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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" \( θ \) } } } {} .

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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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