## 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

1. kittiwitti1
2. anonymous

|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

Question - is this physics-related? I was seriously about to apply my high-school physics knowledge to this xD

4. kittiwitti1

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

5. anonymous

Yes this is physics related, but it uses mathematical concepts of trignometry and vector components

6. kittiwitti1

Ah, alright :D

7. anonymous

8. kittiwitti1

lol

9. kittiwitti1

*attempts to apply physics-esque skills to problem*

10. anonymous

|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

theta=35 here

12. anonymous

sorry... T sin theta

13. kittiwitti1

okay -thumbs up-

14. kittiwitti1

Hold on I need to work this out lol

15. anonymous

Make 2 equations and eliminate the unknown T

16. anonymous

Also for the first part, since you only need magnitude, |-mg sin theta|=mg sin theta

17. kittiwitti1

I think the angles are mixed up on that diagram.

18. anonymous

which one?

19. kittiwitti1

Last

20. anonymous

I think it's good

21. kittiwitti1

This is what I got so far: |dw:1444475980762:dw|

22. kittiwitti1

Wait... instead of 35°, it's 34° right?

23. anonymous

24. kittiwitti1

90-56=34

25. anonymous

for first part

26. kittiwitti1

How is it wrong? And 90-56=34

27. kittiwitti1

Well, unless my calculator malfunctioned. *rechecks*

28. kittiwitti1

|dw:1444476176481:dw|

29. anonymous

Refer to the drawing |dw:1444476245425:dw|

30. kittiwitti1

... what

31. anonymous

That is the correct diagram

32. kittiwitti1

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

33. kittiwitti1

But anyway regardless it's 90-56=34 right? Just making sure

34. kittiwitti1

I saw 35 degrees.

35. anonymous

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

"The slide is inclined at an angle of a = 56.0°"

37. anonymous

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

It never said the ground...

39. kittiwitti1

Man, these things need to be more clear with their descriptions.

40. anonymous

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

-confused-

42. anonymous

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

-_- thx

44. kittiwitti1

45. kittiwitti1

-attempts to revive it-

46. anonymous

|dw:1444477065854:dw|

47. kittiwitti1

Okay

48. kittiwitti1

I'm sorry for annoying you with my reduced intellect u_u;

49. anonymous

Oh no that's absolutely fine, it can frustrating when you can't understand something

50. anonymous

can be*

51. kittiwitti1

Okay ^^;

52. kittiwitti1

I got the weight as 18143.1g

53. kittiwitti1

except the kid has no listed mass

54. anonymous

$|\vec f|=|-m \vec g\sin(56)|=mg\sin(56)$ 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

What.

56. kittiwitti1

then what is g?

57. anonymous

acceleration due to gravity

58. kittiwitti1

wait...

59. kittiwitti1

OH! 9.81?

60. anonymous

yep!

61. kittiwitti1

okay hold on

62. kittiwitti1

Wait... so mg is (x lbs) times "g" ?

63. kittiwitti1

*40 lbs times 9.81?

64. kittiwitti1

I thought it was grams times G

65. anonymous

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

ok ?

67. kittiwitti1

do I convert the 40 lbs o-o

68. anonymous

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

Oh okay

70. anonymous

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

in your previously mentioned equation, f is F force right?

72. anonymous

yep, it's the force you are required to find, but you will find f/m, you will measure it in pounds

73. anonymous

f/g sorry

74. kittiwitti1

Oh okay I kinda get it now

75. kittiwitti1

I got 325.3143435 N

76. anonymous

can u post a screenshot for options of 1st question?

77. kittiwitti1

no options. fill in blanks

78. kittiwitti1

calculation process$\frac{F}{g}=msin56\rightarrow\frac{F}{9.81m/s^{2}}=40\sin56$$F=9.81\times40\sin56\approx325.3$

79. anonymous

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

It will not be Newton!! the unit of mass you are using is pounds, remember that!

81. kittiwitti1

okay not newtons ?

82. anonymous

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

$325.3 \space lbf \neq 325.3 \space N$

84. anonymous

325.3 pound force, not newtons

85. kittiwitti1

-confused-

86. kittiwitti1

I do have a conversion of lbs to g, should I use that?

87. anonymous

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

Since you have taken mass in pounds, your answer will be in pounds force

89. anonymous

I've got to go eat now, maybe someone else can help

90. kittiwitti1

Okay

91. kittiwitti1

OH I see, gotcha

92. kittiwitti1

It says it's wrong -_-...

93. phi

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

Stacey pulls back with a force of 40 sin(56) to balance the force of gravity in the direction down the slide.

95. phi

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

that shows the tension (force) on the rope balances the 42 lbs acting down, and the "pull" acting sideways.

97. anonymous

Preety trickey

98. kittiwitti1

I got both thanks @phi, that was very helpful