VCE Specialist 3/4 Question Thread!

[xZero]

Heres a shitty diagram i drew on mspaint, see if you can connect the dots

[Special At Specialist]

I'm not really sure why you drew that diagram that way...
Did you start with a triangle of forces?

[dc302]

I'm not really sure why you drew that diagram that way...
Did you start with a triangle of forces?

The purpose of the lengths is to find the angle, which will ultimately help find the force.

[xZero]

I'm not really sure why you drew that diagram that way...
Did you start with a triangle of forces?

I never use triangle of forces, i simply drew the question in a 2d diagram and analyze the forces (mind you the triangle of forces dont work in engineering and physics, its all about free body diagram). Also as dc302 said, length of the rope gives you the angle of the tension force, the rope can be 1km or 1m as long as the ratio between opposite and adjacent side remains constant.

[rife168]

Can someone go through the question please? I haven't done problems like that for a while and I'm a bit rusty...

"A particle of mass 2kg hangs from a fixed point, O, by a light inextensible string of length 2.5m. It is pulled aside a horizontal distance of 2m by a force P N inclined at an angle of 75 degrees with the downward vertical, and rests in equilibrium. Find P and the tension of the string."

[abd123]

Can someone go through the question please? I haven't done problems like that for a while and I'm a bit rusty...

"A particle of mass 2kg hangs from a fixed point, O, by a light inextensible string of length 2.5m. It is pulled aside a horizontal distance of 2m by a force P N inclined at an angle of 75 degrees with the downward vertical, and rests in equilibrium. Find P and the tension of the string."

Draw a diagram like xZero has represented, I think he has the 2kg particle missing in the diagram, to make that easier I would have placed the 2kg particle on dot that connects between the vertical force 2.5cm and resultant force which is unknown. Than resolve it as trig, since they are asking us to the find the tension of 'P', Use something formulas that are resolving vertical and horizontal forces that are like and . Also small thing since its a 'fixed point' the particle wouldn't swing and sway around in that incline angle.

[xZero]

Yeh something like that, the black line is the roof, red lines are the length and blue lines are the forces, at the centre of them blue lines is the weight. Sorry not use to drawing the objects in, i usually just represent everything with a dot. What you do is resolve them vertically and horizontally. For vertical you have Tsin(51.xx)=2*9.8 + Psin(75). Do the same thing horizontally and punch them into your cas and solve them simultaneously

[rife168]

Shouldn't the hypoteneuse in xZero's diagram be 2.5m? Hence the vertical component of the triangle would be 1.5m...

[xZero]

bleh my bad, well as long as the the image conveyed the idea its all good

[Special At Specialist]

The purpose of the lengths is to find the angle, which will ultimately help find the force.

Ahh, this makes sense. Thanks!

But the diagram you drew looks a bit different to the "solution", which was this:
http://img848.imageshack.us/img848/4896/mathsproblem12.png

[paulsterio]

there's more than one way to draw it, depending on how you arrange it

[Mr. Study]

Hopefully this isn't a noob question.

Could someone do this equation for me, I get stuck when I square root everything, the book doesn't explain how they got a certain co-efficient.

|z+1|+|z-1|=4, turn to the cartesian equation form.

Thanks!

[brightsky]

let z=x+yi
|(x+1)+yi)| + |(x-1)+yi| = 4
sqrt((x+1)^2 + y^2) + sqrt((x-1)^2 + y^2) = 4
sqrt(x^2 + 2x + 1 + y^2) + sqrt(x^2 - 2x + 1 + y^2) = 4
sqrt(x^2 + 2x + 1 + y^2) = 4 - sqrt(x^2 - 2x + 1 + y^2)
x^2 + 2x + 1 + y^2 = 16 - 8sqrt(x^2 - 2x + 1 + y^2) + x^2 - 2x + 1 + y^2
4x = 16 - 8sqrt(x^2 - 2x + 1 + y^2)
16 - 4x = 8sqrt(x^2 - 2x + 1 + y^2)
256 - 128x + 16x^2 = 64x^2 - 128x + 64 + 64y^2
48x^2 + 64y^2 = 192
3x^2 + 4y^2 = 12
x^2/4 + y^2/3 = 1

[Special At Specialist]

The vector resolute of i + 2j - 3k in the direction perpendicular to vector 2i - j - 2k is given by:
A. 1/3 (-i + 8j - 5k)
B. -2/3 (2i - j - k)
C. 1/3 (7i + 4j - 13k)
D. 1/9 (19i + 28j - 7k)
E. -i + 3j - k

The main problem I have with this question is that I don't know how to find a vector that is perpendicular to a 3D vector. If it was 2D, I would just let the dot product equal 0. But this is 3D. Wouldn't there be heaps of different possibilities, since angles are only defined in 2 dimensions?

[b^3]

Use vector resolutes (refer to the attached image).
u is the vector resolute of a in the direction of b that is parallel to b
w is the vecotr resolute of a in the direction of b that is perpendicular to b

Parallel component
Perpendicular Component

So for this question let
a=i+2j-3k
b=2i-j-2k

So the perpendicular component is






So the answer is A.

(Hopefully I haven't made mistake somewhere, or swapped the vectors around, I'm a bit rusty )