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toxicsugar22

  • 5 years ago

At a concert by The Who in 1976, the sound level 50 meters from the stage registered 120 decibels. The threshold of pain for the human ear is 90 decibles. 1. calculate the sound intensity of the Who concert and the sound intensity of the threshold of pain.

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  1. anonymous
    • 5 years ago
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    Intensity is inversely proportional to the square of the distance, \[I=\frac{k}{r^2}\]k is some constant. You can compare intensities to distance as,\[\frac{I_0}{I_{pain}}=\frac{r_{pain}^2}{r_0^2} \rightarrow r_{pain}=\sqrt{\frac{I_0}{I_{pain}}}r_0=\frac{100}{\sqrt{3}}m\]

  2. toxicsugar22
    • 5 years ago
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    is that the answere

  3. toxicsugar22
    • 5 years ago
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    can you show me

  4. anonymous
    • 5 years ago
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    that's the answer. pain dist. ~ 57.7m

  5. toxicsugar22
    • 5 years ago
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    ok and what about the sound intensity of the concert

  6. toxicsugar22
    • 5 years ago
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    and this time put it in scientific notation

  7. anonymous
    • 5 years ago
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    You need to be using the decibel formula or something. http://hyperphysics.phy-astr.gsu.edu/hbase/sound/intens.html You should post new questions in the question box.

  8. toxicsugar22
    • 5 years ago
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    whait what I dont get that

  9. toxicsugar22
    • 5 years ago
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    i did not understand so that is no the answer you gave me the first on

  10. anonymous
    • 5 years ago
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    The answer for the distance is wrong. I didn't read decibels; I took it as intensity. You have to convert decibels to intensity using the formula\[D=10\log_{10}(10^{12}I)\]

  11. anonymous
    • 5 years ago
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    You should post this question again so someone else can take it. I'm not round for much longer.

  12. toxicsugar22
    • 5 years ago
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    ok but ou ar to good in math

  13. toxicsugar22
    • 5 years ago
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    If you dont mind me asking what grade are you in

  14. anonymous
    • 5 years ago
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    University.

  15. toxicsugar22
    • 5 years ago
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    wow

  16. anonymous
    • 5 years ago
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    Are you doing a timed test or something?

  17. toxicsugar22
    • 5 years ago
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    im in a community collesge

  18. toxicsugar22
    • 5 years ago
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    college

  19. anonymous
    • 5 years ago
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    is this an online assignment?

  20. toxicsugar22
    • 5 years ago
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    yes why

  21. toxicsugar22
    • 5 years ago
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    one more question befroe leaving

  22. anonymous
    • 5 years ago
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    when is it due? i can look at it later...that's all. when i'm rushed me, i make mistakes.

  23. toxicsugar22
    • 5 years ago
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    can you go to the other question ou answerd for me the when about the websites

  24. toxicsugar22
    • 5 years ago
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    here i will post it to you

  25. anonymous
    • 5 years ago
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    i can't find it. if you have the link, post it.

  26. toxicsugar22
    • 5 years ago
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    the number of websites is growing by 15% each month. In June 2005, there were 4000 websites. Write a formula for the number of websites as a function of months since june 2005.38 minutes ago N(t)=4000(1.15) t Now with that question how long will it take fr the number of sites to reach 8000? Give the answer in exact value and decimal form

  27. anonymous
    • 5 years ago
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    8000=4000(1.15)^t so 2=1.15^t \[t=\frac{\log 2}{\log 1.15} \approx 4.96\]months

  28. toxicsugar22
    • 5 years ago
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    and 16000

  29. toxicsugar22
    • 5 years ago
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    how long will it take fr the number of sites to reach 16000

  30. anonymous
    • 5 years ago
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    16000=4000(1.15)^t so 4=(1.15)^t\[\log 4 = t \log 1.15 \rightarrow t=\frac{\log 4}{\log 1.15} \approx 9.92\]months

  31. toxicsugar22
    • 5 years ago
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    thanks

  32. anonymous
    • 5 years ago
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    ok

  33. toxicsugar22
    • 5 years ago
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    you are thhe bomb

  34. anonymous
    • 5 years ago
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    The Who question:using the formula I gave you,\[120=10\log_{10}(10^{12}I) \rightarrow I=1\]i.e. Intensity is 1 Watt/m^2 at 50 meters. To find out the distance for 90 decibels,\[90=10\log _{10}(10^{12}I_{pain}) \rightarrow I_{pain}=10^{-10}\]Now, again, intensity is inversely proportional to the square of the distance, so\[\frac{I_0}{I_{pain}}=\frac{k/r_0^2}{k/r_{pain}^2}=\frac{r_{pain}^2}{r_0^2}\]so that\[r^2_{pain}=r_0^2\frac{I_0}{I_{pain}}=50^2 \times \frac{1}{10^{-3}}=2.5 \times 10^6m^2 \]so the pain distance is\[r_{pain}=\sqrt{2.5 \times 10^6}\approx1581m\]I don't know what level of significance you want for significant figures, so I can't do that part.

