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kittiwitti1

  • one year ago

More trig questions... ONE - Danny and Stacey have gone from the swing to the slide at the park. The slide is inclined at an angle of a = 56.0°. Danny weighs 40.0 pounds. He is sitting in a cardboard box with a piece of wax paper on the bottom. Stacey is at the top of the slide holding on to the cardboard box (see the figure below). Find the magnitude of the force Stacey must pull with, in order to keep Danny from sliding down the slide. (We are assuming that the wax paper makes the slide into a frictionless surface, so that the only force keeping Danny from sliding is the force with which Stacey pulls. Round your answer to one decimal place.) FIGURE A - http://www.webassign.net/mcktrig6/2-5-037.gif TWO - Jadon is 5 years old and weighs 42.0 pounds. He is sitting on a swing when his cousin Allison pulls him and the swing back horizontally through an angle of 35.0° and then stops. Find the magnitude of the force exerted by Allison (see the figure below). FIGURE B - http://www.webassign.net/mcktrig7/2-5-059.gif

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  1. kittiwitti1
    • one year ago
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    FULL: ONE - http://icecream.me/f2de6848075baa9f18c05b853602d902 TWO - http://icecream.me/92037082b8d26d230ff924feb654b4e0

  2. anonymous
    • one year ago
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    |dw:1444474692435:dw| Here theta is 56 degrees The normal reaction and mg cosine theta components balance each other, so to keep Danny in equilibrium, Stacy must apply a force mg sin theta in the opposite direction, or -mg sin theta

  3. kittiwitti1
    • one year ago
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    Question - is this physics-related? I was seriously about to apply my high-school physics knowledge to this xD

  4. kittiwitti1
    • one year ago
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    It's just that I did this once in pre-calculus and got a nonsensical answer, so I'm being cautious.

  5. anonymous
    • one year ago
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    Yes this is physics related, but it uses mathematical concepts of trignometry and vector components

  6. kittiwitti1
    • one year ago
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    Ah, alright :D

  7. anonymous
    • one year ago
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    My bad, trigonometry*

  8. kittiwitti1
    • one year ago
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    lol

  9. kittiwitti1
    • one year ago
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    *attempts to apply physics-esque skills to problem*

  10. anonymous
    • one year ago
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    |dw:1444475098336:dw| The magnitude of the force is simply mg cos theta since the vertical and horizontal components must balance each other for equilibrium

  11. anonymous
    • one year ago
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    theta=35 here

  12. anonymous
    • one year ago
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    sorry... T sin theta

  13. kittiwitti1
    • one year ago
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    okay -thumbs up-

  14. kittiwitti1
    • one year ago
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    Hold on I need to work this out lol

  15. anonymous
    • one year ago
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    Make 2 equations and eliminate the unknown T

  16. anonymous
    • one year ago
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    Also for the first part, since you only need magnitude, |-mg sin theta|=mg sin theta

  17. kittiwitti1
    • one year ago
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    I think the angles are mixed up on that diagram.

  18. anonymous
    • one year ago
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    which one?

  19. kittiwitti1
    • one year ago
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    Last

  20. anonymous
    • one year ago
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    I think it's good

  21. kittiwitti1
    • one year ago
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    This is what I got so far: |dw:1444475980762:dw|

  22. kittiwitti1
    • one year ago
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    Wait... instead of 35°, it's 34° right?

  23. anonymous
    • one year ago
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    Nope, your diagram is wrong

  24. kittiwitti1
    • one year ago
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    90-56=34

  25. anonymous
    • one year ago
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    for first part

  26. kittiwitti1
    • one year ago
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    How is it wrong? And 90-56=34

  27. kittiwitti1
    • one year ago
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    Well, unless my calculator malfunctioned. *rechecks*

  28. kittiwitti1
    • one year ago
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    |dw:1444476176481:dw|

  29. anonymous
    • one year ago
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    Refer to the drawing |dw:1444476245425:dw|

  30. kittiwitti1
    • one year ago
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    ... what

  31. anonymous
    • one year ago
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    That is the correct diagram

  32. kittiwitti1
    • one year ago
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    Well yeah but so is mine :T I was taught that that's step one, then transfer the angle to that part.

