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The internal structure of a finger with tendon, extensor muscle, and flexor muscle is shown. The force in the muscles is shown by arrows pointing along the tendon. In the second figure, part of a bicycle with a brake cable is shown. Three tension vectors are shown by the arrows along the brake cable, starting from the handle to the wheels. The tensions have the same magnitude but different directions.
(a) Tendons in the finger carry force T size 12{T} {} from the muscles to other parts of the finger, usually changing the force’s direction, but not its magnitude (the tendons are relatively friction free). (b) The brake cable on a bicycle carries the tension T size 12{T} {} from the handlebars to the brake mechanism. Again, the direction but not the magnitude of T size 12{T} {} is changed.

What is the tension in a tightrope?

Calculate the tension in the wire supporting the 70.0-kg tightrope walker shown in [link] .

A tightrope walker is walking on a wire. His weight W is acting downward, shown by a vector arrow. The wire sags and makes a five-degree angle with the horizontal at both ends. T sub R, shown by a vector arrow, is toward the right along the wire. T sub L is shown by an arrow toward the left along the wire. All three vectors W, T sub L, and T sub R start from the foot of the person on the wire. In a free-body diagram, W is acting downward, T sub R is acting toward the right with a small inclination, and T sub L is acting toward the left with a small inclination.
The weight of a tightrope walker causes a wire to sag by 5.0 degrees. The system of interest here is the point in the wire at which the tightrope walker is standing.

Strategy

As you can see in the figure, the wire is not perfectly horizontal (it cannot be!), but is bent under the person’s weight. Thus, the tension on either side of the person has an upward component that can support his weight. As usual, forces are vectors represented pictorially by arrows having the same directions as the forces and lengths proportional to their magnitudes. The system is the tightrope walker, and the only external forces acting on him are his weight w size 12{w} {} and the two tensions T L size 12{T rSub { size 8{L} } } {} (left tension) and T R size 12{T rSub { size 8{R} } } {} (right tension), as illustrated. It is reasonable to neglect the weight of the wire itself. The net external force is zero since the system is stationary. A little trigonometry can now be used to find the tensions. One conclusion is possible at the outset—we can see from part (b) of the figure that the magnitudes of the tensions T L size 12{T rSub { size 8{L} } } {} and T R size 12{T rSub { size 8{R} } } {} must be equal. This is because there is no horizontal acceleration in the rope, and the only forces acting to the left and right are T L size 12{T rSub { size 8{L} } } {} and T R size 12{T rSub { size 8{R} } } {} . Thus, the magnitude of those forces must be equal so that they cancel each other out.

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] ,

Practice Key Terms 3

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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