  35. toxicsugar22
    • 5 years ago
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    ok

  36. anonymous
    • 5 years ago
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    you still there

  37. anonymous
    • 5 years ago
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    you still there

  38. anonymous
    • 5 years ago
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    are you looking for toxicsugar?

  39. anonymous
    • 5 years ago
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    yeah, i was going to say

  40. anonymous
    • 5 years ago
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    I = P / 4pi r^2

  41. anonymous
    • 5 years ago
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    Yeah...I skipped it. I think he just wants answers, though.

  42. anonymous
    • 5 years ago
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    :s

  43. anonymous
    • 5 years ago
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    yeah

  44. anonymous
    • 5 years ago
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    im just googling this, learning it

  45. anonymous
    • 5 years ago
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    ah

  46. anonymous
    • 5 years ago
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    apparently it is difficult to change db to hertz

  47. anonymous
    • 5 years ago
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    is it?

  48. anonymous
    • 5 years ago
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    how can you

  49. anonymous
    • 5 years ago
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    ok how many hertz is 90 db

  50. anonymous
    • 5 years ago
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    I don't know. I haven't really looked into it. I just saw a couple of things doing a search. It seems to depend on the context.

  51. anonymous
    • 5 years ago
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    yeah but you solved this problem well. you know your physics

  52. anonymous
    • 5 years ago
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    i have question. two astronauts are pulling on each other with 20 N on a rope. what is the tension in the rope (no gravity here)

  53. anonymous
    • 5 years ago
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    Sorry, was away. It' 20N.

  54. anonymous
    • 5 years ago
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    how did you get that

  55. anonymous
    • 5 years ago
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    give me a minute. i need coffee.

  56. anonymous
    • 5 years ago
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    sure

  57. anonymous
    • 5 years ago
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    hey

  58. anonymous
    • 5 years ago
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    sry, back

  59. anonymous
    • 5 years ago
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    Tension isn't a force, which is what screws people up. Mathematically, it's a rank 2 tensor - a function that takes a vector and spits out another. The input is a direction vector and the output is the force exerted on that part of the rope you're looking at, in the direction of the vector you plugged in. We know the direction up yields 20N, and direction down yields 20N...we *say* the tension in the rope is 20N.

  60. anonymous
    • 5 years ago
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    hmmm

  61. anonymous
    • 5 years ago
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    what if one of the astronauts died, and the live astronaut pulls on the dead one 20 N

  62. anonymous
    • 5 years ago
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    It's a third law of motion thing.

  63. anonymous
    • 5 years ago
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    is the tension still 20 N

  64. anonymous
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    There'd be no tension.

  65. anonymous
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    there is, because the dead astronaut has weight

  66. anonymous
    • 5 years ago
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    err, inertia i mean

  67. anonymous
    • 5 years ago
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    Oh...I thought he floated away :)

  68. anonymous
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    no, he just died when the moment came to pull on it

  69. anonymous
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    lol, are you just coming up with this?

  70. anonymous
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    If he died and is still attached, the tension will be whatever force the live astronaut applies to accelerate him or her.

  71. anonymous
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    The force is 'communicated' through the rope by the third law.

  72. anonymous
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    so its 20 N in both cases

  73. anonymous
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    If that's what the live one is pulling the dead one with.

  74. anonymous
    • 5 years ago
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    just found this - http://physics.stackexchange.com/questions/1220/rope-tension-question it might help

  75. anonymous
    • 5 years ago
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    how come, say there is mass in the rope, 1kg and one guy pulls 30 N and the other pulls 20 N in opposite direction

  76. anonymous
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    so there is a net force of 10 N on the rope. so why doesnt it accelerate 10/1 = a

  77. anonymous
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    10 m/s^2

  78. anonymous
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    I'm not sure what you mean. If there is a net force on the rope, it will accelerate.

  79. anonymous
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    You have to remember that most questions in physics on this matter assume a massless rope...

  80. anonymous
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    right but 10 m/s^2 is crazy fast , it will shoot off

  81. anonymous
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    It's okay if it's fast.

  82. anonymous
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    but the masses are accelerating VERY slow

  83. anonymous
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    I'm not sure what you mean.

