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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 axis horizontal and the other vertical. We call the horizontal the x size 12{x} {} -axis and the vertical the y size 12{y} {} -axis.

Solution

First, we need to resolve the tension vectors into their horizontal and vertical components. It helps to draw a new free-body diagram showing all of the horizontal and vertical components of each force acting on the system.

A vector T sub L making an angle of five degrees with the negative x axis is shown. It has two components, one in the vertical direction, T sub L y, and another horizontal, T sub L x. Another vector is shown making an angle of five degrees with the positive x axis, having two components, one along the y direction, T sub R y, and the other along the x direction, T sub R x. In the free-body diagram, vertical component T sub L y is shown by a vector arrow in the upward direction, T sub R y is shown by a vector arrow in the upward direction, and weight W is shown by a vector arrow in the downward direction. The net force F sub y is equal to zero. In the horizontal direction, T sub R x is shown by a vector arrow pointing toward the right and T sub L x is shown by a vector arrow pointing toward the left, both having the same length so that the net force in the horizontal direction, F sub x, is equal to zero.
When the vectors are projected onto vertical and horizontal axes, their components along those axes must add to zero, since the tightrope walker is stationary. The small angle results in T size 12{T} {} being much greater than w size 12{w} {} .

Consider the horizontal components of the forces (denoted with a subscript x size 12{x} {} ):

F net x = T L x T R x size 12{F rSub { size 8{"net x"} } = T rSub { size 8{"Lx"} } - T rSub { size 8{"Rx"} } } {} .

The net external horizontal force F net x = 0 size 12{F rSub { size 8{"net x"} } = 0} {} , since the person is stationary. Thus,

F net x = 0 = T L x T R x T L x = T R x . alignl { stack { size 12{F rSub { size 8{"net x"} } =0=T rSub { size 8{"LX"} } - T rSub { size 8{"Rx"} } } {} #T rSub { size 8{"Lx"} } = T rSub { size 8{"Rx"} } {} } } {}

Now, observe [link] . You can use trigonometry to determine the magnitude of T L size 12{T rSub { size 8{L} } } {} and T R size 12{T rSub { size 8{R} } } {} . Notice that:

cos ( 5.0º ) = T L x T L T L x = T L cos ( 5.0º ) cos ( 5.0º ) = T R x T R T R x = T R cos ( 5.0º ) . alignl { stack { size 12{"cos" \( 5 "." 0° \) = { {T rSub { size 8{"Lx"} } } over {T rSub { size 8{L} } } } } {} #T rSub { size 8{"Lx"} } =T rSub { size 8{L} } "cos" \( 5 "." 0° \) {} # "cos" \( 5 "." 0° \) = { {T rSub { size 8{"RX"} } } over {T rSub { size 8{R} } } } {} #T rSub { size 8{"Rx"} } =T rSub { size 8{R} } "cos" \( 5 "." 0° \) {} } } {}

Equating T L x size 12{T rSub { size 8{"Lx"} } } {} and T R x size 12{T rSub { size 8{"Rx"} } } {} :

T L cos ( 5.0º ) = T R cos ( 5.0º ) size 12{T rSub { size 8{L} } "cos" \( 5 "." 0° \) =T rSub { size 8{R} } "cos" \( 5 "." 0° \) } {} .

Thus,

T L = T R = T size 12{T rSub { size 8{L} } =T rSub { size 8{R} } =T} {} ,

as predicted. Now, considering the vertical components (denoted by a subscript y size 12{y} {} ), we can solve for T size 12{T} {} . Again, since the person is stationary, Newton’s second law implies that net F y = 0 size 12{F rSub { size 8{y} } =0} {} . Thus, as illustrated in the free-body diagram in [link] ,

F net y = T L y + T R y w = 0 size 12{F rSub { size 8{"net "} rSub { size 8{y} } } =T rSub { size 8{L} rSub { size 8{y} } } +T rSub { size 8{R} rSub { size 8{y} } } - w=0} {} .

