<< Chapter < Page Chapter >> Page >

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)

Extended topic: real forces and inertial frames

There is another distinction among forces in addition to the types already mentioned. Some forces are real, whereas others are not. Real forces are those that have some physical origin, such as the gravitational pull. Contrastingly, fictitious forces are those that arise simply because an observer is in an accelerating frame of reference, such as one that rotates (like a merry-go-round) or undergoes linear acceleration (like a car slowing down). For example, if a satellite is heading due north above Earth’s northern hemisphere, then to an observer on Earth it will appear to experience a force to the west that has no physical origin. Of course, what is happening here is that Earth is rotating toward the east and moves east under the satellite. In Earth’s frame this looks like a westward force on the satellite, or it can be interpreted as a violation of Newton’s first law (the law of inertia). An inertial frame of reference    is one in which all forces are real and, equivalently, one in which Newton’s laws have the simple forms given in this chapter.

Questions & Answers

how did you get 1640
Noor Reply
If auger is pair are the roots of equation x2+5x-3=0
Peter Reply
Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.
MATTHEW Reply
420
Sharon
from theory: distance [miles] = speed [mph] × time [hours] info #1 speed_Dennis × 1.5 = speed_Wayne × 2 => speed_Wayne = 0.75 × speed_Dennis (i) info #2 speed_Dennis = speed_Wayne + 7 [mph] (ii) use (i) in (ii) => [...] speed_Dennis = 28 mph speed_Wayne = 21 mph
George
Let W be Wayne's speed in miles per hour and D be Dennis's speed in miles per hour. We know that W + 7 = D and W * 2 = D * 1.5. Substituting the first equation into the second: W * 2 = (W + 7) * 1.5 W * 2 = W * 1.5 + 7 * 1.5 0.5 * W = 7 * 1.5 W = 7 * 3 or 21 W is 21 D = W + 7 D = 21 + 7 D = 28
Salma
Devon is 32 32​​ years older than his son, Milan. The sum of both their ages is 54 54​. Using the variables d d​ and m m​ to represent the ages of Devon and Milan, respectively, write a system of equations to describe this situation. Enter the equations below, separated by a comma.
Aaron Reply
find product (-6m+6) ( 3m²+4m-3)
SIMRAN Reply
-42m²+60m-18
Salma
what is the solution
bill
how did you arrive at this answer?
bill
-24m+3+3mÁ^2
Susan
i really want to learn
Amira
I only got 42 the rest i don't know how to solve it. Please i need help from anyone to help me improve my solving mathematics please
Amira
Hw did u arrive to this answer.
Aphelele
hi
Bajemah
-6m(3mA²+4m-3)+6(3mA²+4m-3) =-18m²A²-24m²+18m+18mA²+24m-18 Rearrange like items -18m²A²-24m²+42m+18A²-18
Salma
complete the table of valuesfor each given equatio then graph. 1.x+2y=3
Jovelyn Reply
x=3-2y
Salma
y=x+3/2
Salma
Hi
Enock
given that (7x-5):(2+4x)=8:7find the value of x
Nandala
3x-12y=18
Kelvin
please why isn't that the 0is in ten thousand place
Grace Reply
please why is it that the 0is in the place of ten thousand
Grace
Send the example to me here and let me see
Stephen
A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of one of the other legs. Find the lengths of the hypotenuse and the other leg
Marry Reply
how far
Abubakar
cool u
Enock
state in which quadrant or on which axis each of the following angles given measure. in standard position would lie 89°
Abegail Reply
hello
BenJay
hi
Method
I am eliacin, I need your help in maths
Rood
how can I help
Sir
hmm can we speak here?
Amoon
however, may I ask you some questions about Algarba?
Amoon
hi
Enock
what the last part of the problem mean?
Roger
The Jones family took a 15 mile canoe ride down the Indian River in three hours. After lunch, the return trip back up the river took five hours. Find the rate, in mph, of the canoe in still water and the rate of the current.
cameron Reply
Shakir works at a computer store. His weekly pay will be either a fixed amount, $925, or $500 plus 12% of his total sales. How much should his total sales be for his variable pay option to exceed the fixed amount of $925.
mahnoor Reply
I'm guessing, but it's somewhere around $4335.00 I think
Lewis
12% of sales will need to exceed 925 - 500, or 425 to exceed fixed amount option. What amount of sales does that equal? 425 ÷ (12÷100) = 3541.67. So the answer is sales greater than 3541.67. Check: Sales = 3542 Commission 12%=425.04 Pay = 500 + 425.04 = 925.04. 925.04 > 925.00
Munster
difference between rational and irrational numbers
Arundhati Reply
When traveling to Great Britain, Bethany exchanged $602 US dollars into £515 British pounds. How many pounds did she receive for each US dollar?
Jakoiya Reply
how to reduced echelon form
Solomon Reply
Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?
Zack Reply
d=r×t the equation would be 8/r+24/r+4=3 worked out
Sheirtina
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply
Practice Key Terms 3

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'College physics for ap® courses' conversation and receive update notifications?

Ask