  33. kittiwitti1
    • one year ago
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    But anyway regardless it's 90-56=34 right? Just making sure

  34. kittiwitti1
    • one year ago
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    I saw 35 degrees.

  35. anonymous
    • one year ago
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    Idk how you were taught, but that upper angle can't be 56, even in your drawing you posted it's that lower angle that is 56 Also that 35 is for another question, it's not the same slide necessarily

  36. kittiwitti1
    • one year ago
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    "The slide is inclined at an angle of a = 56.0°"

  37. anonymous
    • one year ago
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    Yep, so the lower angle connecting the end of the slide with the ground will be 56, that's the inclination of the slide, at how much angle it is inclined to the ground

  38. kittiwitti1
    • one year ago
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    It never said the ground...

  39. kittiwitti1
    • one year ago
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    Man, these things need to be more clear with their descriptions.

  40. anonymous
    • one year ago
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    They don't need to, they have stated that it's the slide's INCLINATION, it's inclination will of course the angle it makes with the ground

  41. kittiwitti1
    • one year ago
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    -confused-

  42. anonymous
    • one year ago
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    even if you feel confused, refer to your drawing, it clearly shows what they mean by their "inclination", it's the lower angle

  43. anonymous
    • one year ago
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    -_- thx

  44. kittiwitti1
    • one year ago
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    Okay... Sorry, brain half-dead

  45. kittiwitti1
    • one year ago
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    -attempts to revive it-

  46. anonymous
    • one year ago
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    |dw:1444477065854:dw|

  47. kittiwitti1
    • one year ago
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    Okay

  48. kittiwitti1
    • one year ago
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    I'm sorry for annoying you with my reduced intellect u_u;

  49. anonymous
    • one year ago
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    Oh no that's absolutely fine, it can frustrating when you can't understand something

  50. anonymous
    • one year ago
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    can be*

  51. kittiwitti1
    • one year ago
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    Okay ^^;

  52. kittiwitti1
    • one year ago
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    I got the weight as 18143.1g

  53. kittiwitti1
    • one year ago
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    except the kid has no listed mass

  54. anonymous
    • one year ago
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    \[|\vec f|=|-m \vec g\sin(56)|=mg\sin(56)\] In terms of pound force,(lb force) \[\frac{f}{g}=m \sin(56)\] This will give the force in lb force, you are given m as 40 lbs

  55. kittiwitti1
    • one year ago
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    What.

  56. kittiwitti1
    • one year ago
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    then what is g?

  57. anonymous
    • one year ago
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    acceleration due to gravity

  58. kittiwitti1
    • one year ago
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    wait...

  59. kittiwitti1
    • one year ago
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    OH! 9.81?

  60. anonymous
    • one year ago
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    yep!

  61. kittiwitti1
    • one year ago
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    okay hold on

  62. kittiwitti1
    • one year ago
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    Wait... so mg is (x lbs) times "g" ?

  63. kittiwitti1
    • one year ago
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    *40 lbs times 9.81?

  64. kittiwitti1
    • one year ago
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    I thought it was grams times G

  65. anonymous
    • one year ago
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    Weight is measured in Newtons, but for common people, it is much easier for them to identity with kilograms or pounds insteads(hence you see people saying, I weigh 70kilograms) such a statement is absolutely wrong, but is commonly used So although your answer will in units of pounds m/s^2, it is a common to practice to just express the force in units of mass, when measuring the force in units of pounds, it's called pound force and when in kilograms, it's called kilogram force When someone says I weight 70 kilogram, they mean they weight 70 kilogram force, which is 70*9.8 Newtons

  66. kittiwitti1
    • one year ago
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    ok ?

