5.6 Common forces  (Page 5/11)

 Page 5 / 11

Whenever we have two-dimensional vector problems in which no two vectors are parallel, the easiest method of solution is to pick a convenient coordinate system and project the vectors onto its axes. In this case, the best coordinate system has one horizontal axis ( x ) and one vertical axis ( y ).

Solution

First, we need to resolve the tension vectors into their horizontal and vertical components. It helps to look at a new free-body diagram showing all horizontal and vertical components of each force acting on the system ( [link] ). When the vectors are projected onto vertical and horizontal axes, their components along these axes must add to zero, since the tightrope walker is stationary. The small angle results in T being much greater than w .

Consider the horizontal components of the forces (denoted with a subscript x ):

${F}_{\text{net}\phantom{\rule{0.2em}{0ex}}x}={T}_{\text{R}x}-{T}_{\text{L}x}.$

The net external horizontal force ${F}_{\text{net}\phantom{\rule{0.2em}{0ex}}x}=0,$ since the person is stationary. Thus,

$\begin{array}{ccc}\hfill {F}_{\text{net}\phantom{\rule{0.2em}{0ex}}x}& =\hfill & 0={T}_{\text{R}x}-{T}_{\text{L}x}\hfill \\ \hfill {T}_{\text{L}x}& =\hfill & {T}_{\text{R}x}.\hfill \end{array}$

Now observe [link] . You can use trigonometry to determine the magnitude of ${T}_{\text{L}}$ and ${T}_{\text{R}}$ :

$\begin{array}{}\\ \\ \hfill \text{cos}\phantom{\rule{0.2em}{0ex}}5.0\text{°}& =\hfill & \frac{{T}_{\text{L}x}}{{T}_{\text{L}}},\phantom{\rule{0.5em}{0ex}}{T}_{\text{L}x}={T}_{\text{L}}\text{cos}\phantom{\rule{0.2em}{0ex}}5.0\text{°}\hfill \\ \hfill \text{cos}\phantom{\rule{0.2em}{0ex}}5.0\text{°}& =\hfill & \frac{{T}_{\text{R}x}}{{T}_{\text{R}}},\phantom{\rule{0.5em}{0ex}}{T}_{\text{R}x}={T}_{\text{R}}\text{cos}\phantom{\rule{0.2em}{0ex}}5.0\text{°}.\hfill \end{array}$

Equating T L x and T R x :

${T}_{\text{L}}\text{cos}\phantom{\rule{0.2em}{0ex}}5.0\text{°}={T}_{\text{R}}\text{cos}\phantom{\rule{0.2em}{0ex}}5.0\text{°}.$

Thus,

${T}_{\text{L}}={T}_{\text{R}}=T,$

as predicted. Now, considering the vertical components (denoted by a subscript y ), we can solve for T . Again, since the person is stationary, Newton’s second law implies that ${F}_{\text{net}\phantom{\rule{0.2em}{0ex}}y}=0$ . Thus, as illustrated in the free-body diagram,

${F}_{\text{net}\phantom{\rule{0.2em}{0ex}}y}={T}_{\text{L}y}+{T}_{\text{R}y}-w=0.$

We can use trigonometry to determine the relationships among ${T}_{\text{Ly}},{T}_{\text{Ry}},$ and T . As we determined from the analysis in the horizontal direction, ${T}_{\text{L}}={T}_{\text{R}}=T$ :

$\begin{array}{}\\ \hfill \text{sin}\phantom{\rule{0.2em}{0ex}}5.0\text{°}& =\hfill & \frac{{T}_{\text{L}y}}{{T}_{\text{L}}},\phantom{\rule{0.5em}{0ex}}{T}_{\text{L}y}={T}_{\text{L}}\text{sin}\phantom{\rule{0.2em}{0ex}}5.0\text{°}=T\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}5.0\text{°}\hfill \\ \hfill \text{sin}\phantom{\rule{0.2em}{0ex}}5.0\text{°}& =\hfill & \frac{{T}_{\text{R}y}}{{T}_{\text{R}}},\phantom{\rule{0.5em}{0ex}}{T}_{\text{R}y}={T}_{\text{R}}\text{sin}\phantom{\rule{0.2em}{0ex}}5.0\text{°}=T\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}5.0\text{°}.\hfill \end{array}$

Now we can substitute the vales for ${T}_{\text{Ly}}$ and ${T}_{\text{Ry}}$ , into the net force equation in the vertical direction:

$\begin{array}{ccc}\hfill {F}_{\text{net}\phantom{\rule{0.2em}{0ex}}y}& =\hfill & {T}_{\text{L}y}+{T}_{\text{R}y}-w=0\hfill \\ \hfill {F}_{\text{net}\phantom{\rule{0.2em}{0ex}}y}& =\hfill & T\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}5.0\text{°}+T\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}5.0\text{°}-w=0\hfill \\ \hfill 2T\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}5.0\text{°}-w& =\hfill & 0\hfill \\ \hfill 2T\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}5.0\text{°}& =\hfill & w\hfill \end{array}$

and

$T=\frac{w}{2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}5.0\text{°}}=\frac{mg}{2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}5.0\text{°}},$

so

$T=\frac{\left(70.0\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\left(9.80\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}\right)}{2\left(0.0872\right)},$

and the tension is

$T=3930\phantom{\rule{0.2em}{0ex}}\text{N}\text{.}$

Significance

The vertical tension in the wire acts as a force that supports the weight of the tightrope walker. The tension is almost six times the 686-N weight of the tightrope walker. Since the wire is nearly horizontal, the vertical component of its tension is only a fraction of the tension in the wire. The large horizontal components are in opposite directions and cancel, so most of the tension in the wire is not used to support the weight of the tightrope walker.

