelerator, the car will eventually come to a halt as the resistive forces act against the car's forward motion. The total friction effects will bring the car to a halt.

An object can keep moving if it is subjected to a constant force greater than the friction effects e.g. a car will roll down a hill due to gravity.

You should also appreciate that objects will move forever in outer space because there are virtually no molecules to bump into and cause a friction-drag effect.


How does speed affect the drag effect and how can we reduce it?

The faster an object moves through a fluid the greater the rate of particle collisions between the object's surface and the fluid (e.g. air or water.

Therefore the faster the speed of an object the greater the drag effect it experiences.

The greater the speed of a boat in water the greater the drag effect on the surface of the hull.

The greater the speed of an aircraft or skydiver the greater the air resistance on the object's surface.

To reduce the drag effect its not always easy to reduce the surface area, hence reduce friction, but you can design the shape of an object to allow the fluid to flow more easily across the surface.

The hull of a boat is designed to 'cut' through the water to reduce friction. The prow is the forward-most part of a ship's bow that cuts through the water. The 'pointed' sharp shape means the hull-fluid particle collisions occur at a sharper angle than a flat surface at 90o to the ship's movement. The prow can be quite blunt in a slow moving barge efficiently carrying a bulk cargo but not so for a fast moving destroyer class warship!

In the case of cars, trains and aircraft, the streamlined shape of the bodywork is designed to reduce the friction-drag effect of air resistance. You can use a wind tunnel to test different bodywork shapes to find the design of minimum air resistance - the shape that allows the smoothest flow of air across the bodywork.

There are times when we wish to increase the drag effect - see parachuting further down the page.

A parachute is used to slow down military aircraft landing on an aircraft carrier. As soon as the jet touches down on the runway a parachute is ejected from the back of the aircraft to produce a large surface area of friction with the atmosphere. There is a rapid deceleration of the aircraft and reduces the risk of it overshooting into the sea!


Terminal velocity  (balancing forces - rolling down hill and objects falling downwards under gravity in fluids)

(a) Trolley and running board

If you set up a running board and let an object like a trolley roll down it, the 'diluted' effect of gravity gets the object to roll down the incline. At first it speeds up as the force of gravity exceeds the friction of the trolley movement. The force of gravity is constant, but as the speed of the trolley increases, so does the effect of friction e.g. the wheels and axle, wheel contact with running board.

Eventually the resistive forces of friction balance the diluted gravity effect and the trolley continues moving at a constant speed.

This final maximum speed is called the terminal velocity.

You can actually demonstrate this with a ticker tape timer experiment (results illustrated above). You attach a long piece of ticker tape to the trolley and the tape passes through a little 'ticker tape' machine that stamps a dot onto the tape every fraction of a second. The distance between the dots gives you the relative speed. At the start the dots are close together at the lowest speeds, but as the trolley speeds up the distance between the dots increases. Eventually the dots are evenly spaced at their maximum distance apart showing the terminal velocity of the trolley was reached (first at time T).

I presume you can do a similar experiment with a series of light gates to show the same effect, not as much fun though!

The above example illustrate the idea of terminal velocity, but next we consider objects falling down in a fluid due to the force of gravity (air or a liquid).


(b) Small spheres falling down through a liquid


You can demonstrate the effect of resistive forces in a fluid using the experiment illustrated in the diagram above.

Theory: When an object falls through a fluid there are three forces to take into consideration.

W↓ is the weight of the ball bearing due to gravity (Weight = mass x gravity, W = mg)

U↑ is the upthrust experienced by any object in a fluid.

F↑ is the resistive force of friction between the ball and the fluid.

W and U are constant and so W-U is a constant and the resultant force in the downward direction.

Initially F is zero at the point where the ball enters the fluid, but, as it descends and speeds up F increases and when F equals W-U the ball bearing descends with uniform velocity - its maximum velocity = terminal velocity.

Method: A large glass tube, sealed at one end with a rubber bung. is filled with a viscous liquid e.g. oil or glycerine (the latter I think is the best and easier to clean out). You need at least a 50 cm depth of liquid.

The glass tube is marked with suitable depth intervals e.g. every 10 cm.

Small steel ball bearings are carefully dropped down a thinner glass tube to make the entry into the fluid as smooth as possible.

The time it takes for the tiny ball to fall between the distance markers is timed.

You should find the falls relatively slowly at first and then attains a maximum velocity - the terminal velocity when the force of the weight of the object is balanced by the resistive forces of friction as the surface of the ball interacts with the liquid.

The 'theoretical results' are shown in the velocity-time graph below.

At first when the object starts to fall the accelerating force due to gravity, W↓, is greater than the frictional force slowing it down.

You can tell this from the steep positive gradient at the start that at first you get the biggest acceleration.

As the speed increases, friction (force F↑, drag effect) increases and this reduces the acceleration.

But, as time goes on, the acceleration decreases (gradient decreasing) because the value of F↑ is increasing.

When all the forces balanced (W-U) = F, the resultant force is zero and the graph becomes horizontal (at time T) and the horizontal value is the maximum speed or terminal velocity.


The complex behaviour of a parachutist

This is an interesting case because it involves an acceleration, a deceleration and two terminal velocities!

speed/velocity-time graph for a parachute descent

  weight due to gravity W↓ and the drag effect due to air resistance D↑ (I'm ignoring the minor effect of upthrust U↑)

force vectors operating in the jump

(1) The parachutist jumps out of the aircraft and immediately accelerates due to gravity (weight force W↓), but the parachute is not opened yet, so the parachutist is in free fall!

Between (1) to (3) acceleration is taking place, but although W↓ is constant, the drag force D↑ due to air friction (air resistance) is increasing, so the acceleration is decreasing (gradient decreasing).

At (3) the terminal velocity is first reached when the drag force D↑ = weight W of the parachutist.

 Its apparently great to do a spot of sky diving first at ~120 mph (~54 m/s) with arms and legs stretched out (increased surface area) and ~200 mph (~89 m/s) if body compact (minimum surface area). In the graph I've assumed its one or the other!

Between (3) to (5) the descent continues at the (1st) constant terminal speed because the resultant force is zero.

At (5) the parachutist opens the parachute to increase air resistance. The wide area brushing through the air produces a massive air resistance (friction) drag effect so D↑ massively increases, and is far greater than W, so rapid deceleration immediately takes place.

Between (5) to (7) the deceleration continues and as the parachutist decreases in speed, so the drag forces decreases too.

At (7) the drag force once again equals the weight force and the parachutist attains a 2nd, somewhat slower, terminal velocity.

Between (7) and (8) the parachutist descends far more slowly with constant speed to land safely at ~15 mph, again the resultant force is zero.

Other terminal velocity or otherwise situations!

Some 'actual' and 'thought' experiments.

You need to consider the W (weigh, depends on local gravitational field strength), U (upthrust) and F (friction, drag) forces described in the ball bearing - liquid filled tube experiment and the four objects illustrated above.

(a) When the first astronauts landed on the moon they carried a beautifully simple experiment.

One astronaut held up a feather and a hammer and let them both fall simultaneously.

Both objects (i) fell to the ground in the same time experiencing the same gravitational acceleration and (ii) they seemed to fall to the ground more slowly than if they had been dropped on Earth.

(Weight = mass x gravity, W = mg)

(i) Both objects experience the same gravitational acceleration of 1.62 m/s2, BUT the Moon has no atmosphere so there is no air resistance. With no drag effect due to friction both objects, whatever their mass or shape (surface area) are free to fall with maximum acceleration due to the Moon's gravity.

(ii) Acceleration on Earth due to the gravitational field is 9.8 m/s2. The Moon has a much smaller mass and gravity is much weaker and only produces an acceleration of 1.62 m/s2. Therefore objects on the Moon will fall much more slowly than on Earth (ignoring air resistance).


(b) If you drop the feather and hammer simultaneously on Earth the feather takes much longer to fall to the ground because it has a relative greater surface area/mass ratio greatly increasing drag effect.

If it wasn't for air resistance, no drag effect can happen and all objects would fall with the same acceleration.

You can do a laboratory experiment with two objects (one 'light' like a feather and one heavy like a lead weight) by allowing them to fall vertically in a large tube from which the air has been pumped out to create a vacuum.

It seems uncanny when they both hit the bottom of the tube at the same time.

(c) If you simultaneously dropped the dense pool ball and light hollow plastic ball with the same radius (giving same surface area) the pool ball will hit the ground before the lighter hollow plastic ball. This is because the greater weight of the pool ball will overcome the drag effect of air resistance more than the less 'weighty' plastic ball and will accelerate at a greater rate.

On the Moon you would not notice any difference, see (a) for the argument.


(d) If you drop any object (safely!) from a tall building on Earth, it will always achieve a terminal velocity due to the friction effect of air resistance.

Remember the drag effect increases with speed because more air particles are brushing and colliding against the surface of the falling object in the same time interval.




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