  84. anonymous
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    say the guy has a mass of 50 kg

  85. anonymous
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    Oh...I see...the tension would fall off since the rope wouldn't be taught.

  86. anonymous
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    Actually, no...the acceleration would remain the same assuming they're all accelerating at 10m/s^2

  87. anonymous
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    the guy on the right is 50 kg, guy on left is 40 kg, the 50 kg guy is applying 30 N, and the 40 kg guy is applying 10 N , on a rope

  88. anonymous
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    so 30 - T = 50*a T - 20 = 40 *a

  89. anonymous
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    The tension's 20N

  90. anonymous
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    Do a free body diagram for each astronaut

  91. anonymous
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    The net force on each is 20N to the right.

  92. anonymous
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    i changed my problem

  93. anonymous
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    the guy on the right is 50 kg, guy on left is 40 kg, the 50 kg guy is applying 30 N, and the 40 kg guy is applying 10 N , on a rope

  94. anonymous
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    The rope communicates that force.

  95. anonymous
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    but the tension is not 20

  96. anonymous
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    most people think that, but if you draw the force body diagram for the guys you have

  97. anonymous
    • 5 years ago
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    the tension is actually 24.444

  98. anonymous
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    lets call m1 = 50 kg, and m2 = 40 kg

  99. anonymous
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    the forces on m1 is 30 N to the right, and has tension to the left. so 30 - T = 50* a

  100. anonymous
    • 5 years ago
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    I get 18 + 8/9.

  101. anonymous
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    The concept of tension remains the same, except when answering the second question, I got slack and didn't bother to calculate things.

  102. anonymous
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    so do you agree that

  103. anonymous
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    30 - T = 50 * a T - 10 = 40 * a

  104. anonymous
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    Yes, so long as we assume the rope is fully taught.

  105. anonymous
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    I agree with those equations.

  106. anonymous
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    oh so 18.88888 N in tension

  107. anonymous
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    so if the rope has 1kg say, it is accelerating 18.8888 ms^-2

  108. anonymous
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    Actually, I think these equations are for massless rope.

  109. anonymous
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    ok

  110. anonymous
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    suppose that it is massless. and the rope has two forces on it, at either end, so F1 - F2 = 0*a, so F1 = F2

  111. anonymous
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    but the forces at the end are unequal

  112. anonymous
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    Why do you get F1-F2=0? The object's accelerating.

  113. anonymous
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    For it to remain taught and communicate the force, it must be accelerating with the system.

  114. anonymous
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    but yuo said massless

  115. anonymous
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    so m = 0

  116. anonymous
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    F1 - F2 = 0*a

  117. anonymous
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    Sorry, I'm being distracted.

  118. anonymous
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    thats ok

  119. anonymous
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    if the tension is 18.8888 in our example, then 18.888 is the force that is exerted on the rope by both masses being pulled

  120. anonymous
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    does that make sense?

  121. anonymous
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    Oh, I thought you left.

  122. anonymous
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    nope

  123. anonymous
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    Problems are coming in because of the massless thing.

  124. anonymous
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    In reality, the rope will have mass.

  125. anonymous
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    so the rope exerts a force of 18.888 on either end

  126. anonymous
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    The rope carries the force through tension.

  127. anonymous
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    I'm trying to come up with a physical example.

  128. anonymous
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    Sorry, I was preoccupied. Set the problem up like the attachment that's coming.

  129. anonymous
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  130. anonymous
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    The rope is assumed to have a uniform density, so mass per unit length will be constant. At the point x, the rope will have mass *to the right* of x of\[\frac{m}{L}(L-x)\]This mass, times the acceleration of the system (since the rope is taught) *plus* the force F_2, will equal the tension to the left at the point x. So we'd have\[F_2+\frac{m}{L}(L-x)a-T(x)=\left( m+\frac{F_1}{a}+\frac{F_2}{a} \right)a\]where we're dealing with magnitudes (directions of forces are contained in the signs). Solving for T(x), we have,\[T(x)=F_2+\frac{m}{L}(L-x)a\]

  131. anonymous
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    Now, since acceleration is constant throughout, at the point x, we have by equating forces, \[F_2-ma-F_2=Ma\]where M is the total mass of the system. Hence,\[a=\frac{F_1-F_2}{M+m}\]so that\[T(x)=F_2+\frac{m}{L}(L-x)\frac{F_1-F_2}{M+m}\]

  132. anonymous
    • 5 years ago
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    ...and if there's a mistake in there, I wouldn't be surprised. I rarely translate what's on my pad to this site intact.

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