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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: lisafaustina on August 31, 2010, 06:18:11 pm

Title: Essential
Post by: lisafaustina on August 31, 2010, 06:18:11 pm: If you could answer at least one of these questions, thatd be great thx :)

Exercise 13G Q4. How do you do it? And I don't like the resolving vectors crap!

Exercise 13 G Q3c...

Exercise 13F Q14...I cant picture the diagram.

Exercise 13F Q15. CAn I think of the 'tension' as a some sort of force?
Also is triangle of forces actually important for the exam?
Title: Re: Essential
Post by: Chavi on August 31, 2010, 08:37:56 pm: Hey, so here here are the solutions:

13F q14: This seriously needs a diagram. Using it, you can work out all the angles needed, and then resolve the forces into horizontal/vertical components.
Let T be the tension in the string, and P be the force used to pull the particle.

P is applied @ 75 degrees to the vertical, so it is @ 15 degrees to the horizontal
To find the angles for the T triangle, use $\sin(\Theta) = \frac{2}{2.5}$ and $\theta=53.17$

Now resolve vertically and horizontally:
Vertically: $T\cos(53.13)-2g-P\sing(15)=0$
Horizontally: $T\sin(53.13-P\cos(15)=0$

Solve simultaneously [cbf typing this out on LATEX], and you get:
$P=42.094N$ and $T=50.824N$

13F q15 You'll find that the particle rests in the shape of a pythagorean triangle, with side 13cm, 5cm, and 12cm
Find the angles (67.38 and 22.68 degrees {i.e. use tan = 12/5}) and the resolve components:

Let the TEnsion in the 5cm string = A, and the tension in the 12cm = B
Horiz: $A\cos(67.38)-B\cos(22.68)=0$
Vert: $A\sin(67.38)+B\sin(22.68)=5g$
Solve simultaneously to get A = 45.23N and B = 12.18N

13G 3c - okay, so the way this differs from 3b, is that Pushing force now how a downward component that affects the magnitude of the reaction/normal force
REsolve P into horiz and vert components: vert = Psin60 and horiz = Pcos60, and now set up 2 equations:

Vertical forces:
$N-P\sin(60)=100g$ ...(1)
Horizontal:
$P\cos(60)=0.3N$ ... (2)
therefore, $N=\frac{5P}{3}$ , and sub this into (1)

$\frac{5P}{3}-\frac{P\sqrt{3}}{2}=100g$ ...(1)
and P = 1224N

13G 4 The frictional force simply equals the weight component of the particle parallel to the plane (because net force = 0)
so, W(parallel) = 5sin(30) = 2.5N = Fr
Title: Re: Essential
Post by: Chavi on August 31, 2010, 08:50:48 pm: Quote from: lisafaustina on August 31, 2010, 06:18:11 pm
And I don't like the resolving vectors crap!
Such is life, baby
Title: Re: Essential
Post by: Martoman on August 31, 2010, 08:51:39 pm: :smitten:
Title: Re: Essential
Post by: Chavi on August 31, 2010, 08:58:17 pm: Martoman, I must take this opportunity to commend you on your handwriting. The old school (circa ~1950) style is simply breathtaking
Title: Re: Essential
Post by: Martoman on August 31, 2010, 09:24:30 pm: LOL?

How do people usually write?
Title: Re: Essential
Post by: Chavi on August 31, 2010, 09:28:10 pm: without a pen license
Title: Re: Essential
Post by: Martoman on August 31, 2010, 09:32:15 pm: .... i've been told repeatedly in english to not write like a doctor... "there will be plenty of time for that later Laura" :-[
Title: Re: Essential
Post by: Mao on August 31, 2010, 09:53:31 pm: Quote from: Martoman on August 31, 2010, 09:32:15 pm
.... i've been told repeatedly in english to not write like a doctor... "there will be plenty of time for that later Laura" :-[

And so it is, LAURA.

[also, 80²]
Title: Re: Essential
Post by: Martoman on August 31, 2010, 10:00:03 pm: Gah, damn i didn't mean to type that out Emperor MAO.
Title: Re: Essential
Post by: lisafaustina on September 01, 2010, 02:13:04 am: I don't get the explanation for resolution of forced on pg 24...what's with the i and j direction, I don't get how they worked it out
Gah
Title: Re: Essential
Post by: lisafaustina on September 01, 2010, 03:36:25 am: Exercise 13G question 7b. Doesn't the applied force just have to be bigger than the descending force and friction force?
Mine works out to be 27.53 :S
Title: Re: Essential
Post by: lisafaustina on September 01, 2010, 03:58:42 pm: SORRY one moer, on pg 504 Question 6bi it says to find a vector expression for the 'constant resultant force'...how is this different from just the 'resultant force'?
and in this exercise (EXERCISE 13H) when they ask you to find the resultant force at time t, they always seem to let 't' equal 1 and give an answer without the 't' variables...isnt it correct to leave the 't' as it is?

sorry if its a little confusing but you should know what i mean if youve done the questions!