Question 10 - I find your leap of intuition for the expression of the vertical components of the tension rather difficult. I offer this as a more simple minded approach (and one, I suspect, the examiners thought might be followed)

Step1991Paper1Question10.pdf66.4KB

Reply 227

brianeverit

Original post by mikelbird

yes...see attached....

you have misread the question.
The reaction on the particle at B is NOT at angle phi to horizontal.

Reply 228

mortimerch11

Original post by Rabite

III/6

God, this one took ages to type out.
No one's even gonna read it, but hey...

Could someone else at least try the question to see what the real answer is?

Reading it now :smile:

it looks really nice and clean

Reply 229

PADE123

Original post by brianeverit

\text{STEP 1991 Paper II question 13}

\text {For particle we have (1) } R-amg \sin \theta =amx \ddot \theta

\text{ and (2) }F-amg \cos \theta=amx \dot \theta^2

\text{For rod, we have (3) }mk^2 \ddot \theta =-Rx-dmg \sin \theta

\text{From (1) and (3) } mk^2 \ddot \theta+dmg \sin \theta=-ax^2 \ddot \theta-amgx \sin \theta

\text{ i.e. }m(k^2+ax^2) \ddot \theta=-mg \sin \theta(ax+d)

\text{so } \ddot \theta= -\dfrac{g \sin \theta(ax+d)}{k^2+ax^2} \implies 2 \dot \theta \ddot \theta =- \dfrac{2g \sin \theta(ax+d)}{k^2+ax^2 \dot \theta}

Unparseable latex formula:

\implies \dot \theta^2= \dfrac {2g(ax+d)}{k^2+ax^2} \cos \theta}

\text {substituting in (2) gives }F=amg \cos \theta+ \dfrac{2amxg(ax+d)}{k^2+ax^2} \cos \theta

=amg \cos \theta \left( \dfrac{k^2+ax^2+2x(ax+d)}{k^2+ax^2} \right) \text { and } R=amg \sin \theta- \dfrac{amxg \sin \theta(ax+d)}{k^2+ax^2}

=amg \sin \theta \left( \dfrac{k^2+ax^2-x(ax+d)}{k^2+ax^2} \right) = amg \sin \theta \left( \dfrac{k^2-dx}{k^2+ax^2}\right)

\text{hence, } \mu= \dfrac{F}{R}= \dfrac{k^2+ax^2+2x(ax+d)}{(k^2-dx) \tan \theta} \implies \mu \tan \theta= \dfrac{3ax^2+k^2+2xd}{k^2-dx} \text { as required}

Unparseable latex formula:

\text {(i) If }k^2=xd \text { then } \ddot \theta=- \dfrac{g \soin \theta}{x} \text { and particle moves as if it is a simple pendulum of length }x

\text{(ii) If }k^2<xd \text { then particle willm slide upwards towards }O.

Can we use the formula derived for the case k^2>dx for cases (i), (ii)? In my derivation, considering conservation for the ring-rod system and resolving forces in the radial and tangential components for the ring, I didn't make any explicit assumption on the value of k^2. I would really appreciate a clarification. Is there something that I am missing?

Reply 230

Billord

Original post by DFranklin

So we suspect that the axes have equations x+y=1, x-y =2. You could probably stop there, but partly to reassure myself, and partly for the hell of it, we can go further

Sorry to be pedantic, but here you've said x+y=1 and I think you meant x+y=0

Reply 231

Samuel Ding

Original post by Rabite

Well I'll type it up :p:

u^2 + 2u \sinh x - 1 = 0

The quadratic formula involves fractions, and I hate fractions, so let's complete the square:

(u+\sinh x)^2 - 1 - \sinh ^2 x=0

u+\sinh x= ± \cosh x

u = \cosh x -\sinh x

-(\cosh x+\sinh x)

coshx-sinhx is uh...e^{-x} and -(coshx+sinhx) is -e^x.
y`>0 at x=0 so we have to choose the + solution.

\frac{dy}{dx} = e^{-x}

Integrate and find the +C to get

y = 1-e^{-x}

.

For the next bit, let u = dy/dx for a start:

u^2 \sinh x + 2u - \sinh x =0

u^2 + 2u cosech x - 1 =0

(u+cosech x)^2 - 1 - cosech^2 x =0

u+cosech x = ±\coth x

u = \coth x - cosech x

-(\coth x + cosech x)

\coth x - cosech x = \frac{\cosh x - 1}{\sinh x}

= \tanh{½x}

(Consider double "angle" formulae)
But u = dy/dx:

\frac{dy}{dx} = \tanh{½x}

Tanh = sinh / cosh, which is a derivative over its function:

y = 2 \ln \cosh ½ x +C

The solution passes through (0,0); so C =0.

(Taking the negative solution when we square rooted earlier results in something that can't pass through (0,0))

[edit] Is there no /cosech in Latex?

Hi there, I'm a bit confused about why it is possible to divide sinhx because sinh(0) = 0 so wouldnt dividing imply that the set of x doesn't pass through 0?

Reply 232

keithmcnulty

STEP I Q4

Original post by justinsh

STEP I questions 4

Please tell me what you think, I'm not certain of what I did.

I don't think the hint on this question to consider the area he cannot see is the best advice.

So much easier to take the following approach:

Note that there is an L-shaped area that the guard can always see - this has area 80.

In addition, there are two (possibly degenerate) triangles that the guard can see, one with a base from (8,4) to (12,4) and another with a base from (4, 8) to (4, 12). So we need the points at which the sum of the areas of these triangles are at maximum and minimum.

Setting the distance from A as x, some simple trig can tell you that this sum is 4-x + 32/(8-x).
Differentiating and setting to zero we get x = 4(2-sqrt(2)) is the only turning point in the range [0,4]. Taking second derivatives tells you that this is a minimum.

Now since there are no other turning points and this curve is clearly continuous, it must be increasing on either side of this minimum, so maximum would be on one or both of the ends of the interval [0,4]. Easy to see that the max is on both ends.

Hence max view is at x = 0 and x = 4 (where view is a total of 88 units), and min view at x = 4(2-sqrt(2)) where view is a total of 80 + 4(2*sqrt(2) -1) units.