kittiwitti1
  • kittiwitti1
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
kittiwitti1
  • kittiwitti1
FULL: ONE - http://icecream.me/f2de6848075baa9f18c05b853602d902 TWO - http://icecream.me/92037082b8d26d230ff924feb654b4e0
anonymous
  • 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
kittiwitti1
  • kittiwitti1
Question - is this physics-related? I was seriously about to apply my high-school physics knowledge to this xD

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kittiwitti1
  • kittiwitti1
It's just that I did this once in pre-calculus and got a nonsensical answer, so I'm being cautious.
anonymous
  • anonymous
Yes this is physics related, but it uses mathematical concepts of trignometry and vector components
kittiwitti1
  • kittiwitti1
Ah, alright :D
anonymous
  • anonymous
My bad, trigonometry*
kittiwitti1
  • kittiwitti1
lol
kittiwitti1
  • kittiwitti1
*attempts to apply physics-esque skills to problem*
anonymous
  • 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
anonymous
  • anonymous
theta=35 here
anonymous
  • anonymous
sorry... T sin theta
kittiwitti1
  • kittiwitti1
okay -thumbs up-
kittiwitti1
  • kittiwitti1
Hold on I need to work this out lol
anonymous
  • anonymous
Make 2 equations and eliminate the unknown T
anonymous
  • anonymous
Also for the first part, since you only need magnitude, |-mg sin theta|=mg sin theta
kittiwitti1
  • kittiwitti1
I think the angles are mixed up on that diagram.
anonymous
  • anonymous
which one?
kittiwitti1
  • kittiwitti1
Last
anonymous
  • anonymous
I think it's good
kittiwitti1
  • kittiwitti1
This is what I got so far: |dw:1444475980762:dw|
kittiwitti1
  • kittiwitti1
Wait... instead of 35°, it's 34° right?
anonymous
  • anonymous
Nope, your diagram is wrong
kittiwitti1
  • kittiwitti1
90-56=34
anonymous
  • anonymous
for first part
kittiwitti1
  • kittiwitti1
How is it wrong? And 90-56=34
kittiwitti1
  • kittiwitti1
Well, unless my calculator malfunctioned. *rechecks*
kittiwitti1
  • kittiwitti1
|dw:1444476176481:dw|
anonymous
  • anonymous
Refer to the drawing |dw:1444476245425:dw|
kittiwitti1
  • kittiwitti1
... what
anonymous
  • anonymous
That is the correct diagram
kittiwitti1
  • kittiwitti1
Well yeah but so is mine :T I was taught that that's step one, then transfer the angle to that part.
kittiwitti1
  • kittiwitti1
But anyway regardless it's 90-56=34 right? Just making sure
kittiwitti1
  • kittiwitti1
I saw 35 degrees.
anonymous
  • 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
kittiwitti1
  • kittiwitti1
"The slide is inclined at an angle of a = 56.0°"
anonymous
  • 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
kittiwitti1
  • kittiwitti1
It never said the ground...
kittiwitti1
  • kittiwitti1
Man, these things need to be more clear with their descriptions.
anonymous
  • 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
kittiwitti1
  • kittiwitti1
-confused-
anonymous
  • anonymous
even if you feel confused, refer to your drawing, it clearly shows what they mean by their "inclination", it's the lower angle
anonymous
  • anonymous
-_- thx
kittiwitti1
  • kittiwitti1
Okay... Sorry, brain half-dead
kittiwitti1
  • kittiwitti1
-attempts to revive it-
anonymous
  • anonymous
|dw:1444477065854:dw|
kittiwitti1
  • kittiwitti1
Okay
kittiwitti1
  • kittiwitti1
I'm sorry for annoying you with my reduced intellect u_u;
anonymous
  • anonymous
Oh no that's absolutely fine, it can frustrating when you can't understand something
anonymous
  • anonymous
can be*
kittiwitti1
  • kittiwitti1
Okay ^^;
kittiwitti1
  • kittiwitti1
I got the weight as 18143.1g
kittiwitti1
  • kittiwitti1
except the kid has no listed mass
anonymous
  • 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
kittiwitti1
  • kittiwitti1
What.
kittiwitti1
  • kittiwitti1
then what is g?
anonymous
  • anonymous
acceleration due to gravity
kittiwitti1
  • kittiwitti1
wait...
kittiwitti1
  • kittiwitti1
OH! 9.81?
anonymous
  • anonymous
yep!
kittiwitti1
  • kittiwitti1
okay hold on
kittiwitti1
  • kittiwitti1
Wait... so mg is (x lbs) times "g" ?
kittiwitti1
  • kittiwitti1
*40 lbs times 9.81?
kittiwitti1
  • kittiwitti1
I thought it was grams times G
anonymous
  • 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
kittiwitti1
  • kittiwitti1
ok ?
kittiwitti1
  • kittiwitti1
do I convert the 40 lbs o-o
anonymous
  • 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)
kittiwitti1
  • kittiwitti1
Oh okay
anonymous
  • 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
kittiwitti1
  • kittiwitti1
in your previously mentioned equation, f is F force right?
anonymous
  • anonymous
yep, it's the force you are required to find, but you will find f/m, you will measure it in pounds
anonymous
  • anonymous
f/g sorry
kittiwitti1
  • kittiwitti1
Oh okay I kinda get it now
kittiwitti1
  • kittiwitti1
I got 325.3143435 N
anonymous
  • anonymous
can u post a screenshot for options of 1st question?
kittiwitti1
  • kittiwitti1
no options. fill in blanks
kittiwitti1
  • kittiwitti1
calculation process\[\frac{F}{g}=msin56\rightarrow\frac{F}{9.81m/s^{2}}=40\sin56\]\[F=9.81\times40\sin56\approx325.3\]
anonymous
  • 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 \[\frac{f}{g}=18.1437 \times 0.829=15.0411 \space kg\]
anonymous
  • anonymous
It will not be Newton!! the unit of mass you are using is pounds, remember that!
kittiwitti1
  • kittiwitti1
okay not newtons ?
anonymous
  • 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
anonymous
  • anonymous
\[325.3 \space lbf \neq 325.3 \space N\]
anonymous
  • anonymous
325.3 pound force, not newtons
kittiwitti1
  • kittiwitti1
-confused-
kittiwitti1
  • kittiwitti1
I do have a conversion of lbs to g, should I use that?
anonymous
  • 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
anonymous
  • anonymous
Since you have taken mass in pounds, your answer will be in pounds force
anonymous
  • anonymous
I've got to go eat now, maybe someone else can help
kittiwitti1
  • kittiwitti1
Okay
kittiwitti1
  • kittiwitti1
OH I see, gotcha
kittiwitti1
  • kittiwitti1
It says it's wrong -_-...
phi
  • 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|
phi
  • phi
Stacey pulls back with a force of 40 sin(56) to balance the force of gravity in the direction down the slide.
phi
  • 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|
phi
  • phi
that shows the tension (force) on the rope balances the 42 lbs acting down, and the "pull" acting sideways.
anonymous
  • anonymous
Preety trickey
kittiwitti1
  • kittiwitti1
I got both thanks @phi, that was very helpful

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