Observing [link] , we can use trigonometry to determine the relationship between T L y size 12{T rSub { size 8{L} rSub { size 8{y} } } } {} , T R y size 12{T rSub { size 8{R} rSub { size 8{y} } } } {} , and T size 12{T} {} . As we determined from the analysis in the horizontal direction, T L = T R = T size 12{T rSub { size 8{L} } =T rSub { size 8{R} } =T} {} :

sin ( 5.0º ) = T L y T L T L y = T L sin ( 5.0º ) = T sin ( 5.0º ) sin ( 5.0º ) = T R y T R T R y = T R sin ( 5.0º ) = T sin ( 5.0º ) . alignl { stack { size 12{"sin" \( 5 "." 0° \) = { {T rSub { size 8{L} rSub { size 8{y} } } } over {T rSub { size 8{L} } } } } {} #T rSub { size 8{L} rSub { size 8{y} } } =T rSub { size 8{L} } "sin" \( 5 "." 0° \) =T"sin" \( 5 "." 0° \) {} # "sin" \( 5 "." 0° \) = { {T rSub { size 8{R} rSub { size 8{y} } } } over {T rSub { size 8{R} } } } {} #T rSub { size 8{R} rSub { size 8{y} } } =T rSub { size 8{R} } "sin" \( 5 "." 0° \) =T"sin" \( 5 "." 0° \) {} } } {}

Now, we can substitute the values for T L y size 12{T rSub { size 8{L} rSub { size 8{y} } } } {} and T R y size 12{T rSub { size 8{R} rSub { size 8{y} } } } {} , into the net force equation in the vertical direction:

F net y = T L y + T R y w = 0 F net y = T sin ( 5.0º ) + T sin ( 5.0º ) w = 0 2 T sin ( 5.0º ) w = 0 2 T sin ( 5.0º ) = w alignl { stack { size 12{F rSub { size 8{"net "} rSub { size 8{y} } } =T rSub { size 8{L} rSub { size 8{y} } } +T rSub { size 8{R} rSub { size 8{y} } } - w=0} {} #F rSub { size 8{"net "} rSub { size 8{y} } } =T"sin" \( 5 "." 0° \) +T"sin" \( 5 "." 0° \) - w=0 {} # 2T"sin" \( 5 "." 0° \) - w=0 {} #2T"sin" \( 5 "." 0° \) =w {} } } {}

and

T = w 2 sin ( 5.0º ) = mg 2 sin ( 5.0º ) size 12{T= { {w} over {2"sin" \( 5 "." 0° \) } } = { { ital "mg"} over {2"sin" \( 5 "." 0° \) } } } {} ,

so that

T = ( 70 . 0 kg ) ( 9 . 80 m/s 2 ) 2 ( 0 . 0872 ) size 12{T= { { \( "70" "." "0 kg" \) \( 9 "." "80 m/s" rSup { size 8{2} } \) } over {2 \( 0 "." "0872" \) } } = { {"686 N"} over {0 "." "174"} } } {} ,

and the tension is

T = 3900 N size 12{T="3900"" N"} {} .

Discussion

Note that the vertical tension in the wire acts as a normal 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 small fraction of the tension in the wire. The large horizontal components are in opposite directions and cancel, and 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 very large tension, all we have to do is exert a force perpendicular to a flexible connector, as illustrated in [link] . As we saw in the last example, the weight of the tightrope walker acted as a force perpendicular to the rope. We saw that the tension in the roped related to the weight of the tightrope walker in the following way:

T = w 2 sin ( θ ) size 12{T= { {w} over {2"sin" \( θ \) } } } {} .

We can extend this expression to describe the tension T size 12{T} {} created when a perpendicular force ( F size 12{F rSub { size 8{ ortho } } } {} ) is exerted at the middle of a flexible connector:

T = F 2 sin ( θ ) size 12{T= { {F rSub { size 8{ ortho } } } over {2"sin" \( θ \) } } } {} .

Questions & Answers

anyone know any internet site where one can find nanotechnology papers?
Damian Reply
research.net
kanaga
Introduction about quantum dots in nanotechnology
Praveena Reply
what does nano mean?
Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
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Damian Reply
absolutely yes
Daniel
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s. Reply
there is no specific books for beginners but there is book called principle of nanotechnology
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are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
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Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
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Abhijith Reply
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
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SUYASH Reply
for screen printed electrodes ?
SUYASH
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s. Reply
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
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Cied
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how did you get the value of 2000N.What calculations are needed to arrive at it
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Source:  OpenStax, Une: physics for the health professions. OpenStax CNX. Aug 20, 2014 Download for free at http://legacy.cnx.org/content/col11697/1.1
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