  67. kittiwitti1
    • one year ago
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    do I convert the 40 lbs o-o

  68. anonymous
    • one year ago
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    So even if you get the answer is mg sin(theta), we simply divide with g, and get m sin(theta) to express in lbs, I'm saying this because earlier I saw options for your 2nd question, they were all in pounds(or pounds-force more correctly)

  69. kittiwitti1
    • one year ago
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    Oh okay

  70. anonymous
    • one year ago
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    Didn't you find it weird?they are asking for weight(which is a force) and the options are in lbs, which is a unit of mass

  71. kittiwitti1
    • one year ago
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    in your previously mentioned equation, f is F force right?

  72. anonymous
    • one year ago
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    yep, it's the force you are required to find, but you will find f/m, you will measure it in pounds

  73. anonymous
    • one year ago
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    f/g sorry

  74. kittiwitti1
    • one year ago
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    Oh okay I kinda get it now

  75. kittiwitti1
    • one year ago
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    I got 325.3143435 N

  76. anonymous
    • one year ago
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    can u post a screenshot for options of 1st question?

  77. kittiwitti1
    • one year ago
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    no options. fill in blanks

  78. kittiwitti1
    • one year ago
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    calculation process\[\frac{F}{g}=msin56\rightarrow\frac{F}{9.81m/s^{2}}=40\sin56\]\[F=9.81\times40\sin56\approx325.3\]

  79. anonymous
    • one year ago
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    Ok, hmm I don't think you answer is right, let's convert the pounds to kilograms now im using a calc for this \[40 \space lbs=18.1437 \space kgs\] \[g=9.8 ms^{-2}\]\[\sin(56)=0.829\] \[f=mgsin(56)=18.1437 \times 9.8 \times 0.829 = 147.403 \space N\] In terms of kilograms force \[\frac{f}{g}=18.1437 \times 0.829=15.0411 \space kg\]

  80. anonymous
    • one year ago
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    It will not be Newton!! the unit of mass you are using is pounds, remember that!

  81. kittiwitti1
    • one year ago
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    okay not newtons ?

  82. anonymous
    • one year ago
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    Yep numerically value of your answer is correct but unit is not correct, a Newton is when we are talking about kilograms, you just simply write lbs m/s^2 instead

  83. anonymous
    • one year ago
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    \[325.3 \space lbf \neq 325.3 \space N\]

  84. anonymous
    • one year ago
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    325.3 pound force, not newtons

  85. kittiwitti1
    • one year ago
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    -confused-

  86. kittiwitti1
    • one year ago
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    I do have a conversion of lbs to g, should I use that?

  87. anonymous
    • one year ago
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    Look the numerical value of your answer is correct you just need to write pounds force(lbf) instead of Newtons, we use Newton when we take mass of the body in kilograms

  88. anonymous
    • one year ago
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    Since you have taken mass in pounds, your answer will be in pounds force

  89. anonymous
    • one year ago
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    I've got to go eat now, maybe someone else can help

  90. kittiwitti1
    • one year ago
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    Okay

  91. kittiwitti1
    • one year ago
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    OH I see, gotcha

  92. kittiwitti1
    • one year ago
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    It says it's wrong -_-...

  93. phi
    • one year ago
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    lbs are units of force (not mass). You can do the first problem this way. First, looking at the picture they give you, notice they show the angle \(\alpha\) is from the ground to the slide |dw:1444482301947:dw|

  94. phi
    • one year ago
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    Stacey pulls back with a force of 40 sin(56) to balance the force of gravity in the direction down the slide.

  95. phi
    • one year ago
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    For the 2nd problem The rope pulls up (at an angle) so that it balances the force of the weight acting straight down we can write the force of the weight as the sum of any two vectors, in particular |dw:1444482908101:dw|

  96. phi
    • one year ago
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    that shows the tension (force) on the rope balances the 42 lbs acting down, and the "pull" acting sideways.

  97. anonymous
    • one year ago
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    Preety trickey

  98. kittiwitti1
    • one year ago
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    I got both thanks @phi, that was very helpful

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