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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" \( θ \) } } } {} .

Note that θ size 12{θ} {} is the angle between the horizontal and the bent connector. In this case, T size 12{T} {} becomes very large as θ size 12{θ} {} 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., θ = 0 and sin θ = 0 size 12{"sin"θ=0} {} ). (See [link] .)

A car stuck in mud is being pulled out by a chain tied to a tree trunk. A force perpendicular to the length of the chain is applied, represented by an arrow. The tension T along the chain makes an angle with the horizontal line.
We can create a very large tension in the chain by pushing on it perpendicular to its length, as shown. Suppose 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 nearly straight as possible. The tension in the chain is given by T = F 2 sin ( θ ) size 12{T= { {F rSub { size 8{ ortho } } } over {2"sin" \( θ \) } } } {} ; since θ size 12{θ} {} is small, T size 12{T} {} is very large. This situation is analogous to the tightrope walker shown in [link] , except that the tensions shown here are those transmitted to the car and the tree rather than those acting at the point where F size 12{F rSub { size 8{ ortho } } } {} is applied.
A picture of the Golden Gate Bridge.
Unless an infinite tension is exerted, any flexible connector—such as the chain at the bottom of the picture—will sag under its own weight, giving a characteristic curve when the weight is evenly distributed along the length. Suspension bridges—such as the Golden Gate Bridge shown in this image—are essentially very heavy flexible connectors. The weight of the bridge is evenly distributed along the length of flexible connectors, usually cables, which take on the characteristic shape. (credit: Leaflet, Wikimedia Commons)

Section summary

  • When objects rest on a surface, the surface applies a force to the object that supports the weight of the object. This supporting force acts perpendicular to and away from the surface. It is called a normal force, N size 12{N} {} .
  • When objects rest on a non-accelerating horizontal surface, the magnitude of the normal force is equal to the weight of the object:

    N = mg size 12{N= ital "mg"} {} .

  • When objects rest on an inclined plane that makes an angle θ size 12{θ} {} with the horizontal surface, the weight of the object can be resolved into components that act perpendicular ( w ) and parallel ( w size 12{w rSub { size 8{ \lline \lline } } } {} ) to the surface of the plane. These components can be calculated using:

    w = w sin ( θ ) = mg sin ( θ ) size 12{w rSub { size 8{ \lline \lline } } =w"sin" \( θ \) = ital "mg""sin" \( θ \) } {}
    w = w cos ( θ ) = mg cos ( θ ) size 12{w rSub { size 8{ ortho } } =w"cos" \( θ \) = ital "mg""cos" \( θ \) } {} .

  • The pulling force that acts along a stretched flexible connector, such as a rope or cable, is called tension, T size 12{T} {} . When a rope supports the weight of an object that is at rest, the tension in the rope is equal to the weight of the object:

    T = mg size 12{T= ital "mg"} {} .

Problem exercises

Two teams of nine members each engage in a tug of war. Each of the first team’s members has an average mass of 68 kg and exerts an average force of 1350 N horizontally. Each of the second team’s members has an average mass of 73 kg and exerts an average force of 1365 N horizontally. (a) What is magnitude of the acceleration of the two teams? (b) What is the tension in the section of rope between the teams?

  1. 0. 11 m/s 2 size 12{0 "." "11 m/s" rSup { size 8{2} } } {}
  2. 1 . 2 × 10 4 N size 12{1 "." 2 times "10" rSup { size 8{4} } " N"} {}

(a) Calculate the tension in a vertical strand of spider web if a spider of mass 8 . 00 × 10 5 kg size 12{8 "." "00" times "10" rSup { size 8{ - 5} } " kg"} {} hangs motionless on it. (b) Calculate the tension in a horizontal strand of spider web if the same spider sits motionless in the middle of it much like the tightrope walker in [link] . The strand sags at an angle of 12º size 12{"12"°} {} below the horizontal. Compare this with the tension in the vertical strand (find their ratio).

(a) 7 . 84 × 10 -4 N size 12{7 "." "84" times "10" rSup { size 8{4} } " N"} {}

(b) 1 . 89 × 10 –3 N size 12{1 "." "89" times "10" rSup { size 8{"–3"} } " N"} {} . This is 2.41 times the tension in the vertical strand.

Suppose a 60.0-kg gymnast climbs a rope. (a) What is the tension in the rope if he climbs at a constant speed? (b) What is the tension in the rope if he accelerates upward at a rate of 1 . 50 m/s 2 size 12{1 "." "50 m/s" rSup { size 8{2} } } {} ?

Consider the baby being weighed in [link] . (a) What is the mass of the child and basket if a scale reading of 55 N is observed? (b) What is the tension T 1 size 12{T rSub { size 8{1} } } {} in the cord attaching the baby to the scale? (c) What is the tension T 2 size 12{T rSub { size 8{2} } } {} in the cord attaching the scale to the ceiling, if the scale has a mass of 0.500 kg? (d) Draw a sketch of the situation indicating the system of interest used to solve each part. The masses of the cords are negligible.

A vertical spring scale measuring the weight of a baby is shown. The scale is hung from the ceiling by a cord. The weight W of the baby is shown by a vector arrow acting downward and tension T sub one acting in the cord is shown by an arrow upward. The tension in the cord T sub two attached to the ceiling is represented by an arrow upward from the spring scale and downward from the ceiling.
A baby is weighed using a spring scale.

Questions & Answers

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research.net
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Tarell
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Damian
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Source:  OpenStax, Newton's laws. OpenStax CNX. Oct 25, 2015 Download for free at https://legacy.cnx.org/content/col11898/1.1
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