If we wish to create a large tension, all we have to do is exert a force perpendicular to a taut flexible connector, as illustrated in [link] . As we saw in [link] , the weight of the tightrope walker acts as a force perpendicular to the rope. We saw that the tension in the rope is related to the weight of the tightrope walker in the following way:

$T=\frac{w}{2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta }.$

We can extend this expression to describe the tension T created when a perpendicular force $\left({F}_{\perp }\right)$ is exerted at the middle of a flexible connector:

$T=\frac{{F}_{\perp }}{2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta }.$

The angle between the horizontal and the bent connector is represented by $\theta$ . In this case, T becomes large as $\theta$ approaches zero. Even the relatively small weight of any flexible connector will cause it to sag, since an infinite tension would result if it were horizontal (i.e., $\theta =0$ and sin $\theta =0$ ). For example, [link] shows a situation where we wish to pull a car out of the mud when no tow truck is available. Each time the car moves forward, the chain is tightened to keep it as straight as possible. The tension in the chain is given by $T=\frac{{F}_{\perp }}{2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\theta },$ and since $\theta$ is small, T is large. This situation is analogous to the tightrope walker, except that the tensions shown here are those transmitted to the car and the tree rather than those acting at the point where ${F}_{\perp }$ is applied.

What is concept associated with linear motion
what will happen to vapor pressure when you add solute to a solution?
how is freezing point depression different from boiling point elevation?
shane
how is the osmotic pressure affect the blood serum?
shane
what is the example of colligative properties that seen in everyday living?
shane
What is motion
moving place to place
change position with respect to surrounding
to which
to where ?
the phenomenon of an object to changes its position with respect to the reference point with passage of time then it is called as motion
Shubham
it's just a change in position
festus
reference point -it is a fixed point respect to which can say that a object is at rest or motion
Shubham
yes
Shubham
A change in position
Lily
change in position depending on time
bassey
Is there any calculation for line integral in scalar feild?
yes I'm available
Mharsheeraz
what is thrust
when an object is immersed in liquid, it experiences an upward force which is called as upthrust.
Phanindra
@Phanindra Thapa No, that is buoyancy that you're talking about...
Shii
thrust is simply a push
Shii
it is a force that is exerted by liquid.
Phanindra
what is the difference between upthrust and buoyancy?
misbah
The force exerted by a liquid is called buoyancy. not thrust. there are many different types of thrust and I think you should Google it instead of asking here.
Sharath
hey Kumar, don't discourage somebody like that. I think this conversation is all about discussion...remember that the more we discuss the more we know...
festus
thrust is an upward force acting on an object immersed in a liquid.
festus
uptrust and buoyancy are the same
akanbi
the question isn't asking about up thrust. he simply asked what is thrust
Shii
a Thrust is simply a push
Shii
the perpendicular force applied on the body
Shubham
thrust is a force of depression while
bassey
what is friction?
MFON
while upthrust is a force that act on a body when it is fully or partially submerged in a liquid
bassey
mathematically upthrust (u) = Real weight (wr) - Apparent weight (wa) u = wr- wa.
Boay
friction is a force which opposes relative motion.
Boay
how did astromers neasure the mass of earth and sun
wats the simplest and shortest formula to calc. for order of magnitude
papillas
Distinguish between steamline and turbulent flow with at least one example of each
what is newtons first law
It state that an object in rest will continue to remain in rest or an object in motion will continue to remain in motion except resultant(unbalanced force) force act on it
Gerald
Thanks Gerald Fokumla
Theodore
Gerald
it states that a body remains in its state of rest or uniform motion unless acted upon by resultant external force.
festus
it that a body continues to be in a state of rest or in straight line in a motion unless there is an external force acting on it
Usman
state's that a body will continue to maintain it present state of or of uniform unless it's being called upon by an external force
bassey
derive the relation above
formula for find angular velocity
w=v^2/r
Eric
v=wr>2
bassey
Why satellites don't fall on earth? Reason?
because space doesn't have gravity
Evelyn
satellites technically fall to earth but they travel parallel to earth so fast that they orbit instead if falling(plus the gravity is also weaker in the orbit). its a circular motion where the centripetal force is the weight due to gravity
Kameyama
Exactly everyone what is gravity?
the force that attrats a body towards the center of earth,or towards any other physical body having mass
hina
That force which attracts or pulls two objects to each other. A body having mass has gravitational pull. If the object is bigger in mass then it's gravitational pull would be stronger.For Example earth have gravitational pull on other objects that is why we are pulled by earth.
Abdur
Gravity is the force that act on a on body to the center of the earth.
Aguenim
gravity is a force of Attraction
bassey
Qn1(a) Why during the day sky seen blue colour? (b)why during the sunset its seen reddish colour ?
Boay
How the atmosphere reacts with the light from the sun
Grant
what are the application of 2nd law
It's applicable when determining the amount of force needed to make a body to move or to stop a moving body
festus
coplanar force system
how did you get 7.50times
6
Mharsheeraz
what is the formula for frictional force
I believe, correct me if I am wrong, but Ffr=Fn*mu
Grant By    By Prateek Ashtikar By   By 