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A photograph of geese flying in a V formation.
Geese fly in a V formation during their long migratory travels. This shape reduces drag and energy consumption for individual birds, and also allows them a better way to communicate. (credit: “Julo”/Wikimedia Commons)

In lecture demonstrations, we do measurements of the drag force on different objects. The objects are placed in a uniform airstream created by a fan. Calculate the Reynolds number and the drag coefficient.

The calculus of velocity-dependent frictional forces

When a body slides across a surface, the frictional force on it is approximately constant and given by μ k N . Unfortunately, the frictional force on a body moving through a liquid or a gas does not behave so simply. This drag force is generally a complicated function of the body’s velocity. However, for a body moving in a straight line at moderate speeds through a liquid such as water, the frictional force can often be approximated by

f R = b v ,

where b is a constant whose value depends on the dimensions and shape of the body and the properties of the liquid, and v is the velocity of the body. Two situations for which the frictional force can be represented this equation are a motorboat moving through water and a small object falling slowly through a liquid.

Let’s consider the object falling through a liquid. The free-body diagram of this object with the positive direction downward is shown in [link] . Newton’s second law in the vertical direction gives the differential equation

m g b v = m d v d t ,

where we have written the acceleration as d v / d t . As v increases, the frictional force – bv increases until it matches mg . At this point, there is no acceleration and the velocity remains constant at the terminal velocity v T . From the previous equation,

m g b v T = 0 ,


v T = m g b .
The free body diagram shows forces m times vector g pointing vertically down and b times vector v pointing vertically up. The velocity, vector v, is vertically down. The positive y direction is also vertically down.
Free-body diagram of an object falling through a resistive medium.

We can find the object’s velocity by integrating the differential equation for v . First, we rearrange terms in this equation to obtain

d v g ( b / m ) v = d t .

Assuming that v = 0 at t = 0 , integration of this equation yields

0 v d v g ( b / m ) v = 0 t d t ,


m b ln ( g b m v ) | 0 v = t | 0 t ,

where v ' and t ' are dummy variables of integration. With the limits given, we find

m b [ ln ( g b m v ) ln g ] = t .

Since ln A ln B = ln ( A / B ) , and ln ( A / B ) = x implies e x = A / B , we obtain

g ( b v / m ) g = e b t / m ,


v = m g b ( 1 e b t / m ) .

Notice that as t , v m g / b = v T , which is the terminal velocity.

The position at any time may be found by integrating the equation for v . With v = d y / d t ,

d y = m g b ( 1 e b t / m ) d t .

Assuming y = 0 when t = 0 ,

0 y d y = m g b 0 t ( 1 e b t ' / m ) d t ,

which integrates to

y = m g b t + m 2 g b 2 ( e b t / m 1 ) .

Effect of the resistive force on a motorboat

A motorboat is moving across a lake at a speed v 0 when its motor suddenly freezes up and stops. The boat then slows down under the frictional force f R = b v . (a) What are the velocity and position of the boat as functions of time? (b) If the boat slows down from 4.0 to 1.0 m/s in 10 s, how far does it travel before stopping?


  1. With the motor stopped, the only horizontal force on the boat is f R = b v , so from Newton’s second law,
    m d v d t = b v ,

    which we can write as
    d v v = b m d t .

    Integrating this equation between the time zero when the velocity is v 0 and the time t when the velocity is v , we have
    0 v d v v = b m 0 t d t .

    ln v v 0 = b m t ,

    which, since ln A = x implies e x = A , we can write this as
    v = v 0 e b t / m .

    Now from the definition of velocity,
    d x d t = v 0 e b t / m ,

    so we have
    d x = v 0 e b t / m d t .

    With the initial position zero, we have
    0 x d x ' = v 0 0 t e b t ' / m d t ' ,

    x = m v 0 b e b t ' / m | 0 t = m v 0 b ( 1 e b t / m ) .

    As time increases, e b t / m 0 , and the position of the boat approaches a limiting value
    x max = m v 0 b .

    Although this tells us that the boat takes an infinite amount of time to reach x max , the boat effectively stops after a reasonable time. For example, at t = 10 m / b , we have
    v = v 0 e −10 4.5 × 10 −5 v 0 ,

    whereas we also have
    x = x max ( 1 e −10 ) 0.99995 x max .

    Therefore, the boat’s velocity and position have essentially reached their final values.
  2. With v 0 = 4.0 m/s and v = 1.0 m/s, we have 1.0 m/s = ( 4.0 m/s ) e ( b / m ) ( 10 s ) , so
    ln 0.25 = ln 4.0 = b m ( 10 s ) ,

    b m = 1 10 ln 4.0 s -1 = 0.14 s -1 .

    Now the boat’s limiting position is
    x max = m v 0 b = 4.0 m/s 0.14 s −1 = 29 m .

Questions & Answers

plank constant is what
chin Reply
plank constant is a phisical constant that the central quantum
links energy of a photon to it's wave length
What is action of point ?
Bishal Reply
a point . is the spot and the action is what u do when ur at the spot . but the action of a point idk a divider
Quantity which are used in physics simply
Sangram Reply
That's philosophical question.
What is the Physical quantity
Raja Reply
What is Centripetal force
Taiwo Reply
a force of attraction that tends to keep a body moving in a circular path
force pulling the particle towards the center when moving in circular path
good job
y oxygen is shift to first place
Radhika Reply
What is coriolis firce
Shakeel Reply
Angular accelaration force,as the result of the rotatation of earth
Due to the rotation of the earth the winds at the equator get deflected in opposite direction and therefore cause some currents in the Northern and Southern hemisphere which are opposite in there spin.
Thats correct.
if cos = x/y then sec = y/x if sin = y/x then cosec = x/y
Jan Reply
if cosβ=x/y then what is cosecβ
Abubakar Reply
. if cosβ=x/y what is cosecβ
if cosΦ=x/y what is cosecΦ
What are the unknown symbols?
It is defined that cosec = 1/sin and sec = 1/cos
Do you understand this?
of cours
∇(f/g) = (g∇f − f∇g)/g^2 , at points x where g(x) 6= 0 please help me to solve the problem ....
11 Reply
sir please add answer sheet of spigel vector analysis ..
11 Reply
In a pulley system 2 boxes r hanging in both sides of pulley. An other box was joined to the left box through a rope & get accelerated downward. If all the boxes have same mass ,then what will be the acceleration of that system ?
bibek Reply
it may be mg
∇(f/g) = (g∇f − f∇g)/g2 , at points x where g(x) 6= 0 plus solve it
What is momentum
Rika Reply
is the product of mass to it's velocity (mv)
quatity of motion present in a body or product of mass and velocity
show that the kE of a uniform ring of mass m rolling along a smooth horizontal surface so that its centre of mass has a velocity v is mv×v
what is hydration energy
osobase Reply
the energy................................................
what is momentum
Ogwu Reply
dont know want to know the answer
it's the product of mass multiplied by velocity
the quantity of motion possessed by a body is called its momentum.by virtue of which a body can exert a force in the agency which tend to stop it .it is a common experience that stronger force is required to stop more massive body.also faster the body moves harder it is to stop it .this is why
momentum is product of mass and velocity and it is denoted by "P".
Practice Key Terms 2

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