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
${\mu}_{\text{k}}N.$ 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
${f}_{R}=\text{\u2212}bv,$
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
$mg-bv=m\frac{dv}{dt},$
where we have written the acceleration as
$dv\text{/}dt.$ 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
${v}_{\text{T}}.$ From the previous equation,
$mg-b{v}_{\text{T}}=0,$
so
${v}_{\text{T}}=\frac{mg}{b}.$
We can find the object’s velocity by integrating the differential equation for
v . First, we rearrange terms in this equation to obtain
$\frac{dv}{g-(b\text{/}m)v}=dt.$
Assuming that
$v=0\phantom{\rule{0.2em}{0ex}}\text{at}\phantom{\rule{0.2em}{0ex}}t=0,$ integration of this equation yields
where
$v\text{'}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}t\text{'}$ are dummy variables of integration. With the limits given, we find
Since
$\text{ln}A-\text{ln}B=\text{ln}(A\text{/}B),$ and
$\text{ln}(A\text{/}B)=x\phantom{\rule{0.2em}{0ex}}\text{implies}\phantom{\rule{0.2em}{0ex}}{e}^{x}=A\text{/}B,$ we obtain
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
${f}_{R}=\text{\u2212}bv.$ (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?
Solution
With the motor stopped, the only horizontal force on the boat is
${f}_{R}=\text{\u2212}bv,$ so from Newton’s second law,
$m\frac{dv}{dt}=\text{\u2212}bv,$
which we can write as
$\frac{dv}{v}=-\frac{b}{m}dt.$
Integrating this equation between the time zero when the velocity is
${v}_{0}$ and the time
t when the velocity is
$v$ , we have
As time increases,
${e}^{\text{\u2212}bt\text{/}m}\to 0,$ and the position of the boat approaches a limiting value
${x}_{\text{max}}=\frac{m{v}_{0}}{b}.$
Although this tells us that the boat takes an infinite amount of time to reach
${x}_{\text{max}},$ the boat effectively stops after a reasonable time. For example, at
$t=10m\text{/}b,$ we have
Therefore, the boat’s velocity and position have essentially reached their final values.
With
${v}_{0}=4.0\phantom{\rule{0.2em}{0ex}}\text{m/s}$ and
$v=1.0\phantom{\rule{0.2em}{0ex}}\text{m/s,}$ we have
$1.0\phantom{\rule{0.2em}{0ex}}\text{m/s}=(4.0\phantom{\rule{0.2em}{0ex}}\text{m/s}){e}^{\text{\u2212}(b\text{/}m)(10\phantom{\rule{0.2em}{0ex}}\text{s})},$ so
A block (A) of weight 5 kN is to be raised by means of a 20° wedge (B) by the application of a horizontal force (P) as shown in Fig.1. The block A is constrained to move vertically by the application of a horizontal force (S). Find the magnitude of the forces F and S, when the coefficient of fricti
Danilo
A body receives impulses of 24Ns and 35Ns inclined 55 degree to each other. calculate the total impulse
A body receives impulses of 24Ns and 35Ns inclined 55 degree to each other. calculate the total impulse
Previous
twenty four square plus thirty-five square minus to multiple thirty five twenty four and equal answer number square Via this equation defined Total Total impulse
The uniform boom shown below weighs 500 N, and the object hanging from its right end weighs 400 N. The boom is supported by a light cable and by a hinge at the wall. Calculate the tension in the cable and the force on the hinge on the boom. Does the force on the hinge act along the boom?
A 11.0-m boom, AB , of a crane lifting a 3000-kg load is shown below. The center of mass of the boom is at its geometric center, and the mass of the boom is 800 kg. For the position shown, calculate tension T in the cable and the force at the axle A .
a non-uniform boom of a crane 15m long, weighs 2800nts, with its center of gravity at 40% of its lenght from the hingr support. the boom is attached to a hinge at the lower end. rhe boom, which mAKES A 60% ANGLE WITH THE HORIZONTAL IS SUPPORTED BY A HORIZONTAL GUY WIRE AT ITS UPPER END. IF A LOAD OF 5000Nts is hung at the upper end of the boom, find the tension in the guywire and the components of the reaction at the hinge.
centripetal force of attraction that pulls a body that is traversing round the orbit of a circle toward the center of the circle.
Fc = MV²/r
Sampson
centripetal force is the force of attraction that pulls a body that is traversing round the orbit of a circle toward the center of the circle.
Fc = MV²/r
Sampson
I do believe the formula for centripetal force is F=MA or F=m(v^2/r)
John
I mean the formula is Fc= Mass multiplied by square of velocity all over the Radius of the circle
Sampson
Yes
John
The force is equal to the mass times the velocity squared divided by the radius
John
That's the current chapter I'm on in my engineering physics class
John
Centripetal force is a force of attraction which keeps an object round the orbit towards the center of a circle.
Mathematically Fc=mv²/r
Adebileje
In Example, we calculated the final speed of a roller coaster that descended 20 m in height and had an
initial speed of 5 m/s downhill. Suppose the roller coaster had had an initial speed of 5 m/s uphill instead,
and it coasted uphill, stopped, and then rolled back down to a final point 20 m bel
A steel lift column in a service station is 4 meter long and .2 meter in diameter. Young's modulus for steel is 20 X 1010N/m2. By how much does the column shrink when a 5000- kg